diff --git "a/bt_jhumaneval.jsonl" "b/bt_jhumaneval.jsonl" deleted file mode 100644--- "a/bt_jhumaneval.jsonl" +++ /dev/null @@ -1,164 +0,0 @@ -{"task_id":"JHumanEval\/0","prompt_en":"from typing import List\n\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:\n \"\"\" Check if in given list of numbers, are any two numbers closer to each other than\n given threshold.\n >>> has_close_elements([1.0, 2.0, 3.0], 0.5)\n False\n >>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)\n True\n \"\"\"\n","prompt":"リストnumbersの中に、与えられたthresholdより近い２つの数値が存在するか判定する \n >>> has_close_elements([1.0, 2.0, 3.0], 0.5)\n False\n >>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)\n True","entry_point":"has_close_elements","canonical_solution":" for idx, elem in enumerate(numbers):\n for idx2, elem2 in enumerate(numbers):\n if idx != idx2:\n distance = abs(elem - elem2)\n if distance < threshold:\n return True\n\n return False\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([1.0, 2.0, 3.9, 4.0, 5.0, 2.2], 0.3) == True\n assert candidate([1.0, 2.0, 3.9, 4.0, 5.0, 2.2], 0.05) == False\n assert candidate([1.0, 2.0, 5.9, 4.0, 5.0], 0.95) == True\n assert candidate([1.0, 2.0, 5.9, 4.0, 5.0], 0.8) == False\n assert candidate([1.0, 2.0, 3.0, 4.0, 5.0, 2.0], 0.1) == True\n assert candidate([1.1, 2.2, 3.1, 4.1, 5.1], 1.0) == True\n assert candidate([1.1, 2.2, 3.1, 4.1, 5.1], 0.5) == False\n\n","code":"from typing import List\n\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:"} -{"task_id":"JHumanEval\/1","prompt_en":"from typing import List\n\n\ndef separate_paren_groups(paren_string: str) -> List[str]:\n \"\"\" Input to this function is a string containing multiple groups of nested parentheses. Your goal is to\n separate those group into separate strings and return the list of those.\n Separate groups are balanced (each open brace is properly closed) and not nested within each other\n Ignore any spaces in the input string.\n >>> separate_paren_groups('( ) (( )) (( )( ))')\n ['()', '(())', '(()())']\n \"\"\"\n","prompt":"この関数への入力は、入れ子になった括弧が複数含まれる文字列である。\n あなたの目的は、これらの括弧を別々の文字列に分割し、そのリストを返すことである。\n 分離された括弧はバランスがとれ、つまり、開いた括弧はそれぞれ適切に閉じられていて、\n 互いに入れ子になっていない。引数の文字列内の空白は無視せよ。\n >>> separate_paren_groups('( ) (( )) (( )( ))')\n ['()', '(())', '(()())']","entry_point":"separate_paren_groups","canonical_solution":" result = []\n current_string = []\n current_depth = 0\n\n for c in paren_string:\n if c == '(':\n current_depth += 1\n current_string.append(c)\n elif c == ')':\n current_depth -= 1\n current_string.append(c)\n\n if current_depth == 0:\n result.append(''.join(current_string))\n current_string.clear()\n\n return result\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('(()()) ((())) () ((())()())') == [\n '(()())', '((()))', '()', '((())()())'\n ]\n assert candidate('() (()) ((())) (((())))') == [\n '()', '(())', '((()))', '(((())))'\n ]\n assert candidate('(()(())((())))') == [\n '(()(())((())))'\n ]\n assert candidate('( ) (( )) (( )( ))') == ['()', '(())', '(()())']\n","code":"from typing import List\n\n\ndef separate_paren_groups(paren_string: str) -> List[str]:"} -{"task_id":"JHumanEval\/2","prompt_en":"def truncate_number(number: float) -> float:\n \"\"\" Given a positive floating point number, it can be decomposed into\n and integer part (largest integer smaller than given number) and decimals\n (leftover part always smaller than 1).\n\n Return the decimal part of the number.\n >>> truncate_number(3.5)\n 0.5\n \"\"\"\n","prompt":"正の浮動小数点数が与えられると、それを整数部（与えられた数より小さい最大の整数）\n と小数部（常に1より小さい残余部分）に分解することができる。\n \n 関数は、数値の小数部を返す。\n >>> truncate_number(3.5)\n 0.5","entry_point":"truncate_number","canonical_solution":" return number % 1.0\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate(3.5) == 0.5\n assert abs(candidate(1.33) - 0.33) < 1e-6\n assert abs(candidate(123.456) - 0.456) < 1e-6\n","code":"def truncate_number(number: float) -> float:"} -{"task_id":"JHumanEval\/3","prompt_en":"from typing import List\n\n\ndef below_zero(operations: List[int]) -> bool:\n \"\"\" You're given a list of deposit and withdrawal operations on a bank account that starts with\n zero balance. Your task is to detect if at any point the balance of account fallls below zero, and\n at that point function should return True. Otherwise it should return False.\n >>> below_zero([1, 2, 3])\n False\n >>> below_zero([1, 2, -4, 5])\n True\n \"\"\"\n","prompt":"銀行口座に対する入出金操作のリストが与えられます。あなたのタスクは、残高ゼロから\n 始まて、口座の残高がゼロ以下になったかどうかを検出し、その時点で関数がTrueを\n 返すようにすることです。そうでなければFalseを返すようにしてください。\n >>> below_zero([1, 2, 3])\n False\n >>> below_zero([1, 2, -4, 5])\n True","entry_point":"below_zero","canonical_solution":" balance = 0\n\n for op in operations:\n balance += op\n if balance < 0:\n return True\n\n return False\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == False\n assert candidate([1, 2, -3, 1, 2, -3]) == False\n assert candidate([1, 2, -4, 5, 6]) == True\n assert candidate([1, -1, 2, -2, 5, -5, 4, -4]) == False\n assert candidate([1, -1, 2, -2, 5, -5, 4, -5]) == True\n assert candidate([1, -2, 2, -2, 5, -5, 4, -4]) == True\n","code":"from typing import List\n\n\ndef below_zero(operations: List[int]) -> bool:"} -{"task_id":"JHumanEval\/4","prompt_en":"from typing import List\n\n\ndef mean_absolute_deviation(numbers: List[float]) -> float:\n \"\"\" For a given list of input numbers, calculate Mean Absolute Deviation\n around the mean of this dataset.\n Mean Absolute Deviation is the average absolute difference between each\n element and a centerpoint (mean in this case):\n MAD = average | x - x_mean |\n >>> mean_absolute_deviation([1.0, 2.0, 3.0, 4.0])\n 1.0\n \"\"\"\n","prompt":"第一引数の数値リストに対して、このデータセットの平均値を中心とした平均絶対偏差(MAD)を計算する。\n 平均絶対偏差(MAD)とは、各要素と中心点（この場合は平均値）との差の絶対値の平均である:\n MAD = 平均｜x - x_mean｜\n >>> mean_absolute_deviation([1.0, 2.0, 3.0, 4.0])\n 1.0","entry_point":"mean_absolute_deviation","canonical_solution":" mean = sum(numbers) \/ len(numbers)\n return sum(abs(x - mean) for x in numbers) \/ len(numbers)\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert abs(candidate([1.0, 2.0, 3.0]) - 2.0\/3.0) < 1e-6\n assert abs(candidate([1.0, 2.0, 3.0, 4.0]) - 1.0) < 1e-6\n assert abs(candidate([1.0, 2.0, 3.0, 4.0, 5.0]) - 6.0\/5.0) < 1e-6\n\n","code":"from typing import List\n\n\ndef mean_absolute_deviation(numbers: List[float]) -> float:"} -{"task_id":"JHumanEval\/5","prompt_en":"from typing import List\n\n\ndef intersperse(numbers: List[int], delimeter: int) -> List[int]:\n \"\"\" Insert a number 'delimeter' between every two consecutive elements of input list `numbers'\n >>> intersperse([], 4)\n []\n >>> intersperse([1, 2, 3], 4)\n [1, 4, 2, 4, 3]\n \"\"\"\n","prompt":"数値リスト numbers 中の全ての連続する二要素の間に、'delimeterの値を挿入する\n >>> intersperse([], 4)\n []\n >>> intersperse([1, 2, 3], 4)\n [1, 4, 2, 4, 3]","entry_point":"intersperse","canonical_solution":" if not numbers:\n return []\n\n result = []\n\n for n in numbers[:-1]:\n result.append(n)\n result.append(delimeter)\n\n result.append(numbers[-1])\n\n return result\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([], 7) == []\n assert candidate([5, 6, 3, 2], 8) == [5, 8, 6, 8, 3, 8, 2]\n assert candidate([2, 2, 2], 2) == [2, 2, 2, 2, 2]\n","code":"from typing import List\n\n\ndef intersperse(numbers: List[int], delimeter: int) -> List[int]:"} -{"task_id":"JHumanEval\/6","prompt_en":"from typing import List\n\n\ndef parse_nested_parens(paren_string: str) -> List[int]:\n \"\"\" Input to this function is a string represented multiple groups for nested parentheses separated by spaces.\n For each of the group, output the deepest level of nesting of parentheses.\n E.g. (()()) has maximum two levels of nesting while ((())) has three.\n\n >>> parse_nested_parens('(()()) ((())) () ((())()())')\n [2, 3, 1, 3]\n \"\"\"\n","prompt":"この関数の入力は、空白で区切られた複数の入れ子になった括弧のグループを表す文字列です。\n 各グループについて、括弧の最も深い入れ子のレベルを出力します。\n 例えば、'(()())'は最大で2レベルの入れ子になっていますが、'((()))'は3レベルです。\n \n >>> parse_nested_parens('(()()) ((())) () ((())()())')\n [2, 3, 1, 3]","entry_point":"parse_nested_parens","canonical_solution":" def parse_paren_group(s):\n depth = 0\n max_depth = 0\n for c in s:\n if c == '(':\n depth += 1\n max_depth = max(depth, max_depth)\n else:\n depth -= 1\n\n return max_depth\n\n return [parse_paren_group(x) for x in paren_string.split(' ') if x]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('(()()) ((())) () ((())()())') == [2, 3, 1, 3]\n assert candidate('() (()) ((())) (((())))') == [1, 2, 3, 4]\n assert candidate('(()(())((())))') == [4]\n","code":"from typing import List\n\n\ndef parse_nested_parens(paren_string: str) -> List[int]:"} -{"task_id":"JHumanEval\/7","prompt_en":"from typing import List\n\n\ndef filter_by_substring(strings: List[str], substring: str) -> List[str]:\n \"\"\" Filter an input list of strings only for ones that contain given substring\n >>> filter_by_substring([], 'a')\n []\n >>> filter_by_substring(['abc', 'bacd', 'cde', 'array'], 'a')\n ['abc', 'bacd', 'array']\n \"\"\"\n","prompt":"文字列リストstringsを、与えれた部分文字列substringを含むものだけにフィルタする\n >>> filter_by_substring([], 'a')\n []\n >>> filter_by_substring(['abc', 'bacd', 'cde', 'array'], 'a')\n ['abc', 'bacd', 'array']","entry_point":"filter_by_substring","canonical_solution":" return [x for x in strings if substring in x]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([], 'john') == []\n assert candidate(['xxx', 'asd', 'xxy', 'john doe', 'xxxAAA', 'xxx'], 'xxx') == ['xxx', 'xxxAAA', 'xxx']\n assert candidate(['xxx', 'asd', 'aaaxxy', 'john doe', 'xxxAAA', 'xxx'], 'xx') == ['xxx', 'aaaxxy', 'xxxAAA', 'xxx']\n assert candidate(['grunt', 'trumpet', 'prune', 'gruesome'], 'run') == ['grunt', 'prune']\n","code":"from typing import List\n\n\ndef filter_by_substring(strings: List[str], substring: str) -> List[str]:"} -{"task_id":"JHumanEval\/8","prompt_en":"from typing import List, Tuple\n\n\ndef sum_product(numbers: List[int]) -> Tuple[int, int]:\n \"\"\" For a given list of integers, return a tuple consisting of a sum and a product of all the integers in a list.\n Empty sum should be equal to 0 and empty product should be equal to 1.\n >>> sum_product([])\n (0, 1)\n >>> sum_product([1, 2, 3, 4])\n (10, 24)\n \"\"\"\n","prompt":"与えられた整数リストに対して、リスト内のすべての整数の和と積からなるタプルを返す。\n ただし、空の和は0、空の積は1とする。\n >>> sum_product([])\n (0, 1)\n >>> sum_product([1, 2, 3, 4])\n (10, 24)","entry_point":"sum_product","canonical_solution":" sum_value = 0\n prod_value = 1\n\n for n in numbers:\n sum_value += n\n prod_value *= n\n return sum_value, prod_value\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == (0, 1)\n assert candidate([1, 1, 1]) == (3, 1)\n assert candidate([100, 0]) == (100, 0)\n assert candidate([3, 5, 7]) == (3 + 5 + 7, 3 * 5 * 7)\n assert candidate([10]) == (10, 10)\n","code":"from typing import List, Tuple\n\n\ndef sum_product(numbers: List[int]) -> Tuple[int, int]:"} -{"task_id":"JHumanEval\/9","prompt_en":"from typing import List, Tuple\n\n\ndef rolling_max(numbers: List[int]) -> List[int]:\n \"\"\" From a given list of integers, generate a list of rolling maximum element found until given moment\n in the sequence.\n >>> rolling_max([1, 2, 3, 2, 3, 4, 2])\n [1, 2, 3, 3, 3, 4, 4]\n \"\"\"\n","prompt":"与えられた整数リストから、各要素のそこまでの最大値（ローリング最大値）のリストを生成する。\n >>> rolling_max([1, 2, 3, 2, 3, 4, 2])\n [1, 2, 3, 3, 3, 4, 4]","entry_point":"rolling_max","canonical_solution":" running_max = None\n result = []\n\n for n in numbers:\n if running_max is None:\n running_max = n\n else:\n running_max = max(running_max, n)\n\n result.append(running_max)\n\n return result\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == []\n assert candidate([1, 2, 3, 4]) == [1, 2, 3, 4]\n assert candidate([4, 3, 2, 1]) == [4, 4, 4, 4]\n assert candidate([3, 2, 3, 100, 3]) == [3, 3, 3, 100, 100]\n","code":"from typing import List, Tuple\n\n\ndef rolling_max(numbers: List[int]) -> List[int]:"} -{"task_id":"JHumanEval\/10","prompt_en":"def is_palindrome(string: str) -> bool:\n \"\"\" Test if given string is a palindrome \"\"\"\n return string == string[::-1]\n\n\ndef make_palindrome(string: str) -> str:\n \"\"\" Find the shortest palindrome that begins with a supplied string.\n Algorithm idea is simple:\n - Find the longest postfix of supplied string that is a palindrome.\n - Append to the end of the string reverse of a string prefix that comes before the palindromic suffix.\n >>> make_palindrome('')\n ''\n >>> make_palindrome('cat')\n 'catac'\n >>> make_palindrome('cata')\n 'catac'\n \"\"\"\n","prompt":"与えられた文字列で始まる最短の回文を見つけてください。\n アルゴリズムのアイデアは以下の通りです：\n - 与えられた文字列の中で最も長い回文となる接尾辞を見つけます。\n - その回文の接尾辞の前に来る接頭辞を逆順にして、文字列の末尾に追加します。\n >>> make_palindrome('')\n ''\n >>> make_palindrome('cat')\n 'catac'\n >>> make_palindrome('cata')\n 'catac'","entry_point":"make_palindrome","canonical_solution":" if not string:\n return ''\n\n beginning_of_suffix = 0\n\n while not is_palindrome(string[beginning_of_suffix:]):\n beginning_of_suffix += 1\n\n return string + string[:beginning_of_suffix][::-1]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == ''\n assert candidate('x') == 'x'\n assert candidate('xyz') == 'xyzyx'\n assert candidate('xyx') == 'xyx'\n assert candidate('jerry') == 'jerryrrej'\n","code":"def is_palindrome(string: str) -> bool:\n \n return string == string[::-1]\n\n\ndef make_palindrome(string: str) -> str:"} -{"task_id":"JHumanEval\/11","prompt_en":"from typing import List\n\n\ndef string_xor(a: str, b: str) -> str:\n \"\"\" Input are two strings a and b consisting only of 1s and 0s.\n Perform binary XOR on these inputs and return result also as a string.\n >>> string_xor('010', '110')\n '100'\n \"\"\"\n","prompt":"引数は1と0のみからなる文字列aとbである。\n これらの引数に対して排他論理和(XOR)を実行し、結果を文字列として返す。\n >>> string_xor('010', '110')\n '100'","entry_point":"string_xor","canonical_solution":" def xor(i, j):\n if i == j:\n return '0'\n else:\n return '1'\n\n return ''.join(xor(x, y) for x, y in zip(a, b))\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('111000', '101010') == '010010'\n assert candidate('1', '1') == '0'\n assert candidate('0101', '0000') == '0101'\n","code":"from typing import List\n\n\ndef string_xor(a: str, b: str) -> str:"} -{"task_id":"JHumanEval\/12","prompt_en":"from typing import List, Optional\n\n\ndef longest(strings: List[str]) -> Optional[str]:\n \"\"\" Out of list of strings, return the longest one. Return the first one in case of multiple\n strings of the same length. Return None in case the input list is empty.\n >>> longest([])\n\n >>> longest(['a', 'b', 'c'])\n 'a'\n >>> longest(['a', 'bb', 'ccc'])\n 'ccc'\n \"\"\"\n","prompt":"文字列のリストのうち、最も長いものを返す。同じ長さの文字列が\n 複数ある場合は最初のものを返す。入力リストが空の場合は None を返す。\n >>> longest([])\n\n >>> longest(['a', 'b', 'c'])\n 'a'\n >>> longest(['a', 'bb', 'ccc'])\n 'ccc'","entry_point":"longest","canonical_solution":" if not strings:\n return None\n\n maxlen = max(len(x) for x in strings)\n for s in strings:\n if len(s) == maxlen:\n return s\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == None\n assert candidate(['x', 'y', 'z']) == 'x'\n assert candidate(['x', 'yyy', 'zzzz', 'www', 'kkkk', 'abc']) == 'zzzz'\n","code":"from typing import List, Optional\n\n\ndef longest(strings: List[str]) -> Optional[str]:"} -{"task_id":"JHumanEval\/13","prompt_en":"def greatest_common_divisor(a: int, b: int) -> int:\n \"\"\" Return a greatest common divisor of two integers a and b\n >>> greatest_common_divisor(3, 5)\n 1\n >>> greatest_common_divisor(25, 15)\n 5\n \"\"\"\n","prompt":"整数 a と b の最大公約数を返す\n >>> greatest_common_divisor(3, 5)\n 1\n >>> greatest_common_divisor(25, 15)\n 5","entry_point":"greatest_common_divisor","canonical_solution":" while b:\n a, b = b, a % b\n return a\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate(3, 7) == 1\n assert candidate(10, 15) == 5\n assert candidate(49, 14) == 7\n assert candidate(144, 60) == 12\n","code":"def greatest_common_divisor(a: int, b: int) -> int:"} -{"task_id":"JHumanEval\/14","prompt_en":"from typing import List\n\n\ndef all_prefixes(string: str) -> List[str]:\n \"\"\" Return list of all prefixes from shortest to longest of the input string\n >>> all_prefixes('abc')\n ['a', 'ab', 'abc']\n \"\"\"\n","prompt":"引数で与えられた文字列に対して、短いものから長いものへ、全ての接頭辞のリストを返す\n >>> all_prefixes('abc')\n ['a', 'ab', 'abc']","entry_point":"all_prefixes","canonical_solution":" result = []\n\n for i in range(len(string)):\n result.append(string[:i+1])\n return result\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == []\n assert candidate('asdfgh') == ['a', 'as', 'asd', 'asdf', 'asdfg', 'asdfgh']\n assert candidate('WWW') == ['W', 'WW', 'WWW']\n","code":"from typing import List\n\n\ndef all_prefixes(string: str) -> List[str]:"} -{"task_id":"JHumanEval\/15","prompt_en":"def string_sequence(n: int) -> str:\n \"\"\" Return a string containing space-delimited numbers starting from 0 upto n inclusive.\n >>> string_sequence(0)\n '0'\n >>> string_sequence(5)\n '0 1 2 3 4 5'\n \"\"\"\n","prompt":"0からnまでの数字を空白区切りで連結した文字列で返す。\n >>> string_sequence(0)\n '0'\n >>> string_sequence(5)\n '0 1 2 3 4 5'","entry_point":"string_sequence","canonical_solution":" return ' '.join([str(x) for x in range(n + 1)])\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate(0) == '0'\n assert candidate(3) == '0 1 2 3'\n assert candidate(10) == '0 1 2 3 4 5 6 7 8 9 10'\n","code":"def string_sequence(n: int) -> str:"} -{"task_id":"JHumanEval\/16","prompt_en":"def count_distinct_characters(string: str) -> int:\n \"\"\" Given a string, find out how many distinct characters (regardless of case) does it consist of\n >>> count_distinct_characters('xyzXYZ')\n 3\n >>> count_distinct_characters('Jerry')\n 4\n \"\"\"\n","prompt":"文字列が与えられたとき、その文字列が（大文字小文字に関係なく）いくつの異なる文字が含まれているか数える\n >>> count_distinct_characters('xyzXYZ')\n 3\n >>> count_distinct_characters('Jerry')\n 4","entry_point":"count_distinct_characters","canonical_solution":" return len(set(string.lower()))\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == 0\n assert candidate('abcde') == 5\n assert candidate('abcde' + 'cade' + 'CADE') == 5\n assert candidate('aaaaAAAAaaaa') == 1\n assert candidate('Jerry jERRY JeRRRY') == 5\n","code":"def count_distinct_characters(string: str) -> int:"} -{"task_id":"JHumanEval\/17","prompt_en":"from typing import List\n\n\ndef parse_music(music_string: str) -> List[int]:\n \"\"\" Input to this function is a string representing musical notes in a special ASCII format.\n Your task is to parse this string and return list of integers corresponding to how many beats does each\n not last.\n\n Here is a legend:\n 'o' - whole note, lasts four beats\n 'o|' - half note, lasts two beats\n '.|' - quater note, lasts one beat\n\n >>> parse_music('o o| .| o| o| .| .| .| .| o o')\n [4, 2, 1, 2, 2, 1, 1, 1, 1, 4, 4]\n \"\"\"\n","prompt":"この関数の引数は、特別なASCII形式の音符を表す文字列である。あなたの仕事は、この文字列を解析して、それぞれの音符が何拍続くかに対応する整数のリストを返すことである。\n ここに凡例がある：\n o' - 全音符、4拍続く\n o|' - 2分音符、2拍続く\n .|」-4分音符、1拍続く\n \n >>> parse_music('o o| .| o| o| .| .| .| .| o o')\n [4, 2, 1, 2, 2, 1, 1, 1, 1, 4, 4]","entry_point":"parse_music","canonical_solution":" note_map = {'o': 4, 'o|': 2, '.|': 1}\n return [note_map[x] for x in music_string.split(' ') if x]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == []\n assert candidate('o o o o') == [4, 4, 4, 4]\n assert candidate('.| .| .| .|') == [1, 1, 1, 1]\n assert candidate('o| o| .| .| o o o o') == [2, 2, 1, 1, 4, 4, 4, 4]\n assert candidate('o| .| o| .| o o| o o|') == [2, 1, 2, 1, 4, 2, 4, 2]\n","code":"from typing import List\n\n\ndef parse_music(music_string: str) -> List[int]:"} -{"task_id":"JHumanEval\/18","prompt_en":"def how_many_times(string: str, substring: str) -> int:\n \"\"\" Find how many times a given substring can be found in the original string. Count overlaping cases.\n >>> how_many_times('', 'a')\n 0\n >>> how_many_times('aaa', 'a')\n 3\n >>> how_many_times('aaaa', 'aa')\n 3\n \"\"\"\n","prompt":"部分文字列substringが文字列stringの中で何回見つかるか数える。\n 重なるケースもカウントに含まれる。\n >>> how_many_times('', 'a')\n 0\n >>> how_many_times('aaa', 'a')\n 3\n >>> how_many_times('aaaa', 'aa')\n 3","entry_point":"how_many_times","canonical_solution":" times = 0\n\n for i in range(len(string) - len(substring) + 1):\n if string[i:i+len(substring)] == substring:\n times += 1\n\n return times\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('', 'x') == 0\n assert candidate('xyxyxyx', 'x') == 4\n assert candidate('cacacacac', 'cac') == 4\n assert candidate('john doe', 'john') == 1\n","code":"def how_many_times(string: str, substring: str) -> int:"} -{"task_id":"JHumanEval\/19","prompt_en":"from typing import List\n\n\ndef sort_numbers(numbers: str) -> str:\n \"\"\" Input is a space-delimited string of numberals from 'zero' to 'nine'.\n Valid choices are 'zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight' and 'nine'.\n Return the string with numbers sorted from smallest to largest\n >>> sort_numbers('three one five')\n 'one three five'\n \"\"\"\n","prompt":"引数は'zero'から'nine'までの英単語の数を空白で区切った文字列である。\n 有効な英単語は''、'zero', 'one'、'two'、'three'、'four'、'five'、'six'、'seven'、'eight'、'nine'である。\n 関数は、英単語の数を小さい方から大きい方へとソートした文字列を返す。\n >>> sort_numbers('three one five')\n 'one three five'","entry_point":"sort_numbers","canonical_solution":" value_map = {\n 'zero': 0,\n 'one': 1,\n 'two': 2,\n 'three': 3,\n 'four': 4,\n 'five': 5,\n 'six': 6,\n 'seven': 7,\n 'eight': 8,\n 'nine': 9\n }\n return ' '.join(sorted([x for x in numbers.split(' ') if x], key=lambda x: value_map[x]))\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == ''\n assert candidate('three') == 'three'\n assert candidate('three five nine') == 'three five nine'\n assert candidate('five zero four seven nine eight') == 'zero four five seven eight nine'\n assert candidate('six five four three two one zero') == 'zero one two three four five six'\n","code":"from typing import List\n\n\ndef sort_numbers(numbers: str) -> str:"} -{"task_id":"JHumanEval\/20","prompt_en":"from typing import List, Tuple\n\n\ndef find_closest_elements(numbers: List[float]) -> Tuple[float, float]:\n \"\"\" From a supplied list of numbers (of length at least two) select and return two that are the closest to each\n other and return them in order (smaller number, larger number).\n >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])\n (2.0, 2.2)\n >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])\n (2.0, 2.0)\n \"\"\"\n","prompt":"（少なくとも長さ2以上の）リストnumbersから、互いに最も近いものを2つ選び、\n 順番に（小さい数、大きい数）返す。\n >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])\n (2.0, 2.2)\n >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])\n (2.0, 2.0)","entry_point":"find_closest_elements","canonical_solution":" closest_pair = None\n distance = None\n\n for idx, elem in enumerate(numbers):\n for idx2, elem2 in enumerate(numbers):\n if idx != idx2:\n if distance is None:\n distance = abs(elem - elem2)\n closest_pair = tuple(sorted([elem, elem2]))\n else:\n new_distance = abs(elem - elem2)\n if new_distance < distance:\n distance = new_distance\n closest_pair = tuple(sorted([elem, elem2]))\n\n return closest_pair\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([1.0, 2.0, 3.9, 4.0, 5.0, 2.2]) == (3.9, 4.0)\n assert candidate([1.0, 2.0, 5.9, 4.0, 5.0]) == (5.0, 5.9)\n assert candidate([1.0, 2.0, 3.0, 4.0, 5.0, 2.2]) == (2.0, 2.2)\n assert candidate([1.0, 2.0, 3.0, 4.0, 5.0, 2.0]) == (2.0, 2.0)\n assert candidate([1.1, 2.2, 3.1, 4.1, 5.1]) == (2.2, 3.1)\n\n","code":"from typing import List, Tuple\n\n\ndef find_closest_elements(numbers: List[float]) -> Tuple[float, float]:"} -{"task_id":"JHumanEval\/21","prompt_en":"from typing import List\n\n\ndef rescale_to_unit(numbers: List[float]) -> List[float]:\n \"\"\" Given list of numbers (of at least two elements), apply a linear transform to that list,\n such that the smallest number will become 0 and the largest will become 1\n >>> rescale_to_unit([1.0, 2.0, 3.0, 4.0, 5.0])\n [0.0, 0.25, 0.5, 0.75, 1.0]\n \"\"\"\n","prompt":"(少なくとも 2 つ以上の要素からなる) リストnumbersに線形変換を適用し、\n 最小の数値が 0 になり、最大の数値が 1 になるリストを返す\n >>> rescale_to_unit([1.0, 2.0, 3.0, 4.0, 5.0])\n [0.0, 0.25, 0.5, 0.75, 1.0]","entry_point":"rescale_to_unit","canonical_solution":" min_number = min(numbers)\n max_number = max(numbers)\n return [(x - min_number) \/ (max_number - min_number) for x in numbers]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([2.0, 49.9]) == [0.0, 1.0]\n assert candidate([100.0, 49.9]) == [1.0, 0.0]\n assert candidate([1.0, 2.0, 3.0, 4.0, 5.0]) == [0.0, 0.25, 0.5, 0.75, 1.0]\n assert candidate([2.0, 1.0, 5.0, 3.0, 4.0]) == [0.25, 0.0, 1.0, 0.5, 0.75]\n assert candidate([12.0, 11.0, 15.0, 13.0, 14.0]) == [0.25, 0.0, 1.0, 0.5, 0.75]\n","code":"from typing import List\n\n\ndef rescale_to_unit(numbers: List[float]) -> List[float]:"} -{"task_id":"JHumanEval\/22","prompt_en":"from typing import List, Any\n\n\ndef filter_integers(values: List[Any]) -> List[int]:\n \"\"\" Filter given list of any python values only for integers\n >>> filter_integers(['a', 3.14, 5])\n [5]\n >>> filter_integers([1, 2, 3, 'abc', {}, []])\n [1, 2, 3]\n \"\"\"\n","prompt":"任意の種類の値が含まれるリストから整数値のみ抽出する\n >>> filter_integers(['a', 3.14, 5])\n [5]\n >>> filter_integers([1, 2, 3, 'abc', {}, []])\n [1, 2, 3]","entry_point":"filter_integers","canonical_solution":" return [x for x in values if isinstance(x, int)]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == []\n assert candidate([4, {}, [], 23.2, 9, 'adasd']) == [4, 9]\n assert candidate([3, 'c', 3, 3, 'a', 'b']) == [3, 3, 3]\n","code":"from typing import List, Any\n\n\ndef filter_integers(values: List[Any]) -> List[int]:"} -{"task_id":"JHumanEval\/23","prompt_en":"def strlen(string: str) -> int:\n \"\"\" Return length of given string\n >>> strlen('')\n 0\n >>> strlen('abc')\n 3\n \"\"\"\n","prompt":"引数で与えられた文字列の長さを返す\n >>> strlen('')\n 0\n >>> strlen('abc')\n 3","entry_point":"strlen","canonical_solution":" return len(string)\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == 0\n assert candidate('x') == 1\n assert candidate('asdasnakj') == 9\n","code":"def strlen(string: str) -> int:"} -{"task_id":"JHumanEval\/24","prompt_en":"def largest_divisor(n: int) -> int:\n \"\"\" For a given number n, find the largest number that divides n evenly, smaller than n\n >>> largest_divisor(15)\n 5\n \"\"\"\n","prompt":"与えられた数nについて、nの約数のうち、nより小さい最大の数を求める\n >>> largest_divisor(15)\n 5","entry_point":"largest_divisor","canonical_solution":" for i in reversed(range(n)):\n if n % i == 0:\n return i\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate(3) == 1\n assert candidate(7) == 1\n assert candidate(10) == 5\n assert candidate(100) == 50\n assert candidate(49) == 7\n","code":"def largest_divisor(n: int) -> int:"} -{"task_id":"JHumanEval\/25","prompt_en":"from typing import List\n\n\ndef factorize(n: int) -> List[int]:\n \"\"\" Return list of prime factors of given integer in the order from smallest to largest.\n Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.\n Input number should be equal to the product of all factors\n >>> factorize(8)\n [2, 2, 2]\n >>> factorize(25)\n [5, 5]\n >>> factorize(70)\n [2, 5, 7]\n \"\"\"\n","prompt":"与えられた整数の素因数のリストを小さいものから大きいものの順に返す。各因数は、\n 因数分解で現れる回数分、リストに登場する。引数の整数は全ての因数の積に等しくな\n ければならない。\n >>> factorize(8)\n [2, 2, 2]\n >>> factorize(25)\n [5, 5]\n >>> factorize(70)\n [2, 5, 7]","entry_point":"factorize","canonical_solution":" import math\n fact = []\n i = 2\n while i <= int(math.sqrt(n) + 1):\n if n % i == 0:\n fact.append(i)\n n \/\/= i\n else:\n i += 1\n\n if n > 1:\n fact.append(n)\n return fact\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate(2) == [2]\n assert candidate(4) == [2, 2]\n assert candidate(8) == [2, 2, 2]\n assert candidate(3 * 19) == [3, 19]\n assert candidate(3 * 19 * 3 * 19) == [3, 3, 19, 19]\n assert candidate(3 * 19 * 3 * 19 * 3 * 19) == [3, 3, 3, 19, 19, 19]\n assert candidate(3 * 19 * 19 * 19) == [3, 19, 19, 19]\n assert candidate(3 * 2 * 3) == [2, 3, 3]\n","code":"from typing import List\n\n\ndef factorize(n: int) -> List[int]:"} -{"task_id":"JHumanEval\/26","prompt_en":"from typing import List\n\n\ndef remove_duplicates(numbers: List[int]) -> List[int]:\n \"\"\" From a list of integers, remove all elements that occur more than once.\n Keep order of elements left the same as in the input.\n >>> remove_duplicates([1, 2, 3, 2, 4])\n [1, 3, 4]\n \"\"\"\n","prompt":"整数のリストから、複数回出現する要素をすべて取り除く。\n 要素の順序は入力と同じようにする。\n >>> remove_duplicates([1, 2, 3, 2, 4])\n [1, 3, 4]","entry_point":"remove_duplicates","canonical_solution":" import collections\n c = collections.Counter(numbers)\n return [n for n in numbers if c[n] <= 1]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == []\n assert candidate([1, 2, 3, 4]) == [1, 2, 3, 4]\n assert candidate([1, 2, 3, 2, 4, 3, 5]) == [1, 4, 5]\n","code":"from typing import List\n\n\ndef remove_duplicates(numbers: List[int]) -> List[int]:"} -{"task_id":"JHumanEval\/27","prompt_en":"def flip_case(string: str) -> str:\n \"\"\" For a given string, flip lowercase characters to uppercase and uppercase to lowercase.\n >>> flip_case('Hello')\n 'hELLO'\n \"\"\"\n","prompt":"与えられた文字列に対して、英小文字を英大文字に、英大文字を英小文字に変換する。\n >>> flip_case('Hello')\n 'hELLO'","entry_point":"flip_case","canonical_solution":" return string.swapcase()\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate('') == ''\n assert candidate('Hello!') == 'hELLO!'\n assert candidate('These violent delights have violent ends') == 'tHESE VIOLENT DELIGHTS HAVE VIOLENT ENDS'\n","code":"def flip_case(string: str) -> str:"} -{"task_id":"JHumanEval\/28","prompt_en":"from typing import List\n\n\ndef concatenate(strings: List[str]) -> str:\n \"\"\" Concatenate list of strings into a single string\n >>> concatenate([])\n ''\n >>> concatenate(['a', 'b', 'c'])\n 'abc'\n \"\"\"\n","prompt":"文字列のリストを1つの文字列に連結する\n >>> concatenate([])\n ''\n >>> concatenate(['a', 'b', 'c'])\n 'abc'","entry_point":"concatenate","canonical_solution":" return ''.join(strings)\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([]) == ''\n assert candidate(['x', 'y', 'z']) == 'xyz'\n assert candidate(['x', 'y', 'z', 'w', 'k']) == 'xyzwk'\n","code":"from typing import List\n\n\ndef concatenate(strings: List[str]) -> str:"} -{"task_id":"JHumanEval\/29","prompt_en":"from typing import List\n\n\ndef filter_by_prefix(strings: List[str], prefix: str) -> List[str]:\n \"\"\" Filter an input list of strings only for ones that start with a given prefix.\n >>> filter_by_prefix([], 'a')\n []\n >>> filter_by_prefix(['abc', 'bcd', 'cde', 'array'], 'a')\n ['abc', 'array']\n \"\"\"\n","prompt":"文字列のリストから、指定された接頭辞prefixで始まるものだけを取り出す。\n >>> filter_by_prefix([], 'a')\n []\n >>> filter_by_prefix(['abc', 'bcd', 'cde', 'array'], 'a')\n ['abc', 'array']","entry_point":"filter_by_prefix","canonical_solution":" return [x for x in strings if x.startswith(prefix)]\n","test":"\n\nMETADATA = {\n 'author': 'jt',\n 'dataset': 'test'\n}\n\n\ndef check(candidate):\n assert candidate([], 'john') == []\n assert candidate(['xxx', 'asd', 'xxy', 'john doe', 'xxxAAA', 'xxx'], 'xxx') == ['xxx', 'xxxAAA', 'xxx']\n","code":"from typing import List\n\n\ndef filter_by_prefix(strings: List[str], prefix: str) -> List[str]:"} -{"task_id":"JHumanEval\/30","prompt_en":"def get_positive(l: list):\n \"\"\"Return only positive numbers in the list.\n >>> get_positive([-1, 2, -4, 5, 6])\n [2, 5, 6]\n >>> get_positive([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])\n [5, 3, 2, 3, 9, 123, 1]\n \"\"\"\n","prompt":"リスト内の正の数だけを返す。\n >>> get_positive([-1, 2, -4, 5, 6])\n [2, 5, 6]\n >>> get_positive([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])\n [5, 3, 2, 3, 9, 123, 1]","entry_point":"get_positive","canonical_solution":" return [e for e in l if e > 0]\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([-1, -2, 4, 5, 6]) == [4, 5, 6]\n assert candidate([5, 3, -5, 2, 3, 3, 9, 0, 123, 1, -10]) == [5, 3, 2, 3, 3, 9, 123, 1]\n assert candidate([-1, -2]) == []\n assert candidate([]) == []\n\n","code":"def get_positive(l: list):"} -{"task_id":"JHumanEval\/31","prompt_en":"def is_prime(n):\n \"\"\"Return true if a given number is prime, and false otherwise.\n >>> is_prime(6)\n False\n >>> is_prime(101)\n True\n >>> is_prime(11)\n True\n >>> is_prime(13441)\n True\n >>> is_prime(61)\n True\n >>> is_prime(4)\n False\n >>> is_prime(1)\n False\n \"\"\"\n","prompt":"与えられた数が素数であれば真を、そうでなければ偽を返す。\n >>> is_prime(6)\n False\n >>> is_prime(101)\n True\n >>> is_prime(11)\n True\n >>> is_prime(13441)\n True\n >>> is_prime(61)\n True\n >>> is_prime(4)\n False\n >>> is_prime(1)\n False","entry_point":"is_prime","canonical_solution":" if n < 2:\n return False\n for k in range(2, n - 1):\n if n % k == 0:\n return False\n return True\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(6) == False\n assert candidate(101) == True\n assert candidate(11) == True\n assert candidate(13441) == True\n assert candidate(61) == True\n assert candidate(4) == False\n assert candidate(1) == False\n assert candidate(5) == True\n assert candidate(11) == True\n assert candidate(17) == True\n assert candidate(5 * 17) == False\n assert candidate(11 * 7) == False\n assert candidate(13441 * 19) == False\n\n","code":"def is_prime(n):"} -{"task_id":"JHumanEval\/32","prompt_en":"import math\n\n\ndef poly(xs: list, x: float):\n \"\"\"\n Evaluates polynomial with coefficients xs at point x.\n return xs[0] + xs[1] * x + xs[1] * x^2 + .... xs[n] * x^n\n \"\"\"\n return sum([coeff * math.pow(x, i) for i, coeff in enumerate(xs)])\n\n\ndef find_zero(xs: list):\n \"\"\" xs are coefficients of a polynomial.\n find_zero find x such that poly(x) = 0.\n find_zero returns only only zero point, even if there are many.\n Moreover, find_zero only takes list xs having even number of coefficients\n and largest non zero coefficient as it guarantees\n a solution.\n >>> round(find_zero([1, 2]), 2) # f(x) = 1 + 2x\n -0.5\n >>> round(find_zero([-6, 11, -6, 1]), 2) # (x - 1) * (x - 2) * (x - 3) = -6 + 11x - 6x^2 + x^3\n 1.0\n \"\"\"\n","prompt":"xsは多項式の係数である。\n find_zero関数は、poly(x) = 0 となる x を見つける。\n find_zero関数は、たとえ複数解があったとしても解をひとつのみを返す。\n さらに、find_zero関数は、解を持つことを保証するため、偶数個の係数xsと\n 最大係数は常に0でないと想定する。\n >>> round(find_zero([1, 2]), 2) # f(x) = 1 + 2x\n -0.5\n >>> round(find_zero([-6, 11, -6, 1]), 2) # (x - 1) * (x - 2) * (x - 3) = -6 + 11x - 6x^2 + x^3\n 1.0","entry_point":"find_zero","canonical_solution":" begin, end = -1., 1.\n while poly(xs, begin) * poly(xs, end) > 0:\n begin *= 2.0\n end *= 2.0\n while end - begin > 1e-10:\n center = (begin + end) \/ 2.0\n if poly(xs, center) * poly(xs, begin) > 0:\n begin = center\n else:\n end = center\n return begin\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n import math\n import random\n rng = random.Random(42)\n import copy\n for _ in range(100):\n ncoeff = 2 * rng.randint(1, 4)\n coeffs = []\n for _ in range(ncoeff):\n coeff = rng.randint(-10, 10)\n if coeff == 0:\n coeff = 1\n coeffs.append(coeff)\n solution = candidate(copy.deepcopy(coeffs))\n assert math.fabs(poly(coeffs, solution)) < 1e-4\n\n","code":"import math\n\n\ndef poly(xs: list, x: float):\n \n return sum([coeff * math.pow(x, i) for i, coeff in enumerate(xs)])\n\n\ndef find_zero(xs: list):"} -{"task_id":"JHumanEval\/33","prompt_en":"def sort_third(l: list):\n \"\"\"This function takes a list l and returns a list l' such that\n l' is identical to l in the indicies that are not divisible by three, while its values at the indicies that are divisible by three are equal\n to the values of the corresponding indicies of l, but sorted.\n >>> sort_third([1, 2, 3])\n [1, 2, 3]\n >>> sort_third([5, 6, 3, 4, 8, 9, 2])\n [2, 6, 3, 4, 8, 9, 5]\n \"\"\"\n","prompt":"この関数はリストlを受け取り、l'を返す。l'は、インデックスが3で割り\n 切れない場合はlと同じであるが、インデックスが3で割り切れる要素は\n ソートされている。\n >>> sort_third([1, 2, 3])\n [1, 2, 3]\n >>> sort_third([5, 6, 3, 4, 8, 9, 2])\n [2, 6, 3, 4, 8, 9, 5]","entry_point":"sort_third","canonical_solution":" l = list(l)\n l[::3] = sorted(l[::3])\n return l\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert tuple(candidate([1, 2, 3])) == tuple(sort_third([1, 2, 3]))\n assert tuple(candidate([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])) == tuple(sort_third([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10]))\n assert tuple(candidate([5, 8, -12, 4, 23, 2, 3, 11, 12, -10])) == tuple(sort_third([5, 8, -12, 4, 23, 2, 3, 11, 12, -10]))\n assert tuple(candidate([5, 6, 3, 4, 8, 9, 2])) == tuple([2, 6, 3, 4, 8, 9, 5])\n assert tuple(candidate([5, 8, 3, 4, 6, 9, 2])) == tuple([2, 8, 3, 4, 6, 9, 5])\n assert tuple(candidate([5, 6, 9, 4, 8, 3, 2])) == tuple([2, 6, 9, 4, 8, 3, 5])\n assert tuple(candidate([5, 6, 3, 4, 8, 9, 2, 1])) == tuple([2, 6, 3, 4, 8, 9, 5, 1])\n\n","code":"def sort_third(l: list):"} -{"task_id":"JHumanEval\/34","prompt_en":"def unique(l: list):\n \"\"\"Return sorted unique elements in a list\n >>> unique([5, 3, 5, 2, 3, 3, 9, 0, 123])\n [0, 2, 3, 5, 9, 123]\n \"\"\"\n","prompt":"リスト内のユニークな要素をソートして返す\n >>> unique([5, 3, 5, 2, 3, 3, 9, 0, 123])\n [0, 2, 3, 5, 9, 123]","entry_point":"unique","canonical_solution":" return sorted(list(set(l)))\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([5, 3, 5, 2, 3, 3, 9, 0, 123]) == [0, 2, 3, 5, 9, 123]\n\n","code":"def unique(l: list):"} -{"task_id":"JHumanEval\/35","prompt_en":"def max_element(l: list):\n \"\"\"Return maximum element in the list.\n >>> max_element([1, 2, 3])\n 3\n >>> max_element([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])\n 123\n \"\"\"\n","prompt":"リスト内の最大要素を返す。\n >>> max_element([1, 2, 3])\n 3\n >>> max_element([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])\n 123","entry_point":"max_element","canonical_solution":" m = l[0]\n for e in l:\n if e > m:\n m = e\n return m\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 2, 3]) == 3\n assert candidate([5, 3, -5, 2, -3, 3, 9, 0, 124, 1, -10]) == 124\n","code":"def max_element(l: list):"} -{"task_id":"JHumanEval\/36","prompt_en":"def fizz_buzz(n: int):\n \"\"\"Return the number of times the digit 7 appears in integers less than n which are divisible by 11 or 13.\n >>> fizz_buzz(50)\n 0\n >>> fizz_buzz(78)\n 2\n >>> fizz_buzz(79)\n 3\n \"\"\"\n","prompt":"11または13で割り切れるn未満の整数の中に7という数字が現れる回数を返す。\n >>> fizz_buzz(50)\n 0\n >>> fizz_buzz(78)\n 2\n >>> fizz_buzz(79)\n 3","entry_point":"fizz_buzz","canonical_solution":" ns = []\n for i in range(n):\n if i % 11 == 0 or i % 13 == 0:\n ns.append(i)\n s = ''.join(list(map(str, ns)))\n ans = 0\n for c in s:\n ans += (c == '7')\n return ans\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(50) == 0\n assert candidate(78) == 2\n assert candidate(79) == 3\n assert candidate(100) == 3\n assert candidate(200) == 6\n assert candidate(4000) == 192\n assert candidate(10000) == 639\n assert candidate(100000) == 8026\n\n","code":"def fizz_buzz(n: int):"} -{"task_id":"JHumanEval\/37","prompt_en":"def sort_even(l: list):\n \"\"\"This function takes a list l and returns a list l' such that\n l' is identical to l in the odd indicies, while its values at the even indicies are equal\n to the values of the even indicies of l, but sorted.\n >>> sort_even([1, 2, 3])\n [1, 2, 3]\n >>> sort_even([5, 6, 3, 4])\n [3, 6, 5, 4]\n \"\"\"\n","prompt":"この関数はリスト l を受け取り、l' を返す。l'は、インデックスが奇数の\n ときは l と同じで、インデックスが偶数のときはソートされている。\n >>> sort_even([1, 2, 3])\n [1, 2, 3]\n >>> sort_even([5, 6, 3, 4])\n [3, 6, 5, 4]","entry_point":"sort_even","canonical_solution":" evens = l[::2]\n odds = l[1::2]\n evens.sort()\n ans = []\n for e, o in zip(evens, odds):\n ans.extend([e, o])\n if len(evens) > len(odds):\n ans.append(evens[-1])\n return ans\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert tuple(candidate([1, 2, 3])) == tuple([1, 2, 3])\n assert tuple(candidate([5, 3, -5, 2, -3, 3, 9, 0, 123, 1, -10])) == tuple([-10, 3, -5, 2, -3, 3, 5, 0, 9, 1, 123])\n assert tuple(candidate([5, 8, -12, 4, 23, 2, 3, 11, 12, -10])) == tuple([-12, 8, 3, 4, 5, 2, 12, 11, 23, -10])\n\n","code":"def sort_even(l: list):"} -{"task_id":"JHumanEval\/38","prompt_en":"def encode_cyclic(s: str):\n \"\"\"\n returns encoded string by cycling groups of three characters.\n \"\"\"\n # split string to groups. Each of length 3.\n groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) \/\/ 3)]\n # cycle elements in each group. Unless group has fewer elements than 3.\n groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]\n return \"\".join(groups)\n\n\ndef decode_cyclic(s: str):\n \"\"\"\n takes as input string encoded with encode_cyclic function. Returns decoded string.\n \"\"\"\n","prompt":"encode_cyclic関数でエンコードされた文字列を引数に取り、デコードされた文字列を返す。","entry_point":"decode_cyclic","canonical_solution":" return encode_cyclic(encode_cyclic(s))\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n from random import randint, choice\n import string\n\n letters = string.ascii_lowercase\n for _ in range(100):\n str = ''.join(choice(letters) for i in range(randint(10, 20)))\n encoded_str = encode_cyclic(str)\n assert candidate(encoded_str) == str\n\n","code":"def encode_cyclic(s: str):\n \n # 文字列長が3になるように文字列をグループ化\n groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) \/\/ 3)]\n # グループの長さが3未満でない限り、各グループを循環させる\n groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]\n return \"\".join(groups)\n\n\ndef decode_cyclic(s: str):"} -{"task_id":"JHumanEval\/39","prompt_en":"def prime_fib(n: int):\n \"\"\"\n prime_fib returns n-th number that is a Fibonacci number and it's also prime.\n >>> prime_fib(1)\n 2\n >>> prime_fib(2)\n 3\n >>> prime_fib(3)\n 5\n >>> prime_fib(4)\n 13\n >>> prime_fib(5)\n 89\n \"\"\"\n","prompt":"prime_fib はフィボナッチ数で、かつ素数であるn番目の数を返す。\n >>> prime_fib(1)\n 2\n >>> prime_fib(2)\n 3\n >>> prime_fib(3)\n 5\n >>> prime_fib(4)\n 13\n >>> prime_fib(5)\n 89","entry_point":"prime_fib","canonical_solution":" import math\n\n def is_prime(p):\n if p < 2:\n return False\n for k in range(2, min(int(math.sqrt(p)) + 1, p - 1)):\n if p % k == 0:\n return False\n return True\n f = [0, 1]\n while True:\n f.append(f[-1] + f[-2])\n if is_prime(f[-1]):\n n -= 1\n if n == 0:\n return f[-1]\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(1) == 2\n assert candidate(2) == 3\n assert candidate(3) == 5\n assert candidate(4) == 13\n assert candidate(5) == 89\n assert candidate(6) == 233\n assert candidate(7) == 1597\n assert candidate(8) == 28657\n assert candidate(9) == 514229\n assert candidate(10) == 433494437\n\n","code":"def prime_fib(n: int):"} -{"task_id":"JHumanEval\/40","prompt_en":"def triples_sum_to_zero(l: list):\n \"\"\"\n triples_sum_to_zero takes a list of integers as an input.\n it returns True if there are three distinct elements in the list that\n sum to zero, and False otherwise.\n\n >>> triples_sum_to_zero([1, 3, 5, 0])\n False\n >>> triples_sum_to_zero([1, 3, -2, 1])\n True\n >>> triples_sum_to_zero([1, 2, 3, 7])\n False\n >>> triples_sum_to_zero([2, 4, -5, 3, 9, 7])\n True\n >>> triples_sum_to_zero([1])\n False\n \"\"\"\n","prompt":"triples_sum_to_zero は整数のリストを引数に取り、\n リストの中に和が0になる３つの要素があればTrueを、\n そうでなければFalseを返す。\n \n >>> triples_sum_to_zero([1, 3, 5, 0])\n False\n >>> triples_sum_to_zero([1, 3, -2, 1])\n True\n >>> triples_sum_to_zero([1, 2, 3, 7])\n False\n >>> triples_sum_to_zero([2, 4, -5, 3, 9, 7])\n True\n >>> triples_sum_to_zero([1])\n False","entry_point":"triples_sum_to_zero","canonical_solution":" for i in range(len(l)):\n for j in range(i + 1, len(l)):\n for k in range(j + 1, len(l)):\n if l[i] + l[j] + l[k] == 0:\n return True\n return False\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 3, 5, 0]) == False\n assert candidate([1, 3, 5, -1]) == False\n assert candidate([1, 3, -2, 1]) == True\n assert candidate([1, 2, 3, 7]) == False\n assert candidate([1, 2, 5, 7]) == False\n assert candidate([2, 4, -5, 3, 9, 7]) == True\n assert candidate([1]) == False\n assert candidate([1, 3, 5, -100]) == False\n assert candidate([100, 3, 5, -100]) == False\n\n","code":"def triples_sum_to_zero(l: list):"} -{"task_id":"JHumanEval\/41","prompt_en":"def car_race_collision(n: int):\n \"\"\"\n Imagine a road that's a perfectly straight infinitely long line.\n n cars are driving left to right; simultaneously, a different set of n cars\n are driving right to left. The two sets of cars start out being very far from\n each other. All cars move in the same speed. Two cars are said to collide\n when a car that's moving left to right hits a car that's moving right to left.\n However, the cars are infinitely sturdy and strong; as a result, they continue moving\n in their trajectory as if they did not collide.\n\n This function outputs the number of such collisions.\n \"\"\"\n","prompt":"完全な直線で無限に長い道路を想像してほしい。\n n台の車が左から右に向かって走っている。同時に、別のn台の車が\n 右から左に向かって走っている。この2組の車は、最初は互いに非\n 常に離れている。すべての車は同じ速度で動く。2台の車は次のよ\n うに衝突する。左から右に動いている車が、右から左に動いている\n 車にぶつかること。\n しかし、車は限りなく頑丈で強い。あたかも衝突しなかったかのよ\n うに、その軌道を進み続ける。 \n \n この関数は、このような衝突の回数を出力する。","entry_point":"car_race_collision","canonical_solution":" return n**2\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(2) == 4\n assert candidate(3) == 9\n assert candidate(4) == 16\n assert candidate(8) == 64\n assert candidate(10) == 100\n\n","code":"def car_race_collision(n: int):"} -{"task_id":"JHumanEval\/42","prompt_en":"def incr_list(l: list):\n \"\"\"Return list with elements incremented by 1.\n >>> incr_list([1, 2, 3])\n [2, 3, 4]\n >>> incr_list([5, 3, 5, 2, 3, 3, 9, 0, 123])\n [6, 4, 6, 3, 4, 4, 10, 1, 124]\n \"\"\"\n","prompt":"要素を1ずつ増やしたリストを返す。\n >>> incr_list([1, 2, 3])\n [2, 3, 4]\n >>> incr_list([5, 3, 5, 2, 3, 3, 9, 0, 123])\n [6, 4, 6, 3, 4, 4, 10, 1, 124]","entry_point":"incr_list","canonical_solution":" return [(e + 1) for e in l]\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([]) == []\n assert candidate([3, 2, 1]) == [4, 3, 2]\n assert candidate([5, 2, 5, 2, 3, 3, 9, 0, 123]) == [6, 3, 6, 3, 4, 4, 10, 1, 124]\n\n","code":"def incr_list(l: list):"} -{"task_id":"JHumanEval\/43","prompt_en":"def pairs_sum_to_zero(l):\n \"\"\"\n pairs_sum_to_zero takes a list of integers as an input.\n it returns True if there are two distinct elements in the list that\n sum to zero, and False otherwise.\n >>> pairs_sum_to_zero([1, 3, 5, 0])\n False\n >>> pairs_sum_to_zero([1, 3, -2, 1])\n False\n >>> pairs_sum_to_zero([1, 2, 3, 7])\n False\n >>> pairs_sum_to_zero([2, 4, -5, 3, 5, 7])\n True\n >>> pairs_sum_to_zero([1])\n False\n \"\"\"\n","prompt":"pairs_sum_to_zero は整数のリストを引数にとる。\n リストの中に２つの要素の和がゼロになる要素があればTrueを、\n そうでなければFalseを返す。\n >>> pairs_sum_to_zero([1, 3, 5, 0])\n False\n >>> pairs_sum_to_zero([1, 3, -2, 1])\n False\n >>> pairs_sum_to_zero([1, 2, 3, 7])\n False\n >>> pairs_sum_to_zero([2, 4, -5, 3, 5, 7])\n True\n >>> pairs_sum_to_zero([1])\n False","entry_point":"pairs_sum_to_zero","canonical_solution":" for i, l1 in enumerate(l):\n for j in range(i + 1, len(l)):\n if l1 + l[j] == 0:\n return True\n return False\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 3, 5, 0]) == False\n assert candidate([1, 3, -2, 1]) == False\n assert candidate([1, 2, 3, 7]) == False\n assert candidate([2, 4, -5, 3, 5, 7]) == True\n assert candidate([1]) == False\n\n assert candidate([-3, 9, -1, 3, 2, 30]) == True\n assert candidate([-3, 9, -1, 3, 2, 31]) == True\n assert candidate([-3, 9, -1, 4, 2, 30]) == False\n assert candidate([-3, 9, -1, 4, 2, 31]) == False\n\n","code":"def pairs_sum_to_zero(l):"} -{"task_id":"JHumanEval\/44","prompt_en":"def change_base(x: int, base: int):\n \"\"\"Change numerical base of input number x to base.\n return string representation after the conversion.\n base numbers are less than 10.\n >>> change_base(8, 3)\n '22'\n >>> change_base(8, 2)\n '1000'\n >>> change_base(7, 2)\n '111'\n \"\"\"\n","prompt":"引数xの基数をbaseに変換する。\n 返り値は変換後の文字列表現である。\n 基数は10未満である。\n >>> change_base(8, 3)\n '22'\n >>> change_base(8, 2)\n '1000'\n >>> change_base(7, 2)\n '111'","entry_point":"change_base","canonical_solution":" ret = \"\"\n while x > 0:\n ret = str(x % base) + ret\n x \/\/= base\n return ret\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(8, 3) == \"22\"\n assert candidate(9, 3) == \"100\"\n assert candidate(234, 2) == \"11101010\"\n assert candidate(16, 2) == \"10000\"\n assert candidate(8, 2) == \"1000\"\n assert candidate(7, 2) == \"111\"\n for x in range(2, 8):\n assert candidate(x, x + 1) == str(x)\n\n","code":"def change_base(x: int, base: int):"} -{"task_id":"JHumanEval\/45","prompt_en":"def triangle_area(a, h):\n \"\"\"Given length of a side and high return area for a triangle.\n >>> triangle_area(5, 3)\n 7.5\n \"\"\"\n","prompt":"三角形の一辺の長さと高さが与えられたとき、面積を返す。\n >>> triangle_area(5, 3)\n 7.5","entry_point":"triangle_area","canonical_solution":" return a * h \/ 2.0\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(5, 3) == 7.5\n assert candidate(2, 2) == 2.0\n assert candidate(10, 8) == 40.0\n\n","code":"def triangle_area(a, h):"} -{"task_id":"JHumanEval\/46","prompt_en":"def fib4(n: int):\n \"\"\"The Fib4 number sequence is a sequence similar to the Fibbonacci sequnece that's defined as follows:\n fib4(0) -> 0\n fib4(1) -> 0\n fib4(2) -> 2\n fib4(3) -> 0\n fib4(n) -> fib4(n-1) + fib4(n-2) + fib4(n-3) + fib4(n-4).\n Please write a function to efficiently compute the n-th element of the fib4 number sequence. Do not use recursion.\n >>> fib4(5)\n 4\n >>> fib4(6)\n 8\n >>> fib4(7)\n 14\n \"\"\"\n","prompt":"fib4数列はフィボナッチ数列に似た数列で、次のように定義される：\n fib4(0) -> 0\n fib4(1) -> 0\n fib4(2) -> 2\n fib4(3) -> 0\n fib4(n) -> fib4(n-1) + fib4(n-2) + fib4(n-3) + fib4(n-4).\n fib4数列のn番目の要素を効率的に計算する関数を書け。再帰は使わないこと。\n >>> fib4(5)\n 4\n >>> fib4(6)\n 8\n >>> fib4(7)\n 14","entry_point":"fib4","canonical_solution":" results = [0, 0, 2, 0]\n if n < 4:\n return results[n]\n\n for _ in range(4, n + 1):\n results.append(results[-1] + results[-2] + results[-3] + results[-4])\n results.pop(0)\n\n return results[-1]\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(5) == 4\n assert candidate(8) == 28\n assert candidate(10) == 104\n assert candidate(12) == 386\n\n","code":"def fib4(n: int):"} -{"task_id":"JHumanEval\/47","prompt_en":"def median(l: list):\n \"\"\"Return median of elements in the list l.\n >>> median([3, 1, 2, 4, 5])\n 3\n >>> median([-10, 4, 6, 1000, 10, 20])\n 15.0\n \"\"\"\n","prompt":"リスト l の要素の中央値を返す。\n >>> median([3, 1, 2, 4, 5])\n 3\n >>> median([-10, 4, 6, 1000, 10, 20])\n 15.0","entry_point":"median","canonical_solution":" l = sorted(l)\n if len(l) % 2 == 1:\n return l[len(l) \/\/ 2]\n else:\n return (l[len(l) \/\/ 2 - 1] + l[len(l) \/\/ 2]) \/ 2.0\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([3, 1, 2, 4, 5]) == 3\n assert candidate([-10, 4, 6, 1000, 10, 20]) == 8.0\n assert candidate([5]) == 5\n assert candidate([6, 5]) == 5.5\n assert candidate([8, 1, 3, 9, 9, 2, 7]) == 7 \n\n","code":"def median(l: list):"} -{"task_id":"JHumanEval\/48","prompt_en":"def is_palindrome(text: str):\n \"\"\"\n Checks if given string is a palindrome\n >>> is_palindrome('')\n True\n >>> is_palindrome('aba')\n True\n >>> is_palindrome('aaaaa')\n True\n >>> is_palindrome('zbcd')\n False\n \"\"\"\n","prompt":"与えられた文字列が回文かどうかを判定する\n >>> is_palindrome('')\n True\n >>> is_palindrome('aba')\n True\n >>> is_palindrome('aaaaa')\n True\n >>> is_palindrome('zbcd')\n False","entry_point":"is_palindrome","canonical_solution":" for i in range(len(text)):\n if text[i] != text[len(text) - 1 - i]:\n return False\n return True\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate('') == True\n assert candidate('aba') == True\n assert candidate('aaaaa') == True\n assert candidate('zbcd') == False\n assert candidate('xywyx') == True\n assert candidate('xywyz') == False\n assert candidate('xywzx') == False\n\n","code":"def is_palindrome(text: str):"} -{"task_id":"JHumanEval\/49","prompt_en":"def modp(n: int, p: int):\n \"\"\"Return 2^n modulo p (be aware of numerics).\n >>> modp(3, 5)\n 3\n >>> modp(1101, 101)\n 2\n >>> modp(0, 101)\n 1\n >>> modp(3, 11)\n 8\n >>> modp(100, 101)\n 1\n \"\"\"\n","prompt":"2^n を p で割ったモジュロを返す。計算精度に注意。\n >>> modp(3, 5)\n 3\n >>> modp(1101, 101)\n 2\n >>> modp(0, 101)\n 1\n >>> modp(3, 11)\n 8\n >>> modp(100, 101)\n 1","entry_point":"modp","canonical_solution":" ret = 1\n for i in range(n):\n ret = (2 * ret) % p\n return ret\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(3, 5) == 3\n assert candidate(1101, 101) == 2\n assert candidate(0, 101) == 1\n assert candidate(3, 11) == 8\n assert candidate(100, 101) == 1\n assert candidate(30, 5) == 4\n assert candidate(31, 5) == 3\n\n","code":"def modp(n: int, p: int):"} -{"task_id":"JHumanEval\/50","prompt_en":"def encode_shift(s: str):\n \"\"\"\n returns encoded string by shifting every character by 5 in the alphabet.\n \"\"\"\n return \"\".join([chr(((ord(ch) + 5 - ord(\"a\")) % 26) + ord(\"a\")) for ch in s])\n\n\ndef decode_shift(s: str):\n \"\"\"\n takes as input string encoded with encode_shift function. Returns decoded string.\n \"\"\"\n","prompt":"encode_shift関数でエンコードされた文字列を引数に取り、デコードされた文字列を返す。","entry_point":"decode_shift","canonical_solution":" return \"\".join([chr(((ord(ch) - 5 - ord(\"a\")) % 26) + ord(\"a\")) for ch in s])\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n from random import randint, choice\n import copy\n import string\n\n letters = string.ascii_lowercase\n for _ in range(100):\n str = ''.join(choice(letters) for i in range(randint(10, 20)))\n encoded_str = encode_shift(str)\n assert candidate(copy.deepcopy(encoded_str)) == str\n\n","code":"def encode_shift(s: str):\n \n return \"\".join([chr(((ord(ch) + 5 - ord(\"a\")) % 26) + ord(\"a\")) for ch in s])\n\n\ndef decode_shift(s: str):"} -{"task_id":"JHumanEval\/51","prompt_en":"def remove_vowels(text):\n \"\"\"\n remove_vowels is a function that takes string and returns string without vowels.\n >>> remove_vowels('')\n ''\n >>> remove_vowels(\"abcdef\\nghijklm\")\n 'bcdf\\nghjklm'\n >>> remove_vowels('abcdef')\n 'bcdf'\n >>> remove_vowels('aaaaa')\n ''\n >>> remove_vowels('aaBAA')\n 'B'\n >>> remove_vowels('zbcd')\n 'zbcd'\n \"\"\"\n","prompt":"remove_vowelsは文字列を引数に取り、母音を除いた文字列を返す関数である。\n >>> remove_vowels('')\n ''\n >>> remove_vowels(\"abcdef\\nghijklm\")\n 'bcdf\\nghjklm'\n >>> remove_vowels('abcdef')\n 'bcdf'\n >>> remove_vowels('aaaaa')\n ''\n >>> remove_vowels('aaBAA')\n 'B'\n >>> remove_vowels('zbcd')\n 'zbcd'","entry_point":"remove_vowels","canonical_solution":" return \"\".join([s for s in text if s.lower() not in [\"a\", \"e\", \"i\", \"o\", \"u\"]])\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate('') == ''\n assert candidate(\"abcdef\\nghijklm\") == 'bcdf\\nghjklm'\n assert candidate('fedcba') == 'fdcb'\n assert candidate('eeeee') == ''\n assert candidate('acBAA') == 'cB'\n assert candidate('EcBOO') == 'cB'\n assert candidate('ybcd') == 'ybcd'\n\n","code":"def remove_vowels(text):"} -{"task_id":"JHumanEval\/52","prompt_en":"def below_threshold(l: list, t: int):\n \"\"\"Return True if all numbers in the list l are below threshold t.\n >>> below_threshold([1, 2, 4, 10], 100)\n True\n >>> below_threshold([1, 20, 4, 10], 5)\n False\n \"\"\"\n","prompt":"リスト l 内の全ての数値が閾値 t 以下の場合、Trueを返す。\n >>> below_threshold([1, 2, 4, 10], 100)\n True\n >>> below_threshold([1, 20, 4, 10], 5)\n False","entry_point":"below_threshold","canonical_solution":" for e in l:\n if e >= t:\n return False\n return True\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 2, 4, 10], 100)\n assert not candidate([1, 20, 4, 10], 5)\n assert candidate([1, 20, 4, 10], 21)\n assert candidate([1, 20, 4, 10], 22)\n assert candidate([1, 8, 4, 10], 11)\n assert not candidate([1, 8, 4, 10], 10)\n\n","code":"def below_threshold(l: list, t: int):"} -{"task_id":"JHumanEval\/53","prompt_en":"def add(x: int, y: int):\n \"\"\"Add two numbers x and y\n >>> add(2, 3)\n 5\n >>> add(5, 7)\n 12\n \"\"\"\n","prompt":"2つの数xとyを足す\n >>> add(2, 3)\n 5\n >>> add(5, 7)\n 12","entry_point":"add","canonical_solution":" return x + y\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n import random\n\n assert candidate(0, 1) == 1\n assert candidate(1, 0) == 1\n assert candidate(2, 3) == 5\n assert candidate(5, 7) == 12\n assert candidate(7, 5) == 12\n\n for i in range(100):\n x, y = random.randint(0, 1000), random.randint(0, 1000)\n assert candidate(x, y) == x + y\n\n","code":"def add(x: int, y: int):"} -{"task_id":"JHumanEval\/54","prompt_en":"def same_chars(s0: str, s1: str):\n \"\"\"\n Check if two words have the same characters.\n >>> same_chars('eabcdzzzz', 'dddzzzzzzzddeddabc')\n True\n >>> same_chars('abcd', 'dddddddabc')\n True\n >>> same_chars('dddddddabc', 'abcd')\n True\n >>> same_chars('eabcd', 'dddddddabc')\n False\n >>> same_chars('abcd', 'dddddddabce')\n False\n >>> same_chars('eabcdzzzz', 'dddzzzzzzzddddabc')\n False\n \"\"\"\n","prompt":"2つの単語が同じ文字セットから構成されるかどうか判定する。\n >>> same_chars('eabcdzzzz', 'dddzzzzzzzddeddabc')\n True\n >>> same_chars('abcd', 'dddddddabc')\n True\n >>> same_chars('dddddddabc', 'abcd')\n True\n >>> same_chars('eabcd', 'dddddddabc')\n False\n >>> same_chars('abcd', 'dddddddabce')\n False\n >>> same_chars('eabcdzzzz', 'dddzzzzzzzddddabc')\n False","entry_point":"same_chars","canonical_solution":" return set(s0) == set(s1)\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate('eabcdzzzz', 'dddzzzzzzzddeddabc') == True\n assert candidate('abcd', 'dddddddabc') == True\n assert candidate('dddddddabc', 'abcd') == True\n assert candidate('eabcd', 'dddddddabc') == False\n assert candidate('abcd', 'dddddddabcf') == False\n assert candidate('eabcdzzzz', 'dddzzzzzzzddddabc') == False\n assert candidate('aabb', 'aaccc') == False\n\n","code":"def same_chars(s0: str, s1: str):"} -{"task_id":"JHumanEval\/55","prompt_en":"def fib(n: int):\n \"\"\"Return n-th Fibonacci number.\n >>> fib(10)\n 55\n >>> fib(1)\n 1\n >>> fib(8)\n 21\n \"\"\"\n","prompt":"n番目のフィボナッチ数を返す。\n >>> fib(10)\n 55\n >>> fib(1)\n 1\n >>> fib(8)\n 21","entry_point":"fib","canonical_solution":" if n == 0:\n return 0\n if n == 1:\n return 1\n return fib(n - 1) + fib(n - 2)\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(10) == 55\n assert candidate(1) == 1\n assert candidate(8) == 21\n assert candidate(11) == 89\n assert candidate(12) == 144\n\n","code":"def fib(n: int):"} -{"task_id":"JHumanEval\/56","prompt_en":"def correct_bracketing(brackets: str):\n \"\"\" brackets is a string of \"<\" and \">\".\n return True if every opening bracket has a corresponding closing bracket.\n\n >>> correct_bracketing(\"<\")\n False\n >>> correct_bracketing(\"<>\")\n True\n >>> correct_bracketing(\"<<><>>\")\n True\n >>> correct_bracketing(\"><<>\")\n False\n \"\"\"\n","prompt":"引数bracketsは\"<\"と\">\"の文字列である。\n すべての開き括弧が対応する閉じ括弧を持つ場合、Trueを返す。\n \n >>> correct_bracketing(\"<\")\n False\n >>> correct_bracketing(\"<>\")\n True\n >>> correct_bracketing(\"<<><>>\")\n True\n >>> correct_bracketing(\"><<>\")\n False","entry_point":"correct_bracketing","canonical_solution":" depth = 0\n for b in brackets:\n if b == \"<\":\n depth += 1\n else:\n depth -= 1\n if depth < 0:\n return False\n return depth == 0\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(\"<>\")\n assert candidate(\"<<><>>\")\n assert candidate(\"<><><<><>><>\")\n assert candidate(\"<><><<<><><>><>><<><><<>>>\")\n assert not candidate(\"<<<><>>>>\")\n assert not candidate(\"><<>\")\n assert not candidate(\"<\")\n assert not candidate(\"<<<<\")\n assert not candidate(\">\")\n assert not candidate(\"<<>\")\n assert not candidate(\"<><><<><>><>><<>\")\n assert not candidate(\"<><><<><>><>>><>\")\n\n","code":"def correct_bracketing(brackets: str):"} -{"task_id":"JHumanEval\/57","prompt_en":"def monotonic(l: list):\n \"\"\"Return True is list elements are monotonically increasing or decreasing.\n >>> monotonic([1, 2, 4, 20])\n True\n >>> monotonic([1, 20, 4, 10])\n False\n >>> monotonic([4, 1, 0, -10])\n True\n \"\"\"\n","prompt":"リストの要素が単調増加または単調減少する場合にTrueを返す。\n >>> monotonic([1, 2, 4, 20])\n True\n >>> monotonic([1, 20, 4, 10])\n False\n >>> monotonic([4, 1, 0, -10])\n True","entry_point":"monotonic","canonical_solution":" if l == sorted(l) or l == sorted(l, reverse=True):\n return True\n return False\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 2, 4, 10]) == True\n assert candidate([1, 2, 4, 20]) == True\n assert candidate([1, 20, 4, 10]) == False\n assert candidate([4, 1, 0, -10]) == True\n assert candidate([4, 1, 1, 0]) == True\n assert candidate([1, 2, 3, 2, 5, 60]) == False\n assert candidate([1, 2, 3, 4, 5, 60]) == True\n assert candidate([9, 9, 9, 9]) == True\n\n","code":"def monotonic(l: list):"} -{"task_id":"JHumanEval\/58","prompt_en":"def common(l1: list, l2: list):\n \"\"\"Return sorted unique common elements for two lists.\n >>> common([1, 4, 3, 34, 653, 2, 5], [5, 7, 1, 5, 9, 653, 121])\n [1, 5, 653]\n >>> common([5, 3, 2, 8], [3, 2])\n [2, 3]\n\n \"\"\"\n","prompt":"2つのリストについて、ユニークな共通要素をソートして返す。\n >>> common([1, 4, 3, 34, 653, 2, 5], [5, 7, 1, 5, 9, 653, 121])\n [1, 5, 653]\n >>> common([5, 3, 2, 8], [3, 2])\n [2, 3]","entry_point":"common","canonical_solution":" ret = set()\n for e1 in l1:\n for e2 in l2:\n if e1 == e2:\n ret.add(e1)\n return sorted(list(ret))\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([1, 4, 3, 34, 653, 2, 5], [5, 7, 1, 5, 9, 653, 121]) == [1, 5, 653]\n assert candidate([5, 3, 2, 8], [3, 2]) == [2, 3]\n assert candidate([4, 3, 2, 8], [3, 2, 4]) == [2, 3, 4]\n assert candidate([4, 3, 2, 8], []) == []\n\n","code":"def common(l1: list, l2: list):"} -{"task_id":"JHumanEval\/59","prompt_en":"def largest_prime_factor(n: int):\n \"\"\"Return the largest prime factor of n. Assume n > 1 and is not a prime.\n >>> largest_prime_factor(13195)\n 29\n >>> largest_prime_factor(2048)\n 2\n \"\"\"\n","prompt":"nの最大となる素因数を返す。ただし、 n > 1 を前提とし、素数ではないものとする。\n >>> largest_prime_factor(13195)\n 29\n >>> largest_prime_factor(2048)\n 2","entry_point":"largest_prime_factor","canonical_solution":" def is_prime(k):\n if k < 2:\n return False\n for i in range(2, k - 1):\n if k % i == 0:\n return False\n return True\n largest = 1\n for j in range(2, n + 1):\n if n % j == 0 and is_prime(j):\n largest = max(largest, j)\n return largest\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(15) == 5\n assert candidate(27) == 3\n assert candidate(63) == 7\n assert candidate(330) == 11\n assert candidate(13195) == 29\n\n","code":"def largest_prime_factor(n: int):"} -{"task_id":"JHumanEval\/60","prompt_en":"def sum_to_n(n: int):\n \"\"\"sum_to_n is a function that sums numbers from 1 to n.\n >>> sum_to_n(30)\n 465\n >>> sum_to_n(100)\n 5050\n >>> sum_to_n(5)\n 15\n >>> sum_to_n(10)\n 55\n >>> sum_to_n(1)\n 1\n \"\"\"\n","prompt":"sum_to_nは1からnまでの総和を求める関数である。\n >>> sum_to_n(30)\n 465\n >>> sum_to_n(100)\n 5050\n >>> sum_to_n(5)\n 15\n >>> sum_to_n(10)\n 55\n >>> sum_to_n(1)\n 1","entry_point":"sum_to_n","canonical_solution":" return sum(range(n + 1))\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(1) == 1\n assert candidate(6) == 21\n assert candidate(11) == 66\n assert candidate(30) == 465\n assert candidate(100) == 5050\n\n","code":"def sum_to_n(n: int):"} -{"task_id":"JHumanEval\/61","prompt_en":"def correct_bracketing(brackets: str):\n \"\"\" brackets is a string of \"(\" and \")\".\n return True if every opening bracket has a corresponding closing bracket.\n\n >>> correct_bracketing(\"(\")\n False\n >>> correct_bracketing(\"()\")\n True\n >>> correct_bracketing(\"(()())\")\n True\n >>> correct_bracketing(\")(()\")\n False\n \"\"\"\n","prompt":"引数bracketsは\"(\"と\") \"からなる文字列である。\n すべての開き括弧が対応する閉じ括弧を持つ場合、Trueを返す。\n \n >>> correct_bracketing(\"(\")\n False\n >>> correct_bracketing(\"()\")\n True\n >>> correct_bracketing(\"(()())\")\n True\n >>> correct_bracketing(\")(()\")\n False","entry_point":"correct_bracketing","canonical_solution":" depth = 0\n for b in brackets:\n if b == \"(\":\n depth += 1\n else:\n depth -= 1\n if depth < 0:\n return False\n return depth == 0\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(\"()\")\n assert candidate(\"(()())\")\n assert candidate(\"()()(()())()\")\n assert candidate(\"()()((()()())())(()()(()))\")\n assert not candidate(\"((()())))\")\n assert not candidate(\")(()\")\n assert not candidate(\"(\")\n assert not candidate(\"((((\")\n assert not candidate(\")\")\n assert not candidate(\"(()\")\n assert not candidate(\"()()(()())())(()\")\n assert not candidate(\"()()(()())()))()\")\n\n","code":"def correct_bracketing(brackets: str):"} -{"task_id":"JHumanEval\/62","prompt_en":"def derivative(xs: list):\n \"\"\" xs represent coefficients of a polynomial.\n xs[0] + xs[1] * x + xs[2] * x^2 + ....\n Return derivative of this polynomial in the same form.\n >>> derivative([3, 1, 2, 4, 5])\n [1, 4, 12, 20]\n >>> derivative([1, 2, 3])\n [2, 6]\n \"\"\"\n","prompt":"xsは多項式の係数列を表す。\n xs[0] + xs[1] * x + xs[2] * x^2 + ....\n 関数は、この多項式の導関数を同じ形式で返す。\n >>> derivative([3, 1, 2, 4, 5])\n [1, 4, 12, 20]\n >>> derivative([1, 2, 3])\n [2, 6]","entry_point":"derivative","canonical_solution":" return [(i * x) for i, x in enumerate(xs)][1:]\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate([3, 1, 2, 4, 5]) == [1, 4, 12, 20]\n assert candidate([1, 2, 3]) == [2, 6]\n assert candidate([3, 2, 1]) == [2, 2]\n assert candidate([3, 2, 1, 0, 4]) == [2, 2, 0, 16]\n assert candidate([1]) == []\n\n","code":"def derivative(xs: list):"} -{"task_id":"JHumanEval\/63","prompt_en":"def fibfib(n: int):\n \"\"\"The FibFib number sequence is a sequence similar to the Fibbonacci sequnece that's defined as follows:\n fibfib(0) == 0\n fibfib(1) == 0\n fibfib(2) == 1\n fibfib(n) == fibfib(n-1) + fibfib(n-2) + fibfib(n-3).\n Please write a function to efficiently compute the n-th element of the fibfib number sequence.\n >>> fibfib(1)\n 0\n >>> fibfib(5)\n 4\n >>> fibfib(8)\n 24\n \"\"\"\n","prompt":"FibFib数列はフィボナッチ数列に似た数列で、以下のように定義される：\n fibfib(0) == 0\n fibfib(1) == 0\n fibfib(2) == 1\n fibfib(n) == fibfib(n-1) + fibfib(n-2) + fibfib(n-3).\n fibfib数列のn番目の要素を効率よく計算する関数を書いてください。\n >>> fibfib(1)\n 0\n >>> fibfib(5)\n 4\n >>> fibfib(8)\n 24","entry_point":"fibfib","canonical_solution":" if n == 0:\n return 0\n if n == 1:\n return 0\n if n == 2:\n return 1\n return fibfib(n - 1) + fibfib(n - 2) + fibfib(n - 3)\n","test":"\n\nMETADATA = {}\n\n\ndef check(candidate):\n assert candidate(2) == 1\n assert candidate(1) == 0\n assert candidate(5) == 4\n assert candidate(8) == 24\n assert candidate(10) == 81\n assert candidate(12) == 274\n assert candidate(14) == 927\n\n","code":"def fibfib(n: int):"} -{"task_id":"JHumanEval\/64","prompt_en":"\nFIX = \"\"\"\nAdd more test cases.\n\"\"\"def vowels_count(s):\n \"\"\"Write a function vowels_count which takes a string representing\n a word as input and returns the number of vowels in the string.\n Vowels in this case are 'a', 'e', 'i', 'o', 'u'. Here, 'y' is also a\n vowel, but only when it is at the end of the given word.\n\n Example:\n >>> vowels_count(\"abcde\")\n 2\n >>> vowels_count(\"ACEDY\")\n 3\n \"\"\"\n","prompt":"単語を表す文字列を引数とし、その文字列に含まれる母音の数を返す\n 関数 vowels_count を書きなさい。この場合の母音は'a', 'e', 'i', 'o', 'u'である。\n ここで、与えられた単語の末尾にある場合のみ、'y'も母音とする。\n \n 例:\n >>> vowels_count(\"abcde\")\n 2\n >>> vowels_count(\"ACEDY\")\n 3","entry_point":"vowels_count","canonical_solution":" vowels = \"aeiouAEIOU\"\n n_vowels = sum(c in vowels for c in s)\n if s[-1] == 'y' or s[-1] == 'Y':\n n_vowels += 1\n return n_vowels\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"abcde\") == 2, \"Test 1\"\n assert candidate(\"Alone\") == 3, \"Test 2\"\n assert candidate(\"key\") == 2, \"Test 3\"\n assert candidate(\"bye\") == 1, \"Test 4\"\n assert candidate(\"keY\") == 2, \"Test 5\"\n assert candidate(\"bYe\") == 1, \"Test 6\"\n assert candidate(\"ACEDY\") == 3, \"Test 7\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"FIX = def vowels_count(s):"} -{"task_id":"JHumanEval\/65","prompt_en":"def circular_shift(x, shift):\n \"\"\"Circular shift the digits of the integer x, shift the digits right by shift\n and return the result as a string.\n If shift > number of digits, return digits reversed.\n >>> circular_shift(12, 1)\n \"21\"\n >>> circular_shift(12, 2)\n \"12\"\n \"\"\"\n","prompt":"整数 x の桁を循環シフトする。shift 分だけ桁を右にシフトし、結果を文字列として返す。\n もし、shift > 桁数なら、桁を反転して返す。\n >>> circular_shift(12, 1)\n \"21\"\n >>> circular_shift(12, 2)\n \"12\"","entry_point":"circular_shift","canonical_solution":" s = str(x)\n if shift > len(s):\n return s[::-1]\n else:\n return s[len(s) - shift:] + s[:len(s) - shift]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(100, 2) == \"001\"\n assert candidate(12, 2) == \"12\"\n assert candidate(97, 8) == \"79\"\n assert candidate(12, 1) == \"21\", \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(11, 101) == \"11\", \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def circular_shift(x, shift):"} -{"task_id":"JHumanEval\/66","prompt_en":"def digitSum(s):\n \"\"\"Task\n Write a function that takes a string as input and returns the sum of the upper characters only'\n ASCII codes.\n\n Examples:\n digitSum(\"\") => 0\n digitSum(\"abAB\") => 131\n digitSum(\"abcCd\") => 67\n digitSum(\"helloE\") => 69\n digitSum(\"woArBld\") => 131\n digitSum(\"aAaaaXa\") => 153\n \"\"\"\n","prompt":"タスク\n 文字列を引数にとり、英大文字のみのASCIIコードの和を返す関数を書く。\n Examples:\n digitSum(\"\") => 0\n digitSum(\"abAB\") => 131\n digitSum(\"abcCd\") => 67\n digitSum(\"helloE\") => 69\n digitSum(\"woArBld\") => 131\n digitSum(\"aAaaaXa\") => 153","entry_point":"digitSum","canonical_solution":" if s == \"\": return 0\n return sum(ord(char) if char.isupper() else 0 for char in s)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(\"\") == 0, \"Error\"\n assert candidate(\"abAB\") == 131, \"Error\"\n assert candidate(\"abcCd\") == 67, \"Error\"\n assert candidate(\"helloE\") == 69, \"Error\"\n assert candidate(\"woArBld\") == 131, \"Error\"\n assert candidate(\"aAaaaXa\") == 153, \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(\" How are yOu?\") == 151, \"Error\"\n assert candidate(\"You arE Very Smart\") == 327, \"Error\"\n\n","code":"def digitSum(s):"} -{"task_id":"JHumanEval\/67","prompt_en":"def fruit_distribution(s,n):\n \"\"\"\n In this task, you will be given a string that represents a number of apples and oranges \n that are distributed in a basket of fruit this basket contains \n apples, oranges, and mango fruits. Given the string that represents the total number of \n the oranges and apples and an integer that represent the total number of the fruits \n in the basket return the number of the mango fruits in the basket.\n for examble:\n fruit_distribution(\"5 apples and 6 oranges\", 19) ->19 - 5 - 6 = 8\n fruit_distribution(\"0 apples and 1 oranges\",3) -> 3 - 0 - 1 = 2\n fruit_distribution(\"2 apples and 3 oranges\", 100) -> 100 - 2 - 3 = 95\n fruit_distribution(\"100 apples and 1 oranges\",120) -> 120 - 100 - 1 = 19\n \"\"\"\n","prompt":"この課題では、果物の入ったカゴに配られたリンゴとオレンジの数を表す文字列が\n 与えられ、このカゴにはリンゴ、オレンジ、マンゴーの果実が入っている。オレンジ\n とリンゴの総数を表す文字列と、かごの中の果物の総数を表す整数が与えられたら、\n かごの中のマンゴーの果物の数を返しなさい。\n たとえば:\n fruit_distribution(\"5 apples and 6 oranges\", 19) ->19 - 5 - 6 = 8\n fruit_distribution(\"0 apples and 1 oranges\",3) -> 3 - 0 - 1 = 2\n fruit_distribution(\"2 apples and 3 oranges\", 100) -> 100 - 2 - 3 = 95\n fruit_distribution(\"100 apples and 1 oranges\",120) -> 120 - 100 - 1 = 19","entry_point":"fruit_distribution","canonical_solution":" lis = list()\n for i in s.split(' '):\n if i.isdigit():\n lis.append(int(i))\n return n - sum(lis)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"5 apples and 6 oranges\",19) == 8\n assert candidate(\"5 apples and 6 oranges\",21) == 10\n assert candidate(\"0 apples and 1 oranges\",3) == 2\n assert candidate(\"1 apples and 0 oranges\",3) == 2\n assert candidate(\"2 apples and 3 oranges\",100) == 95\n assert candidate(\"2 apples and 3 oranges\",5) == 0\n assert candidate(\"1 apples and 100 oranges\",120) == 19\n","code":"def fruit_distribution(s,n):"} -{"task_id":"JHumanEval\/68","prompt_en":"def pluck(arr):\n \"\"\"\n \"Given an array representing a branch of a tree that has non-negative integer nodes\n your task is to pluck one of the nodes and return it.\n The plucked node should be the node with the smallest even value.\n If multiple nodes with the same smallest even value are found return the node that has smallest index.\n\n The plucked node should be returned in a list, [ smalest_value, its index ],\n If there are no even values or the given array is empty, return [].\n\n Example 1:\n Input: [4,2,3]\n Output: [2, 1]\n Explanation: 2 has the smallest even value, and 2 has the smallest index.\n\n Example 2:\n Input: [1,2,3]\n Output: [2, 1]\n Explanation: 2 has the smallest even value, and 2 has the smallest index. \n\n Example 3:\n Input: []\n Output: []\n \n Example 4:\n Input: [5, 0, 3, 0, 4, 2]\n Output: [0, 1]\n Explanation: 0 is the smallest value, but there are two zeros,\n so we will choose the first zero, which has the smallest index.\n\n Constraints:\n * 1 <= nodes.length <= 10000\n * 0 <= node.value\n \"\"\"\n","prompt":"非負整数のノードを持つ木の枝を表す配列が与えられたとする。あなたの仕事は、\n ノードの1つを抜き取り、それを返すことである。\n 摘出されるノードは、最小偶数値を持つノードでなければならない。\n 同じ最小偶数値を持つノードが複数見つかった場合は、最小のインデックスを持つ\n ノードを返す。\n\n 摘出されたノードは [ smalest_value, its index ] というリストで返されなければならない。\n 偶数値がない場合や与えられた配列が空の場合は [] を返します。\n 例 1:\n 入力: [4,2,3]\n 出力: [2, 1]\n 解説: 2は最小偶数値を持ち、最小インデックスを持つ。\n\n 例 2:\n 入力: [1,2,3]\n 出力: [2, 1]\n 解説: 2が最小偶数値で、2が最小インデックスを持つ。\n\n 例 3:\n 入力: []\n 出力: []\n \n 例 4:\n 入力: [5, 0, 3, 0, 4, 2]\n 出力: [0, 1]\n 解説: 0は最小値だが、0は2つあるので、最小インデックスを持つ最初の0を選ぶ。\n\n 制約:\n * 1 <= ノードの長さ <= 10000\n * 0 <= ノードの値","entry_point":"pluck","canonical_solution":" if(len(arr) == 0): return []\n evens = list(filter(lambda x: x%2 == 0, arr))\n if(evens == []): return []\n return [min(evens), arr.index(min(evens))]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([4,2,3]) == [2, 1], \"Error\"\n assert candidate([1,2,3]) == [2, 1], \"Error\"\n assert candidate([]) == [], \"Error\"\n assert candidate([5, 0, 3, 0, 4, 2]) == [0, 1], \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([1, 2, 3, 0, 5, 3]) == [0, 3], \"Error\"\n assert candidate([5, 4, 8, 4 ,8]) == [4, 1], \"Error\"\n assert candidate([7, 6, 7, 1]) == [6, 1], \"Error\"\n assert candidate([7, 9, 7, 1]) == [], \"Error\"\n\n","code":"def pluck(arr):"} -{"task_id":"JHumanEval\/69","prompt_en":"def search(lst):\n '''\n You are given a non-empty list of positive integers. Return the greatest integer that is greater than \n zero, and has a frequency greater than or equal to the value of the integer itself. \n The frequency of an integer is the number of times it appears in the list.\n If no such a value exist, return -1.\n Examples:\n search([4, 1, 2, 2, 3, 1]) == 2\n search([1, 2, 2, 3, 3, 3, 4, 4, 4]) == 3\n search([5, 5, 4, 4, 4]) == -1\n '''\n","prompt":"正の整数の空でないリストが与えられる。0より大きく、その整数自身の値以上の頻度を\n 持つ最大の整数を返せ。整数の頻度とは、それがリストに現れる回数である���\n そのような値が存在しない場合は -1 を返す。\n 例:\n search([4, 1, 2, 2, 3, 1]) == 2\n search([1, 2, 2, 3, 3, 3, 4, 4, 4]) == 3\n search([5, 5, 4, 4, 4]) == -1","entry_point":"search","canonical_solution":" frq = [0] * (max(lst) + 1)\n for i in lst:\n frq[i] += 1;\n\n ans = -1\n for i in range(1, len(frq)):\n if frq[i] >= i:\n ans = i\n \n return ans\n","test":"def check(candidate):\n\n # manually generated tests\n assert candidate([5, 5, 5, 5, 1]) == 1\n assert candidate([4, 1, 4, 1, 4, 4]) == 4\n assert candidate([3, 3]) == -1\n assert candidate([8, 8, 8, 8, 8, 8, 8, 8]) == 8\n assert candidate([2, 3, 3, 2, 2]) == 2\n\n # automatically generated tests\n assert candidate([2, 7, 8, 8, 4, 8, 7, 3, 9, 6, 5, 10, 4, 3, 6, 7, 1, 7, 4, 10, 8, 1]) == 1\n assert candidate([3, 2, 8, 2]) == 2\n assert candidate([6, 7, 1, 8, 8, 10, 5, 8, 5, 3, 10]) == 1\n assert candidate([8, 8, 3, 6, 5, 6, 4]) == -1\n assert candidate([6, 9, 6, 7, 1, 4, 7, 1, 8, 8, 9, 8, 10, 10, 8, 4, 10, 4, 10, 1, 2, 9, 5, 7, 9]) == 1\n assert candidate([1, 9, 10, 1, 3]) == 1\n assert candidate([6, 9, 7, 5, 8, 7, 5, 3, 7, 5, 10, 10, 3, 6, 10, 2, 8, 6, 5, 4, 9, 5, 3, 10]) == 5\n assert candidate([1]) == 1\n assert candidate([8, 8, 10, 6, 4, 3, 5, 8, 2, 4, 2, 8, 4, 6, 10, 4, 2, 1, 10, 2, 1, 1, 5]) == 4\n assert candidate([2, 10, 4, 8, 2, 10, 5, 1, 2, 9, 5, 5, 6, 3, 8, 6, 4, 10]) == 2\n assert candidate([1, 6, 10, 1, 6, 9, 10, 8, 6, 8, 7, 3]) == 1\n assert candidate([9, 2, 4, 1, 5, 1, 5, 2, 5, 7, 7, 7, 3, 10, 1, 5, 4, 2, 8, 4, 1, 9, 10, 7, 10, 2, 8, 10, 9, 4]) == 4\n assert candidate([2, 6, 4, 2, 8, 7, 5, 6, 4, 10, 4, 6, 3, 7, 8, 8, 3, 1, 4, 2, 2, 10, 7]) == 4\n assert candidate([9, 8, 6, 10, 2, 6, 10, 2, 7, 8, 10, 3, 8, 2, 6, 2, 3, 1]) == 2\n assert candidate([5, 5, 3, 9, 5, 6, 3, 2, 8, 5, 6, 10, 10, 6, 8, 4, 10, 7, 7, 10, 8]) == -1\n assert candidate([10]) == -1\n assert candidate([9, 7, 7, 2, 4, 7, 2, 10, 9, 7, 5, 7, 2]) == 2\n assert candidate([5, 4, 10, 2, 1, 1, 10, 3, 6, 1, 8]) == 1\n assert candidate([7, 9, 9, 9, 3, 4, 1, 5, 9, 1, 2, 1, 1, 10, 7, 5, 6, 7, 6, 7, 7, 6]) == 1\n assert candidate([3, 10, 10, 9, 2]) == -1\n\n","code":"def search(lst):"} -{"task_id":"JHumanEval\/70","prompt_en":"def strange_sort_list(lst):\n '''\n Given list of integers, return list in strange order.\n Strange sorting, is when you start with the minimum value,\n then maximum of the remaining integers, then minimum and so on.\n\n Examples:\n strange_sort_list([1, 2, 3, 4]) == [1, 4, 2, 3]\n strange_sort_list([5, 5, 5, 5]) == [5, 5, 5, 5]\n strange_sort_list([]) == []\n '''\n","prompt":"整数のリストが与えられたとき、リストを奇妙な順序で返す。\n 奇妙なソートとは、最小値から始まり、残りの整数の最大値、最小値の順で\n ソートすることである。\n \n 例:\n strange_sort_list([1, 2, 3, 4]) == [1, 4, 2, 3]\n strange_sort_list([5, 5, 5, 5]) == [5, 5, 5, 5]\n strange_sort_list([]) == []","entry_point":"strange_sort_list","canonical_solution":" res, switch = [], True\n while lst:\n res.append(min(lst) if switch else max(lst))\n lst.remove(res[-1])\n switch = not switch\n return res","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1, 2, 3, 4]) == [1, 4, 2, 3]\n assert candidate([5, 6, 7, 8, 9]) == [5, 9, 6, 8, 7]\n assert candidate([1, 2, 3, 4, 5]) == [1, 5, 2, 4, 3]\n assert candidate([5, 6, 7, 8, 9, 1]) == [1, 9, 5, 8, 6, 7]\n assert candidate([5, 5, 5, 5]) == [5, 5, 5, 5]\n assert candidate([]) == []\n assert candidate([1,2,3,4,5,6,7,8]) == [1, 8, 2, 7, 3, 6, 4, 5]\n assert candidate([0,2,2,2,5,5,-5,-5]) == [-5, 5, -5, 5, 0, 2, 2, 2]\n assert candidate([111111]) == [111111]\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def strange_sort_list(lst):"} -{"task_id":"JHumanEval\/71","prompt_en":"def triangle_area(a, b, c):\n '''\n Given the lengths of the three sides of a triangle. Return the area of\n the triangle rounded to 2 decimal points if the three sides form a valid triangle. \n Otherwise return -1\n Three sides make a valid triangle when the sum of any two sides is greater \n than the third side.\n Example:\n triangle_area(3, 4, 5) == 6.00\n triangle_area(1, 2, 10) == -1\n '''\n","prompt":"三角形の3辺の長さが与えられた。3辺が有効な三角形を形成していれば、\n 三角形の面積を小数点以下2桁で四捨五入して返す。そうでない場合は-1を\n 返す。\n 任意の2辺の和が3辺より大きいとき、3辺は有効な三角形となる。\n 例:\n triangle_area(3, 4, 5) == 6.00\n triangle_area(1, 2, 10) == -1","entry_point":"triangle_area","canonical_solution":" if a + b <= c or a + c <= b or b + c <= a:\n return -1 \n s = (a + b + c)\/2 \n area = (s * (s - a) * (s - b) * (s - c)) ** 0.5\n area = round(area, 2)\n return area\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(3, 4, 5) == 6.00, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(1, 2, 10) == -1\n assert candidate(4, 8, 5) == 8.18\n assert candidate(2, 2, 2) == 1.73\n assert candidate(1, 2, 3) == -1\n assert candidate(10, 5, 7) == 16.25\n assert candidate(2, 6, 3) == -1\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1, 1, 1) == 0.43, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(2, 2, 10) == -1\n\n","code":"def triangle_area(a, b, c):"} -{"task_id":"JHumanEval\/72","prompt_en":"def will_it_fly(q,w):\n '''\n Write a function that returns True if the object q will fly, and False otherwise.\n The object q will fly if it's balanced (it is a palindromic list) and the sum of its elements is less than or equal the maximum possible weight w.\n\n Example:\n will_it_fly([1, 2], 5) ➞ False \n # 1+2 is less than the maximum possible weight, but it's unbalanced.\n\n will_it_fly([3, 2, 3], 1) ➞ False\n # it's balanced, but 3+2+3 is more than the maximum possible weight.\n\n will_it_fly([3, 2, 3], 9) ➞ True\n # 3+2+3 is less than the maximum possible weight, and it's balanced.\n\n will_it_fly([3], 5) ➞ True\n # 3 is less than the maximum possible weight, and it's balanced.\n '''\n","prompt":"物体qが飛べばTrueを、そうでなければFalseを返す関数を書け。\n 物体qはバランスが取れていて（つまり、リストが回文であって）、その要素の和が\n 最大荷重w以下であれば飛ぶ。\n \n 例:\n will_it_fly([1, 2], 5) ➞ False \n # 1+2 は最大荷重以下であるが、バランスが取れていない\n\n will_it_fly([3, 2, 3], 1) ➞ False\n # バランスが取れているが、3+2+3 は最大荷重を超える\n \n will_it_fly([3, 2, 3], 9) ➞ True\n # 3+2+3 は最大荷重以下であり、バランスも取れている\n\n will_it_fly([3], 5) ➞ True\n # 3 は最大荷重以下であり、バランスも取れている","entry_point":"will_it_fly","canonical_solution":" if sum(q) > w:\n return False\n\n i, j = 0, len(q)-1\n while i true\n is_simple_power(2, 2) => true\n is_simple_power(8, 2) => true\n is_simple_power(3, 2) => false\n is_simple_power(3, 1) => false\n is_simple_power(5, 3) => false\n \"\"\"\n","prompt":"あなたのタスクは、ある数xがnの単純なべき乗である場合にtrueを、\n それ以外の場合にfalseを返す関数を書くことである。\n xは、n**int=xのとき、nの単純なべき乗である。\n 例えば:\n is_simple_power(1, 4) => true\n is_simple_power(2, 2) => true\n is_simple_power(8, 2) => true\n is_simple_power(3, 2) => false\n is_simple_power(3, 1) => false\n is_simple_power(5, 3) => false","entry_point":"is_simple_power","canonical_solution":" if (n == 1): \n return (x == 1) \n power = 1\n while (power < x): \n power = power * n \n return (power == x) \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(16, 2)== True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(143214, 16)== False, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(4, 2)==True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(9, 3)==True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(16, 4)==True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(24, 2)==False, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(128, 4)==False, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(12, 6)==False, \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1, 1)==True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(1, 12)==True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def is_simple_power(x, n):"} -{"task_id":"JHumanEval\/77","prompt_en":"def iscube(a):\n '''\n Write a function that takes an integer a and returns True \n if this ingeger is a cube of some integer number.\n Note: you may assume the input is always valid.\n Examples:\n iscube(1) ==> True\n iscube(2) ==> False\n iscube(-1) ==> True\n iscube(64) ==> True\n iscube(0) ==> True\n iscube(180) ==> False\n '''\n","prompt":"整数aを受け取り、この整数がある整数の3乗である場合にTrue\n を返す関数を書きなさい。\n 注意：入力は常に処理可能であると仮定してよい。\n 例:\n iscube(1) ==> True\n iscube(2) ==> False\n iscube(-1) ==> True\n iscube(64) ==> True\n iscube(0) ==> True\n iscube(180) ==> False","entry_point":"iscube","canonical_solution":" a = abs(a)\n return int(round(a ** (1. \/ 3))) ** 3 == a\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(1) == True, \"First test error: \" + str(candidate(1))\n assert candidate(2) == False, \"Second test error: \" + str(candidate(2))\n assert candidate(-1) == True, \"Third test error: \" + str(candidate(-1))\n assert candidate(64) == True, \"Fourth test error: \" + str(candidate(64))\n assert candidate(180) == False, \"Fifth test error: \" + str(candidate(180))\n assert candidate(1000) == True, \"Sixth test error: \" + str(candidate(1000))\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(0) == True, \"1st edge test error: \" + str(candidate(0))\n assert candidate(1729) == False, \"2nd edge test error: \" + str(candidate(1728))\n\n","code":"def iscube(a):"} -{"task_id":"JHumanEval\/78","prompt_en":"def hex_key(num):\n \"\"\"You have been tasked to write a function that receives \n a hexadecimal number as a string and counts the number of hexadecimal \n digits that are primes (prime number, or a prime, is a natural number \n greater than 1 that is not a product of two smaller natural numbers).\n Hexadecimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.\n Prime numbers are 2, 3, 5, 7, 11, 13, 17,...\n So you have to determine a number of the following digits: 2, 3, 5, 7, \n B (=decimal 11), D (=decimal 13).\n Note: you may assume the input is always correct or empty string, \n and symbols A,B,C,D,E,F are always uppercase.\n Examples:\n For num = \"AB\" the output should be 1.\n For num = \"1077E\" the output should be 2.\n For num = \"ABED1A33\" the output should be 4.\n For num = \"123456789ABCDEF0\" the output should be 6.\n For num = \"2020\" the output should be 2.\n \"\"\"\n","prompt":"16進数の数字を文字列として受け取り、その中に含まれる素数である16進数の桁数を\n カウントする関数を作成するタスクが与えられました。素数とは、1より大きく、\n 2つのより小さい自然数の積でない自然数です。\n 16進数の桁には0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, Fがあります。\n 素数としては2, 3, 5, 7, 11, 13, 17,...があります。\n したがって、次の数字のいずれかがいくつあるかを判定する必要があります：\n 2, 3, 5, 7, B（=10進数で11）, D（=10進数で13）\n 注意：入力は常に正確、または空の文字列であり、記号A, B, C, D, E, Fは常に\n 大文字であると仮定してよいです。\n 例：\n num = \"AB\" の場合、出力は 1 です。\n num = \"1077E\" の場合、出力は 2 です。\n num = \"ABED1A33\" の場合、出力は 4 です。\n num = \"123456789ABCDEF0\" の場合、出力は 6 です。\n num = \"2020\" の場合、出力は 2 です。be 2.","entry_point":"hex_key","canonical_solution":" primes = ('2', '3', '5', '7', 'B', 'D')\n total = 0\n for i in range(0, len(num)):\n if num[i] in primes:\n total += 1\n return total\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"AB\") == 1, \"First test error: \" + str(candidate(\"AB\")) \n assert candidate(\"1077E\") == 2, \"Second test error: \" + str(candidate(\"1077E\")) \n assert candidate(\"ABED1A33\") == 4, \"Third test error: \" + str(candidate(\"ABED1A33\")) \n assert candidate(\"2020\") == 2, \"Fourth test error: \" + str(candidate(\"2020\")) \n assert candidate(\"123456789ABCDEF0\") == 6, \"Fifth test error: \" + str(candidate(\"123456789ABCDEF0\")) \n assert candidate(\"112233445566778899AABBCCDDEEFF00\") == 12, \"Sixth test error: \" + str(candidate(\"112233445566778899AABBCCDDEEFF00\")) \n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([]) == 0\n\n","code":"def hex_key(num):"} -{"task_id":"JHumanEval\/79","prompt_en":"def decimal_to_binary(decimal):\n \"\"\"You will be given a number in decimal form and your task is to convert it to\n binary format. The function should return a string, with each character representing a binary\n number. Each character in the string will be '0' or '1'.\n\n There will be an extra couple of characters 'db' at the beginning and at the end of the string.\n The extra characters are there to help with the format.\n\n Examples:\n decimal_to_binary(15) # returns \"db1111db\"\n decimal_to_binary(32) # returns \"db100000db\"\n \"\"\"\n","prompt":"10進数形式の数値が与えられ、あなたのタスクはそれを2進数形式に変換することである。\n この関数は、文字列を返し、その各文字は2進数を表す。文字列の各文字は'0'か'1'である。\n \n なお、文字列の最初と最後には'db'という余分な文字をつける。\n この文字は書式を助けるためにある。\n \n 例:\n decimal_to_binary(15) # \"db1111db\"を返す\n decimal_to_binary(32) # \"db100000db\"を返す","entry_point":"decimal_to_binary","canonical_solution":" return \"db\" + bin(decimal)[2:] + \"db\"\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(0) == \"db0db\"\n assert candidate(32) == \"db100000db\"\n assert candidate(103) == \"db1100111db\"\n assert candidate(15) == \"db1111db\", \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def decimal_to_binary(decimal):"} -{"task_id":"JHumanEval\/80","prompt_en":"def is_happy(s):\n \"\"\"You are given a string s.\n Your task is to check if the string is happy or not.\n A string is happy if its length is at least 3 and every 3 consecutive letters are distinct\n For example:\n is_happy(a) => False\n is_happy(aa) => False\n is_happy(abcd) => True\n is_happy(aabb) => False\n is_happy(adb) => True\n is_happy(xyy) => False\n \"\"\"\n","prompt":"あなたは文字列sが与えられる。\n あなたのタスクは、その文字列が幸せかどうかをチェックすることである。\n 文字列は幸せとは、文字列の長さが少なくとも3以上で、連続する3文字がすべて異なる場合である。\n 例えば:\n is_happy(a) => False\n is_happy(aa) => False\n is_happy(abcd) => True\n is_happy(aabb) => False\n is_happy(adb) => True\n is_happy(xyy) => False","entry_point":"is_happy","canonical_solution":" if len(s) < 3:\n return False\n\n for i in range(len(s) - 2):\n \n if s[i] == s[i+1] or s[i+1] == s[i+2] or s[i] == s[i+2]:\n return False\n return True\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"a\") == False , \"a\"\n assert candidate(\"aa\") == False , \"aa\"\n assert candidate(\"abcd\") == True , \"abcd\"\n assert candidate(\"aabb\") == False , \"aabb\"\n assert candidate(\"adb\") == True , \"adb\"\n assert candidate(\"xyy\") == False , \"xyy\"\n assert candidate(\"iopaxpoi\") == True , \"iopaxpoi\"\n assert candidate(\"iopaxioi\") == False , \"iopaxioi\"\n","code":"def is_happy(s):"} -{"task_id":"JHumanEval\/81","prompt_en":"def numerical_letter_grade(grades):\n \"\"\"It is the last week of the semester and the teacher has to give the grades\n to students. The teacher has been making her own algorithm for grading.\n The only problem is, she has lost the code she used for grading.\n She has given you a list of GPAs for some students and you have to write \n a function that can output a list of letter grades using the following table:\n GPA | Letter grade\n 4.0 A+\n > 3.7 A \n > 3.3 A- \n > 3.0 B+\n > 2.7 B \n > 2.3 B-\n > 2.0 C+\n > 1.7 C\n > 1.3 C-\n > 1.0 D+ \n > 0.7 D \n > 0.0 D-\n 0.0 E\n \n\n Example:\n grade_equation([4.0, 3, 1.7, 2, 3.5]) ==> ['A+', 'B', 'C-', 'C', 'A-']\n \"\"\"\n","prompt":"学期最終週、教師は生徒に成績をつけなければならない。教師は独自のアルゴリズムで採点している。\n 問題は���彼女が成績評価に使ったコードを紛失してしまったことです。\n 彼女は何人かの生徒のGPAのリストをあなたに渡したので、あなたは次の表を使って評点のリストを\n 出力できる関数を書くことになりました。\n\n GPA | 評点\n 4.0 A+\n > 3.7 A \n > 3.3 A- \n > 3.0 B+\n > 2.7 B \n > 2.3 B-\n > 2.0 C+\n > 1.7 C\n > 1.3 C-\n > 1.0 D+ \n > 0.7 D \n > 0.0 D-\n 0.0 E\n \n\n 例:\n grade_equation([4.0, 3, 1.7, 2, 3.5]) ==> ['A+', 'B', 'C-', 'C', 'A-']","entry_point":"numerical_letter_grade","canonical_solution":"\n \n letter_grade = []\n for gpa in grades:\n if gpa == 4.0:\n letter_grade.append(\"A+\")\n elif gpa > 3.7:\n letter_grade.append(\"A\")\n elif gpa > 3.3:\n letter_grade.append(\"A-\")\n elif gpa > 3.0:\n letter_grade.append(\"B+\")\n elif gpa > 2.7:\n letter_grade.append(\"B\")\n elif gpa > 2.3:\n letter_grade.append(\"B-\")\n elif gpa > 2.0:\n letter_grade.append(\"C+\")\n elif gpa > 1.7:\n letter_grade.append(\"C\")\n elif gpa > 1.3:\n letter_grade.append(\"C-\")\n elif gpa > 1.0:\n letter_grade.append(\"D+\")\n elif gpa > 0.7:\n letter_grade.append(\"D\")\n elif gpa > 0.0:\n letter_grade.append(\"D-\")\n else:\n letter_grade.append(\"E\")\n return letter_grade\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([4.0, 3, 1.7, 2, 3.5]) == ['A+', 'B', 'C-', 'C', 'A-']\n assert candidate([1.2]) == ['D+']\n assert candidate([0.5]) == ['D-']\n assert candidate([0.0]) == ['E']\n assert candidate([1, 0.3, 1.5, 2.8, 3.3]) == ['D', 'D-', 'C-', 'B', 'B+']\n assert candidate([0, 0.7]) == ['E', 'D-']\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def numerical_letter_grade(grades):"} -{"task_id":"JHumanEval\/82","prompt_en":"def prime_length(string):\n \"\"\"Write a function that takes a string and returns True if the string\n length is a prime number or False otherwise\n Examples\n prime_length('Hello') == True\n prime_length('abcdcba') == True\n prime_length('kittens') == True\n prime_length('orange') == False\n \"\"\"\n","prompt":"文字列を受け取り、文字列の長さが素数であればTrueを、そうでなければFalseを返す関数を書く。\n 例\n prime_length('Hello') == True\n prime_length('abcdcba') == True\n prime_length('kittens') == True\n prime_length('orange') == False","entry_point":"prime_length","canonical_solution":" l = len(string)\n if l == 0 or l == 1:\n return False\n for i in range(2, l):\n if l % i == 0:\n return False\n return True\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('Hello') == True\n assert candidate('abcdcba') == True\n assert candidate('kittens') == True\n assert candidate('orange') == False\n assert candidate('wow') == True\n assert candidate('world') == True\n assert candidate('MadaM') == True\n assert candidate('Wow') == True\n assert candidate('') == False\n assert candidate('HI') == True\n assert candidate('go') == True\n assert candidate('gogo') == False\n assert candidate('aaaaaaaaaaaaaaa') == False\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate('Madam') == True\n assert candidate('M') == False\n assert candidate('0') == False\n\n","code":"def prime_length(string):"} -{"task_id":"JHumanEval\/83","prompt_en":"def starts_one_ends(n):\n \"\"\"\n Given a positive integer n, return the count of the numbers of n-digit\n positive integers that start or end with 1.\n \"\"\"\n","prompt":"正の整数 n が与えられたとき、n 桁の正の整数で 1 で始まるか\n もしくは終わる数のカウントを返す","entry_point":"starts_one_ends","canonical_solution":" if n == 1: return 1\n return 18 * (10 ** (n - 2))\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(1) == 1\n assert candidate(2) == 18\n assert candidate(3) == 180\n assert candidate(4) == 1800\n assert candidate(5) == 18000\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def starts_one_ends(n):"} -{"task_id":"JHumanEval\/84","prompt_en":"def solve(N):\n \"\"\"Given a positive integer N, return the total sum of its digits in binary.\n \n Example\n For N = 1000, the sum of digits will be 1 the output should be \"1\".\n For N = 150, the sum of digits will be 6 the output should be \"110\".\n For N = 147, the sum of digits will be 12 the output should be \"1100\".\n \n Variables:\n @N integer\n Constraints: 0 ≤ N ≤ 10000.\n Output:\n a string of binary number\n \"\"\"\n","prompt":"正の整数 N が与えられた時、その桁の総和を2進数で返す。\n 例\n N = 1000のとき, 各桁の総和は1、だから返り値は \"1\".\n N = 150のとき,各桁の総和は6、 だから返り値は \"110\".\n N = 147のとき,各桁の総和は12、 だから返り値は \"1100\".\n \n 変数:\n @N 整数\n 制約: 0 ≤ N ≤ 10000.\n 返り値:\n ２進数表記の文字列","entry_point":"solve","canonical_solution":" return bin(sum(int(i) for i in str(N)))[2:]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(1000) == \"1\", \"Error\"\n assert candidate(150) == \"110\", \"Error\"\n assert candidate(147) == \"1100\", \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(333) == \"1001\", \"Error\"\n assert candidate(963) == \"10010\", \"Error\"\n\n","code":"def solve(N):"} -{"task_id":"JHumanEval\/85","prompt_en":"def add(lst):\n \"\"\"Given a non-empty list of integers lst. add the even elements that are at odd indices..\n\n\n Examples:\n add([4, 2, 6, 7]) ==> 2 \n \"\"\"\n","prompt":"空でない整数のリストlstが与えられたとき、奇数のインデックスにある偶数の要素を加える。\n\n 例:\n add([4, 2, 6, 7]) ==> 2","entry_point":"add","canonical_solution":" return sum([lst[i] for i in range(1, len(lst), 2) if lst[i]%2 == 0])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([4, 88]) == 88\n assert candidate([4, 5, 6, 7, 2, 122]) == 122\n assert candidate([4, 0, 6, 7]) == 0\n assert candidate([4, 4, 6, 8]) == 12\n\n # Check some edge cases that are easy to work out by hand.\n \n","code":"def add(lst):"} -{"task_id":"JHumanEval\/86","prompt_en":"def anti_shuffle(s):\n \"\"\"\n Write a function that takes a string and returns an ordered version of it.\n Ordered version of string, is a string where all words (separated by space)\n are replaced by a new word where all the characters arranged in\n ascending order based on ascii value.\n Note: You should keep the order of words and blank spaces in the sentence.\n\n For example:\n anti_shuffle('Hi') returns 'Hi'\n anti_shuffle('hello') returns 'ehllo'\n anti_shuffle('Hello World!!!') returns 'Hello !!!Wdlor'\n \"\"\"\n","prompt":"文字列を引数として受け取り、その「順序付けられたバージョン」を返す関数を作成してください。\n 順序付けられたバージョンとは、各単語（空白で区切られた）の文字がASCII値に基づいて昇順に\n 並べ替えられた新しい単語に置き換えられた文字列です。\n 注意：文章内の単語と空白の順序はそのまま保ってください。\n \n 例えば：\n anti_shuffle('Hi') は 'Hi'を返す\n anti_shuffle('hello') は 'ehllo'返す\n anti_shuffle('Hello World!!!') は 'Hello !!!Wdlor'返す","entry_point":"anti_shuffle","canonical_solution":" return ' '.join([''.join(sorted(list(i))) for i in s.split(' ')])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('Hi') == 'Hi'\n assert candidate('hello') == 'ehllo'\n assert candidate('number') == 'bemnru'\n assert candidate('abcd') == 'abcd'\n assert candidate('Hello World!!!') == 'Hello !!!Wdlor'\n assert candidate('') == ''\n assert candidate('Hi. My name is Mister Robot. How are you?') == '.Hi My aemn is Meirst .Rboot How aer ?ouy'\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def anti_shuffle(s):"} -{"task_id":"JHumanEval\/87","prompt_en":"def get_row(lst, x):\n \"\"\"\n You are given a 2 dimensional data, as a nested lists,\n which is similar to matrix, however, unlike matrices,\n each row may contain a different number of columns.\n Given lst, and integer x, find integers x in the list,\n and return list of tuples, [(x1, y1), (x2, y2) ...] such that\n each tuple is a coordinate - (row, columns), starting with 0.\n Sort coordinates initially by rows in ascending order.\n Also, sort coordinates of the row by columns in descending order.\n \n Examples:\n get_row([\n [1,2,3,4,5,6],\n [1,2,3,4,1,6],\n [1,2,3,4,5,1]\n ], 1) == [(0, 0), (1, 4), (1, 0), (2, 5), (2, 0)]\n get_row([], 1) == []\n get_row([[], [1], [1, 2, 3]], 3) == [(2, 2)]\n \"\"\"\n","prompt":"2次元のデータがネストされたリストとして与えられる。これは行列に似ているが、\n 行列とは異なり、各行は異なる数の列を含むことができる。\n lstと整数xが与えられたとき、リスト内の整数xを見つけ、各タプルが0から始まる\n 座標（行、列）であるようなタプルのリスト[(x1, y1), (x2, y2) ...]を返す。\n 座標を最初は行の昇順でソートする。\n また、行の座標を列の降順でソートする。\n \n 例:\n get_row([\n [1,2,3,4,5,6],\n [1,2,3,4,1,6],\n [1,2,3,4,5,1]\n ], 1) == [(0, 0), (1, 4), (1, 0), (2, 5), (2, 0)]\n get_row([], 1) == []\n get_row([[], [1], [1, 2, 3]], 3) == [(2, 2)]","entry_point":"get_row","canonical_solution":" coords = [(i, j) for i in range(len(lst)) for j in range(len(lst[i])) if lst[i][j] == x]\n return sorted(sorted(coords, key=lambda x: x[1], reverse=True), key=lambda x: x[0])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([\n [1,2,3,4,5,6],\n [1,2,3,4,1,6],\n [1,2,3,4,5,1]\n ], 1) == [(0, 0), (1, 4), (1, 0), (2, 5), (2, 0)]\n assert candidate([\n [1,2,3,4,5,6],\n [1,2,3,4,5,6],\n [1,2,3,4,5,6],\n [1,2,3,4,5,6],\n [1,2,3,4,5,6],\n [1,2,3,4,5,6]\n ], 2) == [(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)]\n assert candidate([\n [1,2,3,4,5,6],\n [1,2,3,4,5,6],\n [1,1,3,4,5,6],\n [1,2,1,4,5,6],\n [1,2,3,1,5,6],\n [1,2,3,4,1,6],\n [1,2,3,4,5,1]\n ], 1) == [(0, 0), (1, 0), (2, 1), (2, 0), (3, 2), (3, 0), (4, 3), (4, 0), (5, 4), (5, 0), (6, 5), (6, 0)]\n assert candidate([], 1) == []\n assert candidate([[1]], 2) == []\n assert candidate([[], [1], [1, 2, 3]], 3) == [(2, 2)]\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def get_row(lst, x):"} -{"task_id":"JHumanEval\/88","prompt_en":"def sort_array(array):\n \"\"\"\n Given an array of non-negative integers, return a copy of the given array after sorting,\n you will sort the given array in ascending order if the sum( first index value, last index value) is odd,\n or sort it in descending order if the sum( first index value, last index value) is even.\n\n Note:\n * don't change the given array.\n\n Examples:\n * sort_array([]) => []\n * sort_array([5]) => [5]\n * sort_array([2, 4, 3, 0, 1, 5]) => [0, 1, 2, 3, 4, 5]\n * sort_array([2, 4, 3, 0, 1, 5, 6]) => [6, 5, 4, 3, 2, 1, 0]\n \"\"\"\n","prompt":"非負の整数からなる配列が与えられた場合、配列をソートしたコピーを返してください。\n 配列の最初の要素と最後の要素の和が奇数であれば、配列を昇順（小さい順）にソートします。\n その和が偶数であれば、配列を降順（大きい順）にソートします。\n\n 注意点：\n * 与えられた配列自体を変更しないでください。\n\n 例:\n * sort_array([]) => []\n * sort_array([5]) => [5]\n * sort_array([2, 4, 3, 0, 1, 5]) => [0, 1, 2, 3, 4, 5]\n * sort_array([2, 4, 3, 0, 1, 5, 6]) => [6, 5, 4, 3, 2, 1, 0]","entry_point":"sort_array","canonical_solution":" return [] if len(array) == 0 else sorted(array, reverse= (array[0]+array[-1]) % 2 == 0) \n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([]) == [], \"Error\"\n assert candidate([5]) == [5], \"Error\"\n assert candidate([2, 4, 3, 0, 1, 5]) == [0, 1, 2, 3, 4, 5], \"Error\"\n assert candidate([2, 4, 3, 0, 1, 5, 6]) == [6, 5, 4, 3, 2, 1, 0], \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([2, 1]) == [1, 2], \"Error\"\n assert candidate([15, 42, 87, 32 ,11, 0]) == [0, 11, 15, 32, 42, 87], \"Error\"\n assert candidate([21, 14, 23, 11]) == [23, 21, 14, 11], \"Error\"\n\n","code":"def sort_array(array):"} -{"task_id":"JHumanEval\/89","prompt_en":"def encrypt(s):\n \"\"\"Create a function encrypt that takes a string as an argument and\n returns a string encrypted with the alphabet being rotated. \n The alphabet should be rotated in a manner such that the letters \n shift down by two multiplied to two places.\n For example:\n encrypt('hi') returns 'lm'\n encrypt('asdfghjkl') returns 'ewhjklnop'\n encrypt('gf') returns 'kj'\n encrypt('et') returns 'ix'\n \"\"\"\n","prompt":"文字列を引数にとり、アルファベットを回転させて暗号化した\n 文字列を返す関数encryptを作成せよ。\n アルファベットは、文字位置を2倍して2つ下にシフトするように\n 回転する。\n 例:\n encrypt('hi') returns 'lm'\n encrypt('asdfghjkl') returns 'ewhjklnop'\n encrypt('gf') returns 'kj'\n encrypt('et') returns 'ix'","entry_point":"encrypt","canonical_solution":" d = 'abcdefghijklmnopqrstuvwxyz'\n out = ''\n for c in s:\n if c in d:\n out += d[(d.index(c)+2*2) % 26]\n else:\n out += c\n return out\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('hi') == 'lm', \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('asdfghjkl') == 'ewhjklnop', \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('gf') == 'kj', \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('et') == 'ix', \"This prints if this assert fails 1 (good for debugging!)\"\n\n assert candidate('faewfawefaewg')=='jeiajeaijeiak', \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('hellomyfriend')=='lippsqcjvmirh', \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate('dxzdlmnilfuhmilufhlihufnmlimnufhlimnufhfucufh')=='hbdhpqrmpjylqmpyjlpmlyjrqpmqryjlpmqryjljygyjl', \"This prints if this assert fails 3 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate('a')=='e', \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def encrypt(s):"} -{"task_id":"JHumanEval\/90","prompt_en":"def next_smallest(lst):\n \"\"\"\n You are given a list of integers.\n Write a function next_smallest() that returns the 2nd smallest element of the list.\n Return None if there is no such element.\n \n next_smallest([1, 2, 3, 4, 5]) == 2\n next_smallest([5, 1, 4, 3, 2]) == 2\n next_smallest([]) == None\n next_smallest([1, 1]) == None\n \"\"\"\n","prompt":"整数のリストが与えられる。\n リストの2番目に小さい要素を返す関数 next_smallest() を書きなさい。\n そのような要素がない場合は None を返す。\n \n next_smallest([1, 2, 3, 4, 5]) == 2\n next_smallest([5, 1, 4, 3, 2]) == 2\n next_smallest([]) == None\n next_smallest([1, 1]) == None","entry_point":"next_smallest","canonical_solution":" lst = sorted(set(lst))\n return None if len(lst) < 2 else lst[1]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1, 2, 3, 4, 5]) == 2\n assert candidate([5, 1, 4, 3, 2]) == 2\n assert candidate([]) == None\n assert candidate([1, 1]) == None\n assert candidate([1,1,1,1,0]) == 1\n assert candidate([1, 0**0]) == None\n assert candidate([-35, 34, 12, -45]) == -35\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def next_smallest(lst):"} -{"task_id":"JHumanEval\/91","prompt_en":"def is_bored(S):\n \"\"\"\n You'll be given a string of words, and your task is to count the number\n of boredoms. A boredom is a sentence that starts with the word \"I\".\n Sentences are delimited by '.', '?' or '!'.\n \n For example:\n >>> is_bored(\"Hello world\")\n 0\n >>> is_bored(\"The sky is blue. The sun is shining. I love this weather\")\n 1\n \"\"\"\n","prompt":"単語の文字列が与えられ、あなたのタスクは退屈指数を数える\n ことである。退屈指数とは、\"I \"で始まる文のことである。\n 文は'.'、’?’、'!'のいずれかで区切られる。 \n \n 例えば:\n >>> is_bored(\"Hello world\")\n 0\n >>> is_bored(\"The sky is blue. The sun is shining. I love this weather\")\n 1","entry_point":"is_bored","canonical_solution":" import re\n sentences = re.split(r'[.?!]\\s*', S)\n return sum(sentence[0:2] == 'I ' for sentence in sentences)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"Hello world\") == 0, \"Test 1\"\n assert candidate(\"Is the sky blue?\") == 0, \"Test 2\"\n assert candidate(\"I love It !\") == 1, \"Test 3\"\n assert candidate(\"bIt\") == 0, \"Test 4\"\n assert candidate(\"I feel good today. I will be productive. will kill It\") == 2, \"Test 5\"\n assert candidate(\"You and I are going for a walk\") == 0, \"Test 6\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def is_bored(S):"} -{"task_id":"JHumanEval\/92","prompt_en":"def any_int(x, y, z):\n '''\n Create a function that takes 3 numbers.\n Returns true if one of the numbers is equal to the sum of the other two, and all numbers are integers.\n Returns false in any other cases.\n \n Examples\n any_int(5, 2, 7) ➞ True\n \n any_int(3, 2, 2) ➞ False\n\n any_int(3, -2, 1) ➞ True\n \n any_int(3.6, -2.2, 2) ➞ False\n \n\n \n '''\n","prompt":"3つの数値を受け取る関数を作る。\n 1つの数値が他の2つの数値の和と等しく、すべての数値が整数である場合にTrueを返す。\n それ以外の場合��Falseを返す。\n \n 例\n any_int(5, 2, 7) ➞ True\n \n any_int(3, 2, 2) ➞ False\n\n any_int(3, -2, 1) ➞ True\n \n any_int(3.6, -2.2, 2) ➞ False","entry_point":"any_int","canonical_solution":" \n if isinstance(x,int) and isinstance(y,int) and isinstance(z,int):\n if (x+y==z) or (x+z==y) or (y+z==x):\n return True\n return False\n return False\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(2, 3, 1)==True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(2.5, 2, 3)==False, \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate(1.5, 5, 3.5)==False, \"This prints if this assert fails 3 (good for debugging!)\"\n assert candidate(2, 6, 2)==False, \"This prints if this assert fails 4 (good for debugging!)\"\n assert candidate(4, 2, 2)==True, \"This prints if this assert fails 5 (good for debugging!)\"\n assert candidate(2.2, 2.2, 2.2)==False, \"This prints if this assert fails 6 (good for debugging!)\"\n assert candidate(-4, 6, 2)==True, \"This prints if this assert fails 7 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(2,1,1)==True, \"This prints if this assert fails 8 (also good for debugging!)\"\n assert candidate(3,4,7)==True, \"This prints if this assert fails 9 (also good for debugging!)\"\n assert candidate(3.0,4,7)==False, \"This prints if this assert fails 10 (also good for debugging!)\"\n\n","code":"def any_int(x, y, z):"} -{"task_id":"JHumanEval\/93","prompt_en":"def encode(message):\n \"\"\"\n Write a function that takes a message, and encodes in such a \n way that it swaps case of all letters, replaces all vowels in \n the message with the letter that appears 2 places ahead of that \n vowel in the english alphabet. \n Assume only letters. \n \n Examples:\n >>> encode('test')\n 'TGST'\n >>> encode('This is a message')\n 'tHKS KS C MGSSCGG'\n \"\"\"\n","prompt":"メッセージを受け取り、すべての文字の大文字と小文字を入れ替え、\n メッセージ中のすべての母音を英語の母音の2つ前に現れる文字に置\n き換えるようにエンコードする関数を書きなさい。 \n 文字だけを想定する。\n\n 例\n >>> encode('test')\n 'TGST'\n >>> encode('This is a message')\n 'tHKS KS C MGSSCGG'","entry_point":"encode","canonical_solution":" vowels = \"aeiouAEIOU\"\n vowels_replace = dict([(i, chr(ord(i) + 2)) for i in vowels])\n message = message.swapcase()\n return ''.join([vowels_replace[i] if i in vowels else i for i in message])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('TEST') == 'tgst', \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('Mudasir') == 'mWDCSKR', \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate('YES') == 'ygs', \"This prints if this assert fails 3 (good for debugging!)\"\n \n # Check some edge cases that are easy to work out by hand.\n assert candidate('This is a message') == 'tHKS KS C MGSSCGG', \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(\"I DoNt KnOw WhAt tO WrItE\") == 'k dQnT kNqW wHcT Tq wRkTg', \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def encode(message):"} -{"task_id":"JHumanEval\/94","prompt_en":"def skjkasdkd(lst):\n \"\"\"You are given a list of integers.\n You need to find the largest prime value and return the sum of its digits.\n\n Examples:\n For lst = [0,3,2,1,3,5,7,4,5,5,5,2,181,32,4,32,3,2,32,324,4,3] the output should be 10\n For lst = [1,0,1,8,2,4597,2,1,3,40,1,2,1,2,4,2,5,1] the output should be 25\n For lst = [1,3,1,32,5107,34,83278,109,163,23,2323,32,30,1,9,3] the output should be 13\n For lst = [0,724,32,71,99,32,6,0,5,91,83,0,5,6] the output should be 11\n For lst = [0,81,12,3,1,21] the output should be 3\n For lst = [0,8,1,2,1,7] the output should be 7\n \"\"\"\n","prompt":"整数のリストが与えらる。\n 最大の素数を求め、その桁数の和を返す必要がある。\n 例:\n lst = [0,3,2,1,3,5,7,4,5,5,5,2,181,32,4,32,3,2,32,324,4,3] のとき、返り値は10\n lst = [1,0,1,8,2,4597,2,1,3,40,1,2,1,2,4,2,5,1]のとき、返り値は 25\n lst = [1,3,1,32,5107,34,83278,109,163,23,2323,32,30,1,9,3]のとき、返り値は13\n lst = [0,724,32,71,99,32,6,0,5,91,83,0,5,6]のとき、返り値は11\n lst = [0,81,12,3,1,21]のとき、返り値は3\n lst = [0,8,1,2,1,7]のとき、返り値は7","entry_point":"skjkasdkd","canonical_solution":" def isPrime(n):\n for i in range(2,int(n**0.5)+1):\n if n%i==0:\n return False\n\n return True\n maxx = 0\n i = 0\n while i < len(lst):\n if(lst[i] > maxx and isPrime(lst[i])):\n maxx = lst[i]\n i+=1\n result = sum(int(digit) for digit in str(maxx))\n return result\n\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([0,3,2,1,3,5,7,4,5,5,5,2,181,32,4,32,3,2,32,324,4,3]) == 10, \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([1,0,1,8,2,4597,2,1,3,40,1,2,1,2,4,2,5,1]) == 25, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([1,3,1,32,5107,34,83278,109,163,23,2323,32,30,1,9,3]) == 13, \"This prints if this assert fails 3 (also good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([0,724,32,71,99,32,6,0,5,91,83,0,5,6]) == 11, \"This prints if this assert fails 4 (also good for debugging!)\"\n \n # Check some edge cases that are easy to work out by hand.\n assert candidate([0,81,12,3,1,21]) == 3, \"This prints if this assert fails 5 (also good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([0,8,1,2,1,7]) == 7, \"This prints if this assert fails 6 (also good for debugging!)\"\n\n assert candidate([8191]) == 19, \"This prints if this assert fails 7 (also good for debugging!)\"\n assert candidate([8191, 123456, 127, 7]) == 19, \"This prints if this assert fails 8 (also good for debugging!)\"\n assert candidate([127, 97, 8192]) == 10, \"This prints if this assert fails 9 (also good for debugging!)\"\n","code":"def skjkasdkd(lst):"} -{"task_id":"JHumanEval\/95","prompt_en":"def check_dict_case(dict):\n \"\"\"\n Given a dictionary, return True if all keys are strings in lower \n case or all keys are strings in upper case, else return False.\n The function should return False is the given dictionary is empty.\n Examples:\n check_dict_case({\"a\":\"apple\", \"b\":\"banana\"}) should return True.\n check_dict_case({\"a\":\"apple\", \"A\":\"banana\", \"B\":\"banana\"}) should return False.\n check_dict_case({\"a\":\"apple\", 8:\"banana\", \"a\":\"apple\"}) should return False.\n check_dict_case({\"Name\":\"John\", \"Age\":\"36\", \"City\":\"Houston\"}) should return False.\n check_dict_case({\"STATE\":\"NC\", \"ZIP\":\"12345\" }) should return True.\n \"\"\"\n","prompt":"辞書が与えられたとき、すべてのキーが小文字であればTrueを、\n すべてのキーが大文字の文字列であればFalseを返す。\n 与えられた辞書が空の場合、この関数は False を返す。\n 例:\n check_dict_case({\"a\":\"apple\", \"b\":\"banana\"}) は、 Trueを返す。\n check_dict_case({\"a\":\"apple\", \"A\":\"banana\", \"B\":\"banana\"}) は、 Falseを返す。\n check_dict_case({\"a\":\"apple\", 8:\"banana\", \"a\":\"apple\"}) は、 Falseを返す。\n check_dict_case({\"Name\":\"John\", \"Age\":\"36\", \"City\":\"Houston\"}) は、 Falseを返す。\n check_dict_case({\"STATE\":\"NC\", \"ZIP\":\"12345\" }) は、 Trueを返す。","entry_point":"check_dict_case","canonical_solution":" if len(dict.keys()) == 0:\n return False\n else:\n state = \"start\"\n for key in dict.keys():\n\n if isinstance(key, str) == False:\n state = \"mixed\"\n break\n if state == \"start\":\n if key.isupper():\n state = \"upper\"\n elif key.islower():\n state = \"lower\"\n else:\n break\n elif (state == \"upper\" and not key.isupper()) or (state == \"lower\" and not key.islower()):\n state = \"mixed\"\n break\n else:\n break\n return state == \"upper\" or state == \"lower\" \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate({\"p\":\"pineapple\", \"b\":\"banana\"}) == True, \"First test error: \" + str(candidate({\"p\":\"pineapple\", \"b\":\"banana\"}))\n assert candidate({\"p\":\"pineapple\", \"A\":\"banana\", \"B\":\"banana\"}) == False, \"Second test error: \" + str(candidate({\"p\":\"pineapple\", \"A\":\"banana\", \"B\":\"banana\"}))\n assert candidate({\"p\":\"pineapple\", 5:\"banana\", \"a\":\"apple\"}) == False, \"Third test error: \" + str(candidate({\"p\":\"pineapple\", 5:\"banana\", \"a\":\"apple\"}))\n assert candidate({\"Name\":\"John\", \"Age\":\"36\", \"City\":\"Houston\"}) == False, \"Fourth test error: \" + str(candidate({\"Name\":\"John\", \"Age\":\"36\", \"City\":\"Houston\"}))\n assert candidate({\"STATE\":\"NC\", \"ZIP\":\"12345\" }) == True, \"Fifth test error: \" + str(candidate({\"STATE\":\"NC\", \"ZIP\":\"12345\" })) \n assert candidate({\"fruit\":\"Orange\", \"taste\":\"Sweet\" }) == True, \"Fourth test error: \" + str(candidate({\"fruit\":\"Orange\", \"taste\":\"Sweet\" })) \n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate({}) == False, \"1st edge test error: \" + str(candidate({}))\n\n","code":"def check_dict_case(dict):"} -{"task_id":"JHumanEval\/96","prompt_en":"def count_up_to(n):\n \"\"\"Implement a function that takes an non-negative integer and returns an array of the first n\n integers that are prime numbers and less than n.\n for example:\n count_up_to(5) => [2,3]\n count_up_to(11) => [2,3,5,7]\n count_up_to(0) => []\n count_up_to(20) => [2,3,5,7,11,13,17,19]\n count_up_to(1) => []\n count_up_to(18) => [2,3,5,7,11,13,17]\n \"\"\"\n","prompt":"非負整数を受け取り、素数でnより小さい最初のn個の\n 整数の配列を返す関数を実装せよ。\n 例えば:\n count_up_to(5) => [2,3]\n count_up_to(11) => [2,3,5,7]\n count_up_to(0) => []\n count_up_to(20) => [2,3,5,7,11,13,17,19]\n count_up_to(1) => []\n count_up_to(18) => [2,3,5,7,11,13,17]","entry_point":"count_up_to","canonical_solution":" primes = []\n for i in range(2, n):\n is_prime = True\n for j in range(2, i):\n if i % j == 0:\n is_prime = False\n break\n if is_prime:\n primes.append(i)\n return primes\n\n","test":"def check(candidate):\n\n assert candidate(5) == [2,3]\n assert candidate(6) == [2,3,5]\n assert candidate(7) == [2,3,5]\n assert candidate(10) == [2,3,5,7]\n assert candidate(0) == []\n assert candidate(22) == [2,3,5,7,11,13,17,19]\n assert candidate(1) == []\n assert candidate(18) == [2,3,5,7,11,13,17]\n assert candidate(47) == [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43]\n assert candidate(101) == [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]\n\n","code":"def count_up_to(n):"} -{"task_id":"JHumanEval\/97","prompt_en":"def multiply(a, b):\n \"\"\"Complete the function that takes two integers and returns \n the product of their unit digits.\n Assume the input is always valid.\n Examples:\n multiply(148, 412) should return 16.\n multiply(19, 28) should return 72.\n multiply(2020, 1851) should return 0.\n multiply(14,-15) should return 20.\n \"\"\"\n","prompt":"2つの整数を受け取り、その１の位の数の積を返す関数を完成させよ。\n 入力は常に有効範囲にあるとする。\n 例:\n multiply(148, 412) は16を返す。\n multiply(19, 28) は72を返す。\n multiply(2020, 1851) は0を返す。\n multiply(14,-15) は20を返す。","entry_point":"multiply","canonical_solution":" return abs(a % 10) * abs(b % 10)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(148, 412) == 16, \"First test error: \" + str(candidate(148, 412)) \n assert candidate(19, 28) == 72, \"Second test error: \" + str(candidate(19, 28)) \n assert candidate(2020, 1851) == 0, \"Third test error: \" + str(candidate(2020, 1851))\n assert candidate(14,-15) == 20, \"Fourth test error: \" + str(candidate(14,-15)) \n assert candidate(76, 67) == 42, \"Fifth test error: \" + str(candidate(76, 67)) \n assert candidate(17, 27) == 49, \"Sixth test error: \" + str(candidate(17, 27)) \n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(0, 1) == 0, \"1st edge test error: \" + str(candidate(0, 1))\n assert candidate(0, 0) == 0, \"2nd edge test error: \" + str(candidate(0, 0))\n\n","code":"def multiply(a, b):"} -{"task_id":"JHumanEval\/98","prompt_en":"def count_upper(s):\n \"\"\"\n Given a string s, count the number of uppercase vowels in even indices.\n \n For example:\n count_upper('aBCdEf') returns 1\n count_upper('abcdefg') returns 0\n count_upper('dBBE') returns 0\n \"\"\"\n","prompt":"文字列 s が与えられたとき、偶数のインデックスに含まれる大文字の母音の数を数える。\n 例えば:\n count_upper('aBCdEf') returns 1\n count_upper('abcdefg') returns 0\n count_upper('dBBE') returns 0","entry_point":"count_upper","canonical_solution":" count = 0\n for i in range(0,len(s),2):\n if s[i] in \"AEIOU\":\n count += 1\n return count\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('aBCdEf') == 1\n assert candidate('abcdefg') == 0\n assert candidate('dBBE') == 0\n assert candidate('B') == 0\n assert candidate('U') == 1\n assert candidate('') == 0\n assert candidate('EEEE') == 2\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def count_upper(s):"} -{"task_id":"JHumanEval\/99","prompt_en":"def closest_integer(value):\n '''\n Create a function that takes a value (string) representing a number\n and returns the closest integer to it. If the number is equidistant\n from two integers, round it away from zero.\n\n Examples\n >>> closest_integer(\"10\")\n 10\n >>> closest_integer(\"15.3\")\n 15\n\n Note:\n Rounding away from zero means that if the given number is equidistant\n from two integers, the one you should return is the one that is the\n farthest from zero. For example closest_integer(\"14.5\") should\n return 15 and closest_integer(\"-14.5\") should return -15.\n '''\n","prompt":"数値を表す文字列valueを受け取り、それに最も近い整数を返す関数を作る。\n その数値が2つの整数から等距離にある場合は、ゼロから四捨五入する。\n \n 例\n >>> closest_integer(\"10\")\n 10\n >>> closest_integer(\"15.3\")\n 15\n \n Note:\n ゼロからの四捨五入とは、与えられた数値が2つの整数から\n 等距離にある場合、ゼロから遠い方を返すという意味である。\n 例えば、 close_integer(\"14.5\")は15を返し、closest_integer(\"-14.5\")は-15を返す。","entry_point":"closest_integer","canonical_solution":" from math import floor, ceil\n\n if value.count('.') == 1:\n # remove trailing zeros\n while (value[-1] == '0'):\n value = value[:-1]\n\n num = float(value)\n if value[-2:] == '.5':\n if num > 0:\n res = ceil(num)\n else:\n res = floor(num)\n elif len(value) > 0:\n res = int(round(num))\n else:\n res = 0\n\n return res\n\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"10\") == 10, \"Test 1\"\n assert candidate(\"14.5\") == 15, \"Test 2\"\n assert candidate(\"-15.5\") == -16, \"Test 3\"\n assert candidate(\"15.3\") == 15, \"Test 3\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"0\") == 0, \"Test 0\"\n\n","code":"def closest_integer(value):"} -{"task_id":"JHumanEval\/100","prompt_en":"def make_a_pile(n):\n \"\"\"\n Given a positive integer n, you have to make a pile of n levels of stones.\n The first level has n stones.\n The number of stones in the next level is:\n - the next odd number if n is odd.\n - the next even number if n is even.\n Return the number of stones in each level in a list, where element at index\n i represents the number of stones in the level (i+1).\n\n Examples:\n >>> make_a_pile(3)\n [3, 5, 7]\n \"\"\"\n","prompt":"正の整数nが与えられたとき、n段の石の山を作らなければならない。\n 最初の段にはn個の石がある。\n 次の段の石の数は\n - nが奇数なら次の奇数。\n - nが偶数なら次の偶数。\n 各段の石の数をリストで返す。インデックス i の要素は、段 (i+1) の石の\n 数を表すものとする。\n 例:\n >>> make_a_pile(3)\n [3, 5, 7]","entry_point":"make_a_pile","canonical_solution":" return [n + 2*i for i in range(n)]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(3) == [3, 5, 7], \"Test 3\"\n assert candidate(4) == [4,6,8,10], \"Test 4\"\n assert candidate(5) == [5, 7, 9, 11, 13]\n assert candidate(6) == [6, 8, 10, 12, 14, 16]\n assert candidate(8) == [8, 10, 12, 14, 16, 18, 20, 22]\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def make_a_pile(n):"} -{"task_id":"JHumanEval\/101","prompt_en":"def words_string(s):\n \"\"\"\n You will be given a string of words separated by commas or spaces. Your task is\n to split the string into words and return an array of the words.\n \n For example:\n words_string(\"Hi, my name is John\") == [\"Hi\", \"my\", \"name\", \"is\", \"John\"]\n words_string(\"One, two, three, four, five, six\") == [\"One\", \"two\", \"three\", \"four\", \"five\", \"six\"]\n \"\"\"\n","prompt":"カンマまたは空白で区切られた単語の文字列が与えられる。あなたのタスクは、\n 文字列を単語に分割し、単語の配列を返すことである。\n \n 例えば:\n words_string(\"Hi, my name is John\") == [\"Hi\", \"my\", \"name\", \"is\", \"John\"]\n words_string(\"One, two, three, four, five, six\") == [\"One\", \"two\", \"three\", \"four\", \"five\", \"six\"]","entry_point":"words_string","canonical_solution":" if not s:\n return []\n\n s_list = []\n\n for letter in s:\n if letter == ',':\n s_list.append(' ')\n else:\n s_list.append(letter)\n\n s_list = \"\".join(s_list)\n return s_list.split()\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(\"Hi, my name is John\") == [\"Hi\", \"my\", \"name\", \"is\", \"John\"]\n assert candidate(\"One, two, three, four, five, six\") == [\"One\", \"two\", \"three\", \"four\", \"five\", \"six\"]\n assert candidate(\"Hi, my name\") == [\"Hi\", \"my\", \"name\"]\n assert candidate(\"One,, two, three, four, five, six,\") == [\"One\", \"two\", \"three\", \"four\", \"five\", \"six\"]\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(\"\") == []\n assert candidate(\"ahmed , gamal\") == [\"ahmed\", \"gamal\"]\n\n","code":"def words_string(s):"} -{"task_id":"JHumanEval\/102","prompt_en":"def choose_num(x, y):\n \"\"\"This function takes two positive numbers x and y and returns the\n biggest even integer number that is in the range [x, y] inclusive. If \n there's no such number, then the function should return -1.\n\n For example:\n choose_num(12, 15) = 14\n choose_num(13, 12) = -1\n \"\"\"\n","prompt":"この関数は2つの正の数xとyを受け取り、範囲[x, y]（両端を含む）に含まれる\n 最大の偶数整数を返す。そのような数がない場合、関数は-1を返す。\n \n 例えば:\n choose_num(12, 15) = 14\n choose_num(13, 12) = -1","entry_point":"choose_num","canonical_solution":" if x > y:\n return -1\n if y % 2 == 0:\n return y\n if x == y:\n return -1\n return y - 1\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(12, 15) == 14\n assert candidate(13, 12) == -1\n assert candidate(33, 12354) == 12354\n assert candidate(5234, 5233) == -1\n assert candidate(6, 29) == 28\n assert candidate(27, 10) == -1\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(7, 7) == -1\n assert candidate(546, 546) == 546\n\n","code":"def choose_num(x, y):"} -{"task_id":"JHumanEval\/103","prompt_en":"def rounded_avg(n, m):\n \"\"\"You are given two positive integers n and m, and your task is to compute the\n average of the integers from n through m (including n and m). \n Round the answer to the nearest integer and convert that to binary.\n If n is greater than m, return -1.\n Example:\n rounded_avg(1, 5) => \"0b11\"\n rounded_avg(7, 5) => -1\n rounded_avg(10, 20) => \"0b1111\"\n rounded_avg(20, 33) => \"0b11010\"\n \"\"\"\n","prompt":"2つの正の整数nとmが与えられており、あなたのタスクはnからmまでの\n 整数（nとmを含む）の平均を計算することである。\n 答えを最も近い整数に丸め、2進数に変換せよ。\n nがmより大きい場合は-1を返す。\n 例:\n rounded_avg(1, 5) => \"0b11\"\n rounded_avg(7, 5) => -1\n rounded_avg(10, 20) => \"0b1111\"\n rounded_avg(20, 33) => \"0b11010\"","entry_point":"rounded_avg","canonical_solution":" if m < n:\n return -1\n summation = 0\n for i in range(n, m+1):\n summation += i\n return bin(round(summation\/(m - n + 1)))\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(1, 5) == \"0b11\"\n assert candidate(7, 13) == \"0b1010\"\n assert candidate(964,977) == \"0b1111001010\"\n assert candidate(996,997) == \"0b1111100100\"\n assert candidate(560,851) == \"0b1011000010\"\n assert candidate(185,546) == \"0b101101110\"\n assert candidate(362,496) == \"0b110101101\"\n assert candidate(350,902) == \"0b1001110010\"\n assert candidate(197,233) == \"0b11010111\"\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(7, 5) == -1\n assert candidate(5, 1) == -1\n assert candidate(5, 5) == \"0b101\"\n\n","code":"def rounded_avg(n, m):"} -{"task_id":"JHumanEval\/104","prompt_en":"def unique_digits(x):\n \"\"\"Given a list of positive integers x. return a sorted list of all \n elements that hasn't any even digit.\n\n Note: Returned list should be sorted in increasing order.\n \n For example:\n >>> unique_digits([15, 33, 1422, 1])\n [1, 15, 33]\n >>> unique_digits([152, 323, 1422, 10])\n []\n \"\"\"\n","prompt":"正の整数xのリストが与えられたとき、偶数桁の要素を持たない全ての\n 要素をソートしたリストを返す。\n \n 注意: 返されるリストは昇順にソートされていなければならない。\n \n 例えば:\n >>> unique_digits([15, 33, 1422, 1])\n [1, 15, 33]\n >>> unique_digits([152, 323, 1422, 10])\n []","entry_point":"unique_digits","canonical_solution":" odd_digit_elements = []\n for i in x:\n if all (int(c) % 2 == 1 for c in str(i)):\n odd_digit_elements.append(i)\n return sorted(odd_digit_elements)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([15, 33, 1422, 1]) == [1, 15, 33]\n assert candidate([152, 323, 1422, 10]) == []\n assert candidate([12345, 2033, 111, 151]) == [111, 151]\n assert candidate([135, 103, 31]) == [31, 135]\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def unique_digits(x):"} -{"task_id":"JHumanEval\/105","prompt_en":"def by_length(arr):\n \"\"\"\n Given an array of integers, sort the integers that are between 1 and 9 inclusive,\n reverse the resulting array, and then replace each digit by its corresponding name from\n \"One\", \"Two\", \"Three\", \"Four\", \"Five\", \"Six\", \"Seven\", \"Eight\", \"Nine\".\n\n For example:\n arr = [2, 1, 1, 4, 5, 8, 2, 3] \n -> sort arr -> [1, 1, 2, 2, 3, 4, 5, 8] \n -> reverse arr -> [8, 5, 4, 3, 2, 2, 1, 1]\n return [\"Eight\", \"Five\", \"Four\", \"Three\", \"Two\", \"Two\", \"One\", \"One\"]\n \n If the array is empty, return an empty array:\n arr = []\n return []\n \n If the array has any strange number ignore it:\n arr = [1, -1 , 55] \n -> sort arr -> [-1, 1, 55]\n -> reverse arr -> [55, 1, -1]\n return = ['One']\n \"\"\"\n","prompt":"整数の配列が与えられたとき、1から9までの整数をソートし、\n 得られた配列を逆順にし、各桁を以下の数字に相当する名前に置き換える。\n \"One\"、\"Two\"、\"Three\"、\"Four\"、\"Five\"、\"Six\"、\"Seven\"、\"Eight\"、\"Nine \"\n 例えば:\n arr = [2, 1, 1, 4, 5, 8, 2, 3] \n -> arrをソート -> [1, 1, 2, 2, 3, 4, 5, 8] \n -> arrを逆順 -> [8, 5, 4, 3, 2, 2, 1, 1]\n [\"Eight\", \"Five\", \"Four\", \"Three\", \"Two\", \"Two\", \"One\", \"One\"]を返す。\n \n もし空配列なら、空配列を返す:\n arr = []\n return []\n \n もし変な数値が配列に含まれていたら無視せよ:\n arr = [1, -1 , 55] \n -> arrをソート-> [-1, 1, 55]\n -> rarrを逆順 -> [55, 1, -1]\n ['One']を返す。","entry_point":"by_length","canonical_solution":" dic = {\n 1: \"One\",\n 2: \"Two\",\n 3: \"Three\",\n 4: \"Four\",\n 5: \"Five\",\n 6: \"Six\",\n 7: \"Seven\",\n 8: \"Eight\",\n 9: \"Nine\",\n }\n sorted_arr = sorted(arr, reverse=True)\n new_arr = []\n for var in sorted_arr:\n try:\n new_arr.append(dic[var])\n except:\n pass\n return new_arr\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([2, 1, 1, 4, 5, 8, 2, 3]) == [\"Eight\", \"Five\", \"Four\", \"Three\", \"Two\", \"Two\", \"One\", \"One\"], \"Error\"\n assert candidate([]) == [], \"Error\"\n assert candidate([1, -1 , 55]) == ['One'], \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([1, -1, 3, 2]) == [\"Three\", \"Two\", \"One\"]\n assert candidate([9, 4, 8]) == [\"Nine\", \"Eight\", \"Four\"]\n\n","code":"def by_length(arr):"} -{"task_id":"JHumanEval\/106","prompt_en":"def f(n):\n \"\"\" Implement the function f that takes n as a parameter,\n and returns a list of size n, such that the value of the element at index i is the factorial of i if i is even\n or the sum of numbers from 1 to i otherwise.\n i starts from 1.\n the factorial of i is the multiplication of the numbers from 1 to i (1 * 2 * ... * i).\n Example:\n f(5) == [1, 2, 6, 24, 15]\n \"\"\"\n","prompt":"iは1から始まる。iの階乗は1からiまでの数の掛け算（1 * 2 * ... * i）である。\n Example:\n f(5) == [1, 2, 6, 24, 15]","entry_point":"f","canonical_solution":" ret = []\n for i in range(1,n+1):\n if i%2 == 0:\n x = 1\n for j in range(1,i+1): x *= j\n ret += [x]\n else:\n x = 0\n for j in range(1,i+1): x += j\n ret += [x]\n return ret\n","test":"def check(candidate):\n\n assert candidate(5) == [1, 2, 6, 24, 15]\n assert candidate(7) == [1, 2, 6, 24, 15, 720, 28]\n assert candidate(1) == [1]\n assert candidate(3) == [1, 2, 6]\n","code":"def f(n):"} -{"task_id":"JHumanEval\/107","prompt_en":"def even_odd_palindrome(n):\n \"\"\"\n Given a positive integer n, return a tuple that has the number of even and odd\n integer palindromes that fall within the range(1, n), inclusive.\n\n Example 1:\n\n Input: 3\n Output: (1, 2)\n Explanation:\n Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.\n\n Example 2:\n\n Input: 12\n Output: (4, 6)\n Explanation:\n Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.\n\n Note:\n 1. 1 <= n <= 10^3\n 2. returned tuple has the number of even and odd integer palindromes respectively.\n \"\"\"\n","prompt":"与えられた正の整数 nに対して、範囲 1 から n まで（両端を含む）に存在する\n 偶数の回文数(integer palindrome)と奇数の回文数の個数をタプル形式で返す。\n \n 例 1:\n\n 入力: 3\n 出力: (1, 2)\n 解説:\n 回文数は1、2、3であり、そのうち1つは偶数、2つは奇数である。\n\n 例 2:\n\n 入力: 12\n 出力: (4, 6)\n 解説:\n 回文���は、1、2、3、4、5、6、7、8、9、11であり、そのうち4つは偶数、6つは奇数である。\n\n ノート:\n 1. 1 <= n <= 10^3\n 2. 返されるタプルは、それぞれ偶数と奇数の回文数を持つ。","entry_point":"even_odd_palindrome","canonical_solution":" def is_palindrome(n):\n return str(n) == str(n)[::-1]\n\n even_palindrome_count = 0\n odd_palindrome_count = 0\n\n for i in range(1, n+1):\n if i%2 == 1 and is_palindrome(i):\n odd_palindrome_count += 1\n elif i%2 == 0 and is_palindrome(i):\n even_palindrome_count += 1\n return (even_palindrome_count, odd_palindrome_count)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(123) == (8, 13)\n assert candidate(12) == (4, 6)\n assert candidate(3) == (1, 2)\n assert candidate(63) == (6, 8)\n assert candidate(25) == (5, 6)\n assert candidate(19) == (4, 6)\n assert candidate(9) == (4, 5), \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1) == (0, 1), \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def even_odd_palindrome(n):"} -{"task_id":"JHumanEval\/108","prompt_en":"def count_nums(arr):\n \"\"\"\n Write a function count_nums which takes an array of integers and returns\n the number of elements which has a sum of digits > 0.\n If a number is negative, then its first signed digit will be negative:\n e.g. -123 has signed digits -1, 2, and 3.\n >>> count_nums([]) == 0\n >>> count_nums([-1, 11, -11]) == 1\n >>> count_nums([1, 1, 2]) == 3\n \"\"\"\n","prompt":"count_nums 関数は、整数の配列を引数として受け取り、その配列内の各整数の各桁の合計が \n >0 となるような整数の個数を返す。負の数に関しては、最初の桁（符号付き桁）は負となる。\n たとえば、−123 の符号付き桁は −1, 2, 3 である。\n >>> count_nums([]) == 0\n >> count_nums([-1, 11, -11]) == 1\n >> count_nums([1, 1, 2]) == 3","entry_point":"count_nums","canonical_solution":" def digits_sum(n):\n neg = 1\n if n < 0: n, neg = -1 * n, -1 \n n = [int(i) for i in str(n)]\n n[0] = n[0] * neg\n return sum(n)\n return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([]) == 0\n assert candidate([-1, -2, 0]) == 0\n assert candidate([1, 1, 2, -2, 3, 4, 5]) == 6\n assert candidate([1, 6, 9, -6, 0, 1, 5]) == 5\n assert candidate([1, 100, 98, -7, 1, -1]) == 4\n assert candidate([12, 23, 34, -45, -56, 0]) == 5\n assert candidate([-0, 1**0]) == 1\n assert candidate([1]) == 1\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def count_nums(arr):"} -{"task_id":"JHumanEval\/109","prompt_en":"def move_one_ball(arr):\n \"\"\"We have an array 'arr' of N integers arr[1], arr[2], ..., arr[N].The\n numbers in the array will be randomly ordered. Your task is to determine if\n it is possible to get an array sorted in non-decreasing order by performing \n the following operation on the given array:\n You are allowed to perform right shift operation any number of times.\n \n One right shift operation means shifting all elements of the array by one\n position in the right direction. The last element of the array will be moved to\n the starting position in the array i.e. 0th index. \n\n If it is possible to obtain the sorted array by performing the above operation\n then return True else return False.\n If the given array is empty then return True.\n\n Note: The given list is guaranteed to have unique elements.\n\n For Example:\n \n move_one_ball([3, 4, 5, 1, 2])==>True\n Explanation: By performin 2 right shift operations, non-decreasing order can\n be achieved for the given array.\n move_one_ball([3, 5, 4, 1, 2])==>False\n Explanation:It is not possible to get non-decreasing order for the given\n array by performing any number of right shift operations.\n \n \"\"\"\n","prompt":"N個の整数arr[1], arr[2], ..., arr[N]なる配列 'arr' があります。\n この配列の数字はランダムな順番に並んでいます。あなたの課題は、以下の操作を何度でも行うことで、\n 配列を非減少.の順番にソートできるかどうかを判断することです。\n 操作として許されているのは「右シフト」です。\n \n 一回の「右シフト」操作とは、配列のすべての要素を右方向に一つずつずらすことを意味します。\n 配列の最後の要素は配列の先頭、すなわち0番目のインデックスに移動します。\n \n 上記の操作を行って��ートされた配列を得られる場合は True を、そうでない場合は False を返してください。\n 与えられた配列が空の場合は True を返してください。\n \n 注意：与えられたリストには一意の要素しか含まれていないことが保証されています。\n \n 例：\n \n move_one_ball([3, 4, 5, 1, 2]) => True\n 説明：2回の右シフト操作を行うことで、与えられた配列を非減少の順序にすることができます。\n \n move_one_ball([3, 5, 4, 1, 2]) => False\n 説明：どれだけ右シフト操作を行っても、与えられた配列を非減少の順序にすることはできません。","entry_point":"move_one_ball","canonical_solution":" if len(arr)==0:\n return True\n sorted_array=sorted(arr)\n my_arr=[]\n \n min_value=min(arr)\n min_index=arr.index(min_value)\n my_arr=arr[min_index:]+arr[0:min_index]\n for i in range(len(arr)):\n if my_arr[i]!=sorted_array[i]:\n return False\n return True\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([3, 4, 5, 1, 2])==True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([3, 5, 10, 1, 2])==True\n assert candidate([4, 3, 1, 2])==False\n # Check some edge cases that are easy to work out by hand.\n assert candidate([3, 5, 4, 1, 2])==False, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([])==True\n","code":"def move_one_ball(arr):"} -{"task_id":"JHumanEval\/110","prompt_en":"def exchange(lst1, lst2):\n \"\"\"In this problem, you will implement a function that takes two lists of numbers,\n and determines whether it is possible to perform an exchange of elements\n between them to make lst1 a list of only even numbers.\n There is no limit on the number of exchanged elements between lst1 and lst2.\n If it is possible to exchange elements between the lst1 and lst2 to make\n all the elements of lst1 to be even, return \"YES\".\n Otherwise, return \"NO\".\n For example:\n exchange([1, 2, 3, 4], [1, 2, 3, 4]) => \"YES\"\n exchange([1, 2, 3, 4], [1, 5, 3, 4]) => \"NO\"\n It is assumed that the input lists will be non-empty.\n \"\"\"\n","prompt":"この問題では、2つの数のリストを受け取り、lst1を偶数のみのリストに\n するために、それらの間で要素の交換を行うことが可能かどうかを判断す\n る関数を実装する。\n lst1とlst2の間で交換される要素の数に制限はない。\n lst1とlst2の間で要素の交換を行い、lst1の要素をすべて偶数にすることが\n 可能であれば、\"YES \"を返す。\n そうでなければ \"NO \"を返す。\n 例えば:\n exchange([1, 2, 3, 4], [1, 2, 3, 4]) => \"YES\"\n exchange([1, 2, 3, 4], [1, 5, 3, 4]) => \"NO\"\n 受け取るリストは空でないと前提してよい。","entry_point":"exchange","canonical_solution":" odd = 0\n even = 0\n for i in lst1:\n if i%2 == 1:\n odd += 1\n for i in lst2:\n if i%2 == 0:\n even += 1\n if even >= odd:\n return \"YES\"\n return \"NO\"\n \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1, 2, 3, 4], [1, 2, 3, 4]) == \"YES\"\n assert candidate([1, 2, 3, 4], [1, 5, 3, 4]) == \"NO\"\n assert candidate([1, 2, 3, 4], [2, 1, 4, 3]) == \"YES\" \n assert candidate([5, 7, 3], [2, 6, 4]) == \"YES\"\n assert candidate([5, 7, 3], [2, 6, 3]) == \"NO\" \n assert candidate([3, 2, 6, 1, 8, 9], [3, 5, 5, 1, 1, 1]) == \"NO\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([100, 200], [200, 200]) == \"YES\"\n\n","code":"def exchange(lst1, lst2):"} -{"task_id":"JHumanEval\/111","prompt_en":"def histogram(test):\n \"\"\"Given a string representing a space separated lowercase letters, return a dictionary\n of the letter with the most repetition and containing the corresponding count.\n If several letters have the same occurrence, return all of them.\n \n Example:\n histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}\n histogram('a b b a') == {'a': 2, 'b': 2}\n histogram('a b c a b') == {'a': 2, 'b': 2}\n histogram('b b b b a') == {'b': 4}\n histogram('') == {}\n\n \"\"\"\n","prompt":"空白で区切られた小文字を表す文字列が与えられる。最も出現回数が多い文字と\n 対応するカウントの辞書を返す。\n 複数の文字が同じ出現回数を持つ場合、それらすべてを返す。\n \n 例:\n histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}\n histogram('a b b a') == {'a': 2, 'b': 2}\n histogram('a b c a b') == {'a': 2, 'b': 2}\n histogram('b b b b a') == {'b': 4}\n histogram('') == {}","entry_point":"histogram","canonical_solution":" dict1={}\n list1=test.split(\" \")\n t=0\n\n for i in list1:\n if(list1.count(i)>t) and i!='':\n t=list1.count(i)\n if t>0:\n for i in list1:\n if(list1.count(i)==t):\n \n dict1[i]=t\n return dict1\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('a b b a') == {'a':2,'b': 2}, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('a b c a b') == {'a': 2, 'b': 2}, \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate('a b c d g') == {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'g': 1}, \"This prints if this assert fails 3 (good for debugging!)\"\n assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, \"This prints if this assert fails 4 (good for debugging!)\"\n assert candidate('b b b b a') == {'b': 4}, \"This prints if this assert fails 5 (good for debugging!)\"\n assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, \"This prints if this assert fails 6 (good for debugging!)\"\n \n \n # Check some edge cases that are easy to work out by hand.\n assert candidate('') == {}, \"This prints if this assert fails 7 (also good for debugging!)\"\n assert candidate('a') == {'a': 1}, \"This prints if this assert fails 8 (also good for debugging!)\"\n\n","code":"def histogram(test):"} -{"task_id":"JHumanEval\/112","prompt_en":"def reverse_delete(s,c):\n \"\"\"Task\n We are given two strings s and c, you have to deleted all the characters in s that are equal to any character in c\n then check if the result string is palindrome.\n A string is called palindrome if it reads the same backward as forward.\n You should return a tuple containing the result string and True\/False for the check.\n Example\n For s = \"abcde\", c = \"ae\", the result should be ('bcd',False)\n For s = \"abcdef\", c = \"b\" the result should be ('acdef',False)\n For s = \"abcdedcba\", c = \"ab\", the result should be ('cdedc',True)\n \"\"\"\n","prompt":"課題\n sとcの2つの文字列が与えられる。sに含まれる文字のうち、cに含まれる文字と\n 等しいものをすべて削除し、その結果の文字列が回文かどうかをチェックする。\n 文字列は、後ろから読んでも前から読んでも同じであれば回文と呼ばれる。\n 結果文字列とチェックのためのTrue\/Falseを含むタプルを返す必要がある。\n 例\n s = \"abcde\", c = \"ae\"のとき、結果は ('bcd',False)\n s = \"abcdef\", c = \"b\" のとき、結果は ('acdef',False)\n s = \"abcdedcba\", c = \"ab\", のとき、結果は ('cdedc',True)","entry_point":"reverse_delete","canonical_solution":" s = ''.join([char for char in s if char not in c])\n return (s,s[::-1] == s)\n","test":"def check(candidate):\n\n assert candidate(\"abcde\",\"ae\") == ('bcd',False)\n assert candidate(\"abcdef\", \"b\") == ('acdef',False)\n assert candidate(\"abcdedcba\",\"ab\") == ('cdedc',True)\n assert candidate(\"dwik\",\"w\") == ('dik',False)\n assert candidate(\"a\",\"a\") == ('',True)\n assert candidate(\"abcdedcba\",\"\") == ('abcdedcba',True)\n assert candidate(\"abcdedcba\",\"v\") == ('abcdedcba',True)\n assert candidate(\"vabba\",\"v\") == ('abba',True)\n assert candidate(\"mamma\", \"mia\") == (\"\", True)\n","code":"def reverse_delete(s,c):"} -{"task_id":"JHumanEval\/113","prompt_en":"def odd_count(lst):\n \"\"\"Given a list of strings, where each string consists of only digits, return a list.\n Each element i of the output should be \"the number of odd elements in the\n string i of the input.\" where all the i's should be replaced by the number\n of odd digits in the i'th string of the input.\n\n >>> odd_count(['1234567'])\n [\"the number of odd elements 4n the str4ng 4 of the 4nput.\"]\n >>> odd_count(['3',\"11111111\"])\n [\"the number of odd elements 1n the str1ng 1 of the 1nput.\",\n \"the number of odd elements 8n the str8ng 8 of the 8nput.\"]\n \"\"\"\n","prompt":"数字のみで構成された文字列のリストを引数として受け取り、新しいリストを返します。\n 出力される新しいリストの各要素は、\"the number of odd elements in the string i of the \n input.\"となりますが、この文字列内のすべての 'i' は、入力リストのi番目の文字列に含ま\n れる奇数の数に置き換えられます。\n\n >>> odd_count(['1234567'])\n [\"the number of odd elements 4n the str4ng 4 of the 4nput.\"]\n >>> odd_count(['3',\"11111111\"])\n [\"the number of odd elements 1n the str1ng 1 of the 1nput.\",\n \"the number of odd elements 8n the str8ng 8 of the 8nput.\"]","entry_point":"odd_count","canonical_solution":" res = []\n for arr in lst:\n n = sum(int(d)%2==1 for d in arr)\n res.append(\"the number of odd elements \" + str(n) + \"n the str\"+ str(n) +\"ng \"+ str(n) +\" of the \"+ str(n) +\"nput.\")\n return res\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(['1234567']) == [\"the number of odd elements 4n the str4ng 4 of the 4nput.\"], \"Test 1\"\n assert candidate(['3',\"11111111\"]) == [\"the number of odd elements 1n the str1ng 1 of the 1nput.\", \"the number of odd elements 8n the str8ng 8 of the 8nput.\"], \"Test 2\"\n assert candidate(['271', '137', '314']) == [\n 'the number of odd elements 2n the str2ng 2 of the 2nput.',\n 'the number of odd elements 3n the str3ng 3 of the 3nput.',\n 'the number of odd elements 2n the str2ng 2 of the 2nput.'\n ]\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def odd_count(lst):"} -{"task_id":"JHumanEval\/114","prompt_en":"def minSubArraySum(nums):\n \"\"\"\n Given an array of integers nums, find the minimum sum of any non-empty sub-array\n of nums.\n Example\n minSubArraySum([2, 3, 4, 1, 2, 4]) == 1\n minSubArraySum([-1, -2, -3]) == -6\n \"\"\"\n","prompt":"整数の配列 nums が与えられたとき、nums の空でない部分配列の最小和を求めよ。\n 例:\n minSubArraySum([2, 3, 4, 1, 2, 4]) == 1\n minSubArraySum([-1, -2, -3]) == -6","entry_point":"minSubArraySum","canonical_solution":" max_sum = 0\n s = 0\n for num in nums:\n s += -num\n if (s < 0):\n s = 0\n max_sum = max(s, max_sum)\n if max_sum == 0:\n max_sum = max(-i for i in nums)\n min_sum = -max_sum\n return min_sum\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([2, 3, 4, 1, 2, 4]) == 1, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([-1, -2, -3]) == -6\n assert candidate([-1, -2, -3, 2, -10]) == -14\n assert candidate([-9999999999999999]) == -9999999999999999\n assert candidate([0, 10, 20, 1000000]) == 0\n assert candidate([-1, -2, -3, 10, -5]) == -6\n assert candidate([100, -1, -2, -3, 10, -5]) == -6\n assert candidate([10, 11, 13, 8, 3, 4]) == 3\n assert candidate([100, -33, 32, -1, 0, -2]) == -33\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([-10]) == -10, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([7]) == 7\n assert candidate([1, -1]) == -1\n","code":"def minSubArraySum(nums):"} -{"task_id":"JHumanEval\/115","prompt_en":"def max_fill(grid, capacity):\n import math\n \"\"\"\n You are given a rectangular grid of wells. Each row represents a single well,\n and each 1 in a row represents a single unit of water.\n Each well has a corresponding bucket that can be used to extract water from it, \n and all buckets have the same capacity.\n Your task is to use the buckets to empty the wells.\n Output the number of times you need to lower the buckets.\n\n Example 1:\n Input: \n grid : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]\n bucket_capacity : 1\n Output: 6\n\n Example 2:\n Input: \n grid : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]\n bucket_capacity : 2\n Output: 5\n \n Example 3:\n Input: \n grid : [[0,0,0], [0,0,0]]\n bucket_capacity : 5\n Output: 0\n\n Constraints:\n * all wells have the same length\n * 1 <= grid.length <= 10^2\n * 1 <= grid[:,1].length <= 10^2\n * grid[i][j] -> 0 | 1\n * 1 <= capacity <= 10\n \"\"\"\n","prompt":"長方形のグリッド状(grid)の井戸が与えられる。各行が1つの井戸を表し、\n 行の1が1単位の水を表す。\n 各井戸には，そこから水を汲み上げるのに使える対応するバケツがあり，\n すべてのバケツ容量(capacity)は同じである．\n あなたの仕事は，バケツを使って井戸を空にすることである．\n バケツを降ろす回数を出力せよ．\n \n 例 1:\n 入力: \n グリッド : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]\n バケツ容量 : 1\n 出力: 6\n\n 例 2:\n 入力: \n グリッド : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]\n バケツ容量 : 2\n 出力: 5\n \n 例 3:\n 入力: \n グリッド : [[0,0,0], [0,0,0]]\n バケツ容量 : 5\n 出力: 0\n\n 制約:\n * すべての井戸が同じ長さ\n * 1 <= grid.length <= 10^2\n * 1 <= grid[:,1].length <= 10^2\n * grid[i][j] -> 0 | 1\n * 1 <= capacity <= 10","entry_point":"max_fill","canonical_solution":" return sum([math.ceil(sum(arr)\/capacity) for arr in grid])\n","test":"def check(candidate):\n\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([[0,0,1,0], [0,1,0,0], [1,1,1,1]], 1) == 6, \"Error\"\n assert candidate([[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]], 2) == 5, \"Error\"\n assert candidate([[0,0,0], [0,0,0]], 5) == 0, \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([[1,1,1,1], [1,1,1,1]], 2) == 4, \"Error\"\n assert candidate([[1,1,1,1], [1,1,1,1]], 9) == 2, \"Error\"\n\n","code":"def max_fill(grid, capacity):\n import math"} -{"task_id":"JHumanEval\/116","prompt_en":"def sort_array(arr):\n \"\"\"\n In this Kata, you have to sort an array of non-negative integers according to\n number of ones in their binary representation in ascending order.\n For similar number of ones, sort based on decimal value.\n\n It must be implemented like this:\n >>> sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]\n >>> sort_array([-2, -3, -4, -5, -6]) == [-6, -5, -4, -3, -2]\n >>> sort_array([1, 0, 2, 3, 4]) [0, 1, 2, 3, 4]\n \"\"\"\n","prompt":"この問題では、非負整数の配列を2進数表現における\"1\"の個数を昇順でソートする。\n \"1\"の個数が同じ場合は，10進数に基づいてソートする。\n\n 次のように実装する： \n >>> sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]\n >>> sort_array([-2, -3, -4, -5, -6]) == [-6, -5, -4, -3, -2]\n >>> sort_array([1, 0, 2, 3, 4]) [0, 1, 2, 3, 4]","entry_point":"sort_array","canonical_solution":" return sorted(sorted(arr), key=lambda x: bin(x)[2:].count('1'))\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1,5,2,3,4]) == [1, 2, 4, 3, 5]\n assert candidate([-2,-3,-4,-5,-6]) == [-4, -2, -6, -5, -3]\n assert candidate([1,0,2,3,4]) == [0, 1, 2, 4, 3]\n assert candidate([]) == []\n assert candidate([2,5,77,4,5,3,5,7,2,3,4]) == [2, 2, 4, 4, 3, 3, 5, 5, 5, 7, 77]\n assert candidate([3,6,44,12,32,5]) == [32, 3, 5, 6, 12, 44]\n assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]\n assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def sort_array(arr):"} -{"task_id":"JHumanEval\/117","prompt_en":"def select_words(s, n):\n \"\"\"Given a string s and a natural number n, you have been tasked to implement \n a function that returns a list of all words from string s that contain exactly \n n consonants, in order these words appear in the string s.\n If the string s is empty then the function should return an empty list.\n Note: you may assume the input string contains only letters and spaces.\n Examples:\n select_words(\"Mary had a little lamb\", 4) ==> [\"little\"]\n select_words(\"Mary had a little lamb\", 3) ==> [\"Mary\", \"lamb\"]\n select_words(\"simple white space\", 2) ==> []\n select_words(\"Hello world\", 4) ==> [\"world\"]\n select_words(\"Uncle sam\", 3) ==> [\"Uncle\"]\n \"\"\"\n","prompt":"ある文字列sと自然数nが与えらる。あなたに課せられたタスクは、文字列s\n の中から、ちょうどn個の子音を含むすべての単語のリストを現れる順に返す\n 関数を実装することである。\n 注意：入力文字列には英文字と空白しか含まれないと仮定してもよい。\n 例:\n select_words(\"Mary had a little lamb\", 4) ==> [\"little\"]\n select_words(\"Mary had a little lamb\", 3) ==> [\"Mary\", \"lamb\"]\n select_words(\"simple white space\", 2) ==> []\n select_words(\"Hello world\", 4) ==> [\"world\"]\n select_words(\"Uncle sam\", 3) ==> [\"Uncle\"]","entry_point":"select_words","canonical_solution":" result = []\n for word in s.split():\n n_consonants = 0\n for i in range(0, len(word)):\n if word[i].lower() not in [\"a\",\"e\",\"i\",\"o\",\"u\"]:\n n_consonants += 1 \n if n_consonants == n:\n result.append(word)\n return result\n\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"Mary had a little lamb\", 4) == [\"little\"], \"First test error: \" + str(candidate(\"Mary had a little lamb\", 4)) \n assert candidate(\"Mary had a little lamb\", 3) == [\"Mary\", \"lamb\"], \"Second test error: \" + str(candidate(\"Mary had a little lamb\", 3)) \n assert candidate(\"simple white space\", 2) == [], \"Third test error: \" + str(candidate(\"simple white space\", 2)) \n assert candidate(\"Hello world\", 4) == [\"world\"], \"Fourth test error: \" + str(candidate(\"Hello world\", 4)) \n assert candidate(\"Uncle sam\", 3) == [\"Uncle\"], \"Fifth test error: \" + str(candidate(\"Uncle sam\", 3))\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"\", 4) == [], \"1st edge test error: \" + str(candidate(\"\", 4))\n assert candidate(\"a b c d e f\", 1) == [\"b\", \"c\", \"d\", \"f\"], \"2nd edge test error: \" + str(candidate(\"a b c d e f\", 1))\n\n","code":"def select_words(s, n):"} -{"task_id":"JHumanEval\/118","prompt_en":"def get_closest_vowel(word):\n \"\"\"You are given a word. Your task is to find the closest vowel that stands between \n two consonants from the right side of the word (case sensitive).\n \n Vowels in the beginning and ending doesn't count. Return empty string if you didn't\n find any vowel met the above condition. \n\n You may assume that the given string contains English letter only.\n\n Example:\n get_closest_vowel(\"yogurt\") ==> \"u\"\n get_closest_vowel(\"FULL\") ==> \"U\"\n get_closest_vowel(\"quick\") ==> \"\"\n get_closest_vowel(\"ab\") ==> \"\"\n \"\"\"\n","prompt":"単語が与えられる。あなたの仕事は、単語の右側から2つの子音（大文字と\n 小文字を区別）の間に立っている最も近い母音を見つけることである。\n \n 最初と最後の母音はカウントされない。上記の条件を満たす母音が見つから\n なかった場合は、空の文字列を返せ。\n \n 指定された文字列は英字のみを含むとみなしてよい。\n \n 例:\n get_closest_vowel(\"yogurt\") ==> \"u\"\n get_closest_vowel(\"FULL\") ==> \"U\"\n get_closest_vowel(\"quick\") ==> \"\"\n get_closest_vowel(\"ab\") ==> \"\"","entry_point":"get_closest_vowel","canonical_solution":" if len(word) < 3:\n return \"\"\n\n vowels = {\"a\", \"e\", \"i\", \"o\", \"u\", \"A\", \"E\", 'O', 'U', 'I'}\n for i in range(len(word)-2, 0, -1):\n if word[i] in vowels:\n if (word[i+1] not in vowels) and (word[i-1] not in vowels):\n return word[i]\n return \"\"\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"yogurt\") == \"u\"\n assert candidate(\"full\") == \"u\"\n assert candidate(\"easy\") == \"\"\n assert candidate(\"eAsy\") == \"\"\n assert candidate(\"ali\") == \"\"\n assert candidate(\"bad\") == \"a\"\n assert candidate(\"most\") == \"o\"\n assert candidate(\"ab\") == \"\"\n assert candidate(\"ba\") == \"\"\n assert candidate(\"quick\") == \"\"\n assert candidate(\"anime\") == \"i\"\n assert candidate(\"Asia\") == \"\"\n assert candidate(\"Above\") == \"o\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def get_closest_vowel(word):"} -{"task_id":"JHumanEval\/119","prompt_en":"def match_parens(lst):\n '''\n You are given a list of two strings, both strings consist of open\n parentheses '(' or close parentheses ')' only.\n Your job is to check if it is possible to concatenate the two strings in\n some order, that the resulting string will be good.\n A string S is considered to be good if and only if all parentheses in S\n are balanced. For example: the string '(())()' is good, while the string\n '())' is not.\n Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.\n\n Examples:\n match_parens(['()(', ')']) == 'Yes'\n match_parens([')', ')']) == 'No'\n '''\n","prompt":"2つの文字列からなるリストが与えられます。両方の文字列は開き括弧 '(' または\n 閉じ括弧 ')' のみで構成されています。\n あなたの仕事は、2つの文字列を何らかの順序で結合して、「良い」文字列にすることが\n 可能かどうかを確認することです。\n \n 文字列Sが「良い」とは、文字列内のすべての括弧がバランスしている場合に限ります。\n 例えば、文字列 '(())()' は良いですが、文字列 '())' は良くありません。\n 良い文字列を作る方法がある場合は 'Yes' を返し、そうでない場合は 'No' を返してください。\n \n 例\n match_parens(['()(', ')']) == 'Yes'\n match_parens([')', ')']) == 'No'","entry_point":"match_parens","canonical_solution":" def check(s):\n val = 0\n for i in s:\n if i == '(':\n val = val + 1\n else:\n val = val - 1\n if val < 0:\n return False\n return True if val == 0 else False\n\n S1 = lst[0] + lst[1]\n S2 = lst[1] + lst[0]\n return 'Yes' if check(S1) or check(S2) else 'No'\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(['()(', ')']) == 'Yes'\n assert candidate([')', ')']) == 'No'\n assert candidate(['(()(())', '())())']) == 'No'\n assert candidate([')())', '(()()(']) == 'Yes'\n assert candidate(['(())))', '(()())((']) == 'Yes'\n assert candidate(['()', '())']) == 'No'\n assert candidate(['(()(', '()))()']) == 'Yes'\n assert candidate(['((((', '((())']) == 'No'\n assert candidate([')(()', '(()(']) == 'No'\n assert candidate([')(', ')(']) == 'No'\n \n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(['(', ')']) == 'Yes'\n assert candidate([')', '(']) == 'Yes' \n\n","code":"def match_parens(lst):"} -{"task_id":"JHumanEval\/120","prompt_en":"def maximum(arr, k):\n \"\"\"\n Given an array arr of integers and a positive integer k, return a sorted list \n of length k with the maximum k numbers in arr.\n\n Example 1:\n\n Input: arr = [-3, -4, 5], k = 3\n Output: [-4, -3, 5]\n\n Example 2:\n\n Input: arr = [4, -4, 4], k = 2\n Output: [4, 4]\n\n Example 3:\n\n Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1\n Output: [2]\n\n Note:\n 1. The length of the array will be in the range of [1, 1000].\n 2. The elements in the array will be in the range of [-1000, 1000].\n 3. 0 <= k <= len(arr)\n \"\"\"\n","prompt":"整数の配列 arr と正の整数 k が与えられる。arr に含まれる大きい方から k 個の数を含む\n 長さ k のソート済みリストを返す。\n\n 例 1:\n\n 入力: arr = [-3, -4, 5], k = 3\n 出力: [-4, -3, 5]\n\n 例 2:\n\n 入力: arr = [4, -4, 4], k = 2\n 出力: [4, 4]\n\n 例 3:\n\n 入力: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1\n 出力: [2]\n\n ノート:\n 1. 配列の長さは[1, 1000]の範囲とする。\n 2. 配列の要素は [-1000, 1000] の範囲にある。\n 3. 0 <= k <= len(arr)","entry_point":"maximum","canonical_solution":" if k == 0:\n return []\n arr.sort()\n ans = arr[-k:]\n return ans\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([-3, -4, 5], 3) == [-4, -3, 5]\n assert candidate([4, -4, 4], 2) == [4, 4]\n assert candidate([-3, 2, 1, 2, -1, -2, 1], 1) == [2]\n assert candidate([123, -123, 20, 0 , 1, 2, -3], 3) == [2, 20, 123]\n assert candidate([-123, 20, 0 , 1, 2, -3], 4) == [0, 1, 2, 20]\n assert candidate([5, 15, 0, 3, -13, -8, 0], 7) == [-13, -8, 0, 0, 3, 5, 15]\n assert candidate([-1, 0, 2, 5, 3, -10], 2) == [3, 5]\n assert candidate([1, 0, 5, -7], 1) == [5]\n assert candidate([4, -4], 2) == [-4, 4]\n assert candidate([-10, 10], 2) == [-10, 10]\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([1, 2, 3, -23, 243, -400, 0], 0) == []\n\n","code":"def maximum(arr, k):"} -{"task_id":"JHumanEval\/121","prompt_en":"def solution(lst):\n \"\"\"Given a non-empty list of integers, return the sum of all of the odd elements that are in even positions.\n \n\n Examples\n solution([5, 8, 7, 1]) ==> 12\n solution([3, 3, 3, 3, 3]) ==> 9\n solution([30, 13, 24, 321]) ==>0\n \"\"\"\n","prompt":"整数の空でないリストが与えられた時、偶数の位置にある奇数の要素の合計を返す。\n 例\n solution([5, 8, 7, 1]) ==> 12\n solution([3, 3, 3, 3, 3]) ==> 9\n solution([30, 13, 24, 321]) ==>0","entry_point":"solution","canonical_solution":" return sum([x for idx, x in enumerate(lst) if idx%2==0 and x%2==1])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([5, 8, 7, 1]) == 12\n assert candidate([3, 3, 3, 3, 3]) == 9\n assert candidate([30, 13, 24, 321]) == 0\n assert candidate([5, 9]) == 5\n assert candidate([2, 4, 8]) == 0\n assert candidate([30, 13, 23, 32]) == 23\n assert candidate([3, 13, 2, 9]) == 3\n\n # Check some edge cases that are easy to work out by hand.\n\n","code":"def solution(lst):"} -{"task_id":"JHumanEval\/122","prompt_en":"def add_elements(arr, k):\n \"\"\"\n Given a non-empty array of integers arr and an integer k, return\n the sum of the elements with at most two digits from the first k elements of arr.\n\n Example:\n\n Input: arr = [111,21,3,4000,5,6,7,8,9], k = 4\n Output: 24 # sum of 21 + 3\n\n Constraints:\n 1. 1 <= len(arr) <= 100\n 2. 1 <= k <= len(arr)\n \"\"\"\n","prompt":"整数の空でない配列 arr と整数 k が与えられたとき、\n arr の最初の k 個の要素から高々 2 桁までの要素の和を返す。\n\n 例:\n\n 入力: arr = [111,21,3,4000,5,6,7,8,9], k = 4\n 出力: 24 # 21 + 3 の話\n\n 制約:\n 1. 1 <= len(arr) <= 100\n 2. 1 <= k <= len(arr)","entry_point":"add_elements","canonical_solution":" return sum(elem for elem in arr[:k] if len(str(elem)) <= 2)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1,-2,-3,41,57,76,87,88,99], 3) == -4\n assert candidate([111,121,3,4000,5,6], 2) == 0\n assert candidate([11,21,3,90,5,6,7,8,9], 4) == 125\n assert candidate([111,21,3,4000,5,6,7,8,9], 4) == 24, \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([1], 1) == 1, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def add_elements(arr, k):"} -{"task_id":"JHumanEval\/123","prompt_en":"def get_odd_collatz(n):\n \"\"\"\n Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.\n\n The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined\n as follows: start with any positive integer n. Then each term is obtained from the \n previous term as follows: if the previous term is even, the next term is one half of \n the previous term. If the previous term is odd, the next term is 3 times the previous\n term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.\n\n Note: \n 1. Collatz(1) is [1].\n 2. returned list sorted in increasing order.\n\n For example:\n get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.\n \"\"\"\n","prompt":"正の整数nが与えられたとき、コラッツ数列の奇数を持つソートされたリストを返す。\n コラッツ予想とは数学の予想で、次のように定義される数列に関するものである： \n 任意の正の整数nから始め、各項は前の項から次のように求められる。\n 前の項が偶数なら、次の項は前の項の2分の1である。前の項が奇数の場合、次の項は前の項の3倍＋1である。\n 予想では、nがどのような値であっても、数列は必ず1に達する。\n\n 注：\n       1. Collatz(1)は[1]である。\n 2. 返されるリストは昇順にソートされている。\n \n 例えば:\n get_odd_collatz(5) は [1, 5]を返す。つまり、5に対するコラッツ数列 は、[5, 16, 8, 4, 2, 1]であり、 奇数は 1 と 5 である。","entry_point":"get_odd_collatz","canonical_solution":" if n%2==0:\n odd_collatz = [] \n else:\n odd_collatz = [n]\n while n > 1:\n if n % 2 == 0:\n n = n\/2\n else:\n n = n*3 + 1\n \n if n%2 == 1:\n odd_collatz.append(int(n))\n\n return sorted(odd_collatz)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(14) == [1, 5, 7, 11, 13, 17]\n assert candidate(5) == [1, 5]\n assert candidate(12) == [1, 3, 5], \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1) == [1], \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def get_odd_collatz(n):"} -{"task_id":"JHumanEval\/124","prompt_en":"def valid_date(date):\n \"\"\"You have to write a function which validates a given date string and\n returns True if the date is valid otherwise False.\n The date is valid if all of the following rules are satisfied:\n 1. The date string is not empty.\n 2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.\n 3. The months should not be less than 1 or higher than 12.\n 4. The date should be in the format: mm-dd-yyyy\n\n for example: \n valid_date('03-11-2000') => True\n\n valid_date('15-01-2012') => False\n\n valid_date('04-0-2040') => False\n\n valid_date('06-04-2020') => True\n\n valid_date('06\/04\/2020') => False\n \"\"\"\n","prompt":"与えられた日付文字列を検証し、その日付が有効であればTrueを、そうでなければFalseを返す関数を書く必要がある。\n 日付が有効であるのは、以下のルールがすべて満たされている場合である：\n 1. 日付文字列が空でない。\n 2. 日数が、月1,3,5,7,8,10,12の場合、1日以上31日以下である。また、月4,6,9,11については、日数が1以上30日以下である。また、月2については、日数が1以上29以下であること。\n 3. 月は1未満または12以上であってはならない。\n 4. 日付はmm-dd-yyyyの形式でなければならない。\n\n 例えば: \n valid_date('03-11-2000') => True\n\n valid_date('15-01-2012') => False\n\n valid_date('04-0-2040') => False\n\n valid_date('06-04-2020') => True\n\n valid_date('06\/04\/2020') => False","entry_point":"valid_date","canonical_solution":" try:\n date = date.strip()\n month, day, year = date.split('-')\n month, day, year = int(month), int(day), int(year)\n if month < 1 or month > 12:\n return False\n if month in [1,3,5,7,8,10,12] and day < 1 or day > 31:\n return False\n if month in [4,6,9,11] and day < 1 or day > 30:\n return False\n if month == 2 and day < 1 or day > 29:\n return False\n except:\n return False\n\n return True\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('03-11-2000') == True\n\n assert candidate('15-01-2012') == False\n\n assert candidate('04-0-2040') == False\n\n assert candidate('06-04-2020') == True\n\n assert candidate('01-01-2007') == True\n\n assert candidate('03-32-2011') == False\n\n assert candidate('') == False\n\n assert candidate('04-31-3000') == False\n\n assert candidate('06-06-2005') == True\n\n assert candidate('21-31-2000') == False\n\n assert candidate('04-12-2003') == True\n\n assert candidate('04122003') == False\n\n assert candidate('20030412') == False\n\n assert candidate('2003-04') == False\n\n assert candidate('2003-04-12') == False\n\n assert candidate('04-2003') == False\n","code":"def valid_date(date):"} -{"task_id":"JHumanEval\/125","prompt_en":"def split_words(txt):\n '''\n Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you\n should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the\n alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25\n Examples\n split_words(\"Hello world!\") ➞ [\"Hello\", \"world!\"]\n split_words(\"Hello,world!\") ➞ [\"Hello\", \"world!\"]\n split_words(\"abcdef\") == 3 \n '''\n","prompt":"単語の文字列が与えられた場合、空白で分割された単語のリストを返す。\n テキスト中に空白が存在しない場合は、カンマ ',' で分割する必要がある。カンマが存在しない場合は、\n アルファベットの奇数順の小文字の数を返す必要がある。ord('a') = 0, ord('b') = 1, ... ord('z') = 25\n 例\n split_words(\"Hello world!\") ➞ [\"Hello\", \"world!\"]\n split_words(\"Hello,world!\") ➞ [\"Hello\", \"world!\"]\n split_words(\"abcdef\") == 3","entry_point":"split_words","canonical_solution":" if \" \" in txt:\n return txt.split()\n elif \",\" in txt:\n return txt.replace(',',' ').split()\n else:\n return len([i for i in txt if i.islower() and ord(i)%2 == 0])\n","test":"def check(candidate):\n\n assert candidate(\"Hello world!\") == [\"Hello\",\"world!\"]\n assert candidate(\"Hello,world!\") == [\"Hello\",\"world!\"]\n assert candidate(\"Hello world,!\") == [\"Hello\",\"world,!\"]\n assert candidate(\"Hello,Hello,world !\") == [\"Hello,Hello,world\",\"!\"]\n assert candidate(\"abcdef\") == 3\n assert candidate(\"aaabb\") == 2\n assert candidate(\"aaaBb\") == 1\n assert candidate(\"\") == 0\n","code":"def split_words(txt):"} -{"task_id":"JHumanEval\/126","prompt_en":"def is_sorted(lst):\n '''\n Given a list of numbers, return whether or not they are sorted\n in ascending order. If list has more than 1 duplicate of the same\n number, return False. Assume no negative numbers and only integers.\n\n Examples\n is_sorted([5]) ➞ True\n is_sorted([1, 2, 3, 4, 5]) ➞ True\n is_sorted([1, 3, 2, 4, 5]) ➞ False\n is_sorted([1, 2, 3, 4, 5, 6]) ➞ True\n is_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True\n is_sorted([1, 3, 2, 4, 5, 6, 7]) ➞ False\n is_sorted([1, 2, 2, 3, 3, 4]) ➞ True\n is_sorted([1, 2, 2, 2, 3, 4]) ➞ False\n '''\n","prompt":"数字のリストが与えられたとき、昇順に整列されているかどうかを返す。\n リストに同じ数の重複が1つ以上ある場合は、Falseを返す。\n 負の数はなく、整数のみであると仮定する。\n \n 例\n is_sorted([5]) ➞ True\n is_sorted([1, 2, 3, 4, 5]) ➞ True\n is_sorted([1, 3, 2, 4, 5]) ➞ False\n is_sorted([1, 2, 3, 4, 5, 6]) ➞ True\n is_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True\n is_sorted([1, 3, 2, 4, 5, 6, 7]) ➞ False\n is_sorted([1, 2, 2, 3, 3, 4]) ➞ True\n is_sorted([1, 2, 2, 2, 3, 4]) ➞ False","entry_point":"is_sorted","canonical_solution":" count_digit = dict([(i, 0) for i in lst])\n for i in lst:\n count_digit[i]+=1 \n if any(count_digit[i] > 2 for i in lst):\n return False\n if all(lst[i-1] <= lst[i] for i in range(1, len(lst))):\n return True\n else:\n return False\n \n \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([5]) == True\n assert candidate([1, 2, 3, 4, 5]) == True\n assert candidate([1, 3, 2, 4, 5]) == False\n assert candidate([1, 2, 3, 4, 5, 6]) == True\n assert candidate([1, 2, 3, 4, 5, 6, 7]) == True\n assert candidate([1, 3, 2, 4, 5, 6, 7]) == False, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([]) == True, \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate([1]) == True, \"This prints if this assert fails 3 (good for debugging!)\"\n assert candidate([3, 2, 1]) == False, \"This prints if this assert fails 4 (good for debugging!)\"\n \n # Check some edge cases that are easy to work out by hand.\n assert candidate([1, 2, 2, 2, 3, 4]) == False, \"This prints if this assert fails 5 (good for debugging!)\"\n assert candidate([1, 2, 3, 3, 3, 4]) == False, \"This prints if this assert fails 6 (good for debugging!)\"\n assert candidate([1, 2, 2, 3, 3, 4]) == True, \"This prints if this assert fails 7 (good for debugging!)\"\n assert candidate([1, 2, 3, 4]) == True, \"This prints if this assert fails 8 (good for debugging!)\"\n\n","code":"def is_sorted(lst):"} -{"task_id":"JHumanEval\/127","prompt_en":"def intersection(interval1, interval2):\n \"\"\"You are given two intervals,\n where each interval is a pair of integers. For example, interval = (start, end) = (1, 2).\n The given intervals are closed which means that the interval (start, end)\n includes both start and end.\n For each given interval, it is assumed that its start is less or equal its end.\n Your task is to determine whether the length of intersection of these two \n intervals is a prime number.\n Example, the intersection of the intervals (1, 3), (2, 4) is (2, 3)\n which its length is 1, which not a prime number.\n If the length of the intersection is a prime number, return \"YES\",\n otherwise, return \"NO\".\n If the two intervals don't intersect, return \"NO\".\n\n\n [input\/output] samples:\n intersection((1, 2), (2, 3)) ==> \"NO\"\n intersection((-1, 1), (0, 4)) ==> \"NO\"\n intersection((-3, -1), (-5, 5)) ==> \"YES\"\n \"\"\"\n","prompt":"2つの区間が与えられます。\n それぞれの区間は整数のペアで示されます。例えば、区間 = (start, end ) = (1, 2) です。\n 与えられた区間は閉区間であり、start と end の両端が含まれます。\n 各区間について、start は end 以下であると仮定します。\n \n あなたの仕事は、これら2つの区間の交差部分の長さが素数であるかどうかを判断することです。\n \n 例えば、区間 (1, 3) と (2, 4) の交差部分は (2, 3) で、その長さは1ですが、これは素数ではありません。\n \n 交差部分の長さが素数であれば \"YES\" を返し、そうでなければ \"NO\" を返してください。\n もし2つの区間が交差しない場合も \"NO\" を返してください。\n\n [input\/output] サンプル:\n intersection((1, 2), (2, 3)) ==> \"NO\"\n intersection((-1, 1), (0, 4)) ==> \"NO\"\n intersection((-3, -1), (-5, 5)) ==> \"YES\"","entry_point":"intersection","canonical_solution":" def is_prime(num):\n if num == 1 or num == 0:\n return False\n if num == 2:\n return True\n for i in range(2, num):\n if num%i == 0:\n return False\n return True\n\n l = max(interval1[0], interval2[0])\n r = min(interval1[1], interval2[1])\n length = r - l\n if length > 0 and is_prime(length):\n return \"YES\"\n return \"NO\"\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate((1, 2), (2, 3)) == \"NO\"\n assert candidate((-1, 1), (0, 4)) == \"NO\"\n assert candidate((-3, -1), (-5, 5)) == \"YES\"\n assert candidate((-2, 2), (-4, 0)) == \"YES\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate((-11, 2), (-1, -1)) == \"NO\"\n assert candidate((1, 2), (3, 5)) == \"NO\"\n assert candidate((1, 2), (1, 2)) == \"NO\"\n assert candidate((-2, -2), (-3, -2)) == \"NO\"\n\n","code":"def intersection(interval1, interval2):"} -{"task_id":"JHumanEval\/128","prompt_en":"def prod_signs(arr):\n \"\"\"\n You are given an array arr of integers and you need to return\n sum of magnitudes of integers multiplied by product of all signs\n of each number in the array, represented by 1, -1 or 0.\n Note: return None for empty arr.\n\n Example:\n >>> prod_signs([1, 2, 2, -4]) == -9\n >>> prod_signs([0, 1]) == 0\n >>> prod_signs([]) == None\n \"\"\"\n","prompt":"整数の配列 arr が与えられます。この配列に含まれる各数値の絶対値の合計と、\n 各数値の符号（プラスは1、マイナスは-1、ゼロは0）の積を掛け合わせた値を\n 返してください。\n 注意：配列`arr`が空の場合は`None`を返してください。\n \n 例:\n >>> prod_signs([1, 2, 2, -4]) == -9\n >>> prod_signs([0, 1]) == 0\n >>> prod_signs([]) == None","entry_point":"prod_signs","canonical_solution":" if not arr: return None\n prod = 0 if 0 in arr else (-1) ** len(list(filter(lambda x: x < 0, arr)))\n return prod * sum([abs(i) for i in arr])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1, 2, 2, -4]) == -9\n assert candidate([0, 1]) == 0\n assert candidate([1, 1, 1, 2, 3, -1, 1]) == -10\n assert candidate([]) == None\n assert candidate([2, 4,1, 2, -1, -1, 9]) == 20\n assert candidate([-1, 1, -1, 1]) == 4\n assert candidate([-1, 1, 1, 1]) == -4\n assert candidate([-1, 1, 1, 0]) == 0\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def prod_signs(arr):"} -{"task_id":"JHumanEval\/129","prompt_en":"def minPath(grid, k):\n \"\"\"\n Given a grid with N rows and N columns (N >= 2) and a positive integer k, \n each cell of the grid contains a value. Every integer in the range [1, N * N]\n inclusive appears exactly once on the cells of the grid.\n\n You have to find the minimum path of length k in the grid. You can start\n from any cell, and in each step you can move to any of the neighbor cells,\n in other words, you can go to cells which share an edge with you current\n cell.\n Please note that a path of length k means visiting exactly k cells (not\n necessarily distinct).\n You CANNOT go off the grid.\n A path A (of length k) is considered less than a path B (of length k) if\n after making the ordered lists of the values on the cells that A and B go\n through (let's call them lst_A and lst_B), lst_A is lexicographically less\n than lst_B, in other words, there exist an integer index i (1 <= i <= k)\n such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have\n lst_A[j] = lst_B[j].\n It is guaranteed that the answer is unique.\n Return an ordered list of the values on the cells that the minimum path go through.\n\n Examples:\n\n Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3\n Output: [1, 2, 1]\n\n Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1\n Output: [1]\n \"\"\"\n","prompt":"N行とN列 (N >= 2)) のグリッドと正の整数kが与えられた場合、各セルには値が含まれている。\n 範囲[1, N * N]（両端を含む）のすべての整数は、グリッドのセルに一度だけ表れる。\n\n このグリッド内で長さkの最短の経路を見つける必要がある。任意のセルからスタートでき、\n 各ステップで隣接するセルに移動できる。言い換えれば、現在のセルと辺を共有するセルに\n 移動できる。長さkの経路とは、正確にk個のセル（必ずしも異なるとは限らない）を訪れる\n ことを意味する。ただし、グリッドから出ることはない。\n\n 長さkの2つの経路AとBがある場合、AとBが通るセルの値を順番にリスト化したものを\n それぞれlst_A、lst_Bと呼ぶ。lst_Aがlst_Bより辞書順で小さい場合、経路Aは経路Bよりも\n 小さいとする。つまり、整数インデックスi（ 1 <= i <= k ) が存在して、lst_A[i] < lst_B[i] となり、\n 任意の j（ 1 <= j < i ）に対して lst_A[j] = lst_B[j] が成立する。\n\n 答えは一意であることが保証されている。\n 最短の経路が通るセルの値の順番に並べたリストを返すようにせよ。\n \n 例：\n \n 入力：grid =[ [1,2,3], [4,5,6], [7,8,9]]. k = 3\n 出力：[1, 2, 1]\n \n 入力：grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1\n 出力：[1]","entry_point":"minPath","canonical_solution":" n = len(grid)\n val = n * n + 1\n for i in range(n):\n for j in range(n):\n if grid[i][j] == 1:\n temp = []\n if i != 0:\n temp.append(grid[i - 1][j])\n\n if j != 0:\n temp.append(grid[i][j - 1])\n\n if i != n - 1:\n temp.append(grid[i + 1][j])\n\n if j != n - 1:\n temp.append(grid[i][j + 1])\n\n val = min(temp)\n\n ans = []\n for i in range(k):\n if i % 2 == 0:\n ans.append(1)\n else:\n ans.append(val)\n return ans\n","test":"def check(candidate):\n\n # Check some simple cases\n print\n assert candidate([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3) == [1, 2, 1]\n assert candidate([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1) == [1]\n assert candidate([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]], 4) == [1, 2, 1, 2]\n assert candidate([[6, 4, 13, 10], [5, 7, 12, 1], [3, 16, 11, 15], [8, 14, 9, 2]], 7) == [1, 10, 1, 10, 1, 10, 1]\n assert candidate([[8, 14, 9, 2], [6, 4, 13, 15], [5, 7, 1, 12], [3, 10, 11, 16]], 5) == [1, 7, 1, 7, 1]\n assert candidate([[11, 8, 7, 2], [5, 16, 14, 4], [9, 3, 15, 6], [12, 13, 10, 1]], 9) == [1, 6, 1, 6, 1, 6, 1, 6, 1]\n assert candidate([[12, 13, 10, 1], [9, 3, 15, 6], [5, 16, 14, 4], [11, 8, 7, 2]], 12) == [1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6]\n assert candidate([[2, 7, 4], [3, 1, 5], [6, 8, 9]], 8) == [1, 3, 1, 3, 1, 3, 1, 3]\n assert candidate([[6, 1, 5], [3, 8, 9], [2, 7, 4]], 8) == [1, 5, 1, 5, 1, 5, 1, 5]\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([[1, 2], [3, 4]], 10) == [1, 2, 1, 2, 1, 2, 1, 2, 1, 2]\n assert candidate([[1, 3], [3, 2]], 10) == [1, 3, 1, 3, 1, 3, 1, 3, 1, 3]\n\n","code":"def minPath(grid, k):"} -{"task_id":"JHumanEval\/130","prompt_en":"def tri(n):\n \"\"\"Everyone knows Fibonacci sequence, it was studied deeply by mathematicians in \n the last couple centuries. However, what people don't know is Tribonacci sequence.\n Tribonacci sequence is defined by the recurrence:\n tri(1) = 3\n tri(n) = 1 + n \/ 2, if n is even.\n tri(n) = tri(n - 1) + tri(n - 2) + tri(n + 1), if n is odd.\n For example:\n tri(2) = 1 + (2 \/ 2) = 2\n tri(4) = 3\n tri(3) = tri(2) + tri(1) + tri(4)\n = 2 + 3 + 3 = 8 \n You are given a non-negative integer number n, you have to a return a list of the \n first n + 1 numbers of the Tribonacci sequence.\n Examples:\n tri(3) = [1, 3, 2, 8]\n \"\"\"\n","prompt":"フィボナッチ数列は、ここ数世紀の間に数学者によって深く研究され、誰もが知っている。\n しかし、人々が知らないのはトリボナッチ数列である。\n トリボナッチ数列は再帰によって定義される:\n tri(1) = 3\n tri(n) = 1 + n \/ 2, n が偶数の場合.\n tri(n) = tri(n - 1) + tri(n - 2) + tri(n + 1), n が奇数の場合d.\n 例えば:\n tri(2) = 1 + (2 \/ 2) = 2\n tri(4) = 3\n tri(3) = tri(2) + tri(1) + tri(4)\n = 2 + 3 + 3 = 8 \n あなたは非負の整数nが与えられるので、トリボナッチ数列の最初のn + 1個の数の\n リストを返さなければならない。\n 例:\n tri(3) = [1, 3, 2, 8]","entry_point":"tri","canonical_solution":" if n == 0:\n return [1]\n my_tri = [1, 3]\n for i in range(2, n + 1):\n if i % 2 == 0:\n my_tri.append(i \/ 2 + 1)\n else:\n my_tri.append(my_tri[i - 1] + my_tri[i - 2] + (i + 3) \/ 2)\n return my_tri\n","test":"def check(candidate):\n\n # Check some simple cases\n \n assert candidate(3) == [1, 3, 2.0, 8.0]\n assert candidate(4) == [1, 3, 2.0, 8.0, 3.0]\n assert candidate(5) == [1, 3, 2.0, 8.0, 3.0, 15.0]\n assert candidate(6) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0]\n assert candidate(7) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0]\n assert candidate(8) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0]\n assert candidate(9) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0, 35.0]\n assert candidate(20) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0, 35.0, 6.0, 48.0, 7.0, 63.0, 8.0, 80.0, 9.0, 99.0, 10.0, 120.0, 11.0]\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(0) == [1]\n assert candidate(1) == [1, 3]\n","code":"def tri(n):"} -{"task_id":"JHumanEval\/131","prompt_en":"def digits(n):\n \"\"\"Given a positive integer n, return the product of the odd digits.\n Return 0 if all digits are even.\n For example:\n digits(1) == 1\n digits(4) == 0\n digits(235) == 15\n \"\"\"\n","prompt":"正の整数 n が与えられた時、奇数桁数の積を返す。\n 全ての桁が偶数の場合は0を返す。\n 例えば:\n digits(1) == 1\n digits(4) == 0\n digits(235) == 15","entry_point":"digits","canonical_solution":" product = 1\n odd_count = 0\n for digit in str(n):\n int_digit = int(digit)\n if int_digit%2 == 1:\n product= product*int_digit\n odd_count+=1\n if odd_count ==0:\n return 0\n else:\n return product\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(5) == 5\n assert candidate(54) == 5\n assert candidate(120) ==1\n assert candidate(5014) == 5\n assert candidate(98765) == 315\n assert candidate(5576543) == 2625\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(2468) == 0\n\n","code":"def digits(n):"} -{"task_id":"JHumanEval\/132","prompt_en":"def is_nested(string):\n '''\n Create a function that takes a string as input which contains only square brackets.\n The function should return True if and only if there is a valid subsequence of brackets \n where at least one bracket in the subsequence is nested.\n\n is_nested('[[]]') ➞ True\n is_nested('[]]]]]]][[[[[]') ➞ False\n is_nested('[][]') ➞ False\n is_nested('[]') ➞ False\n is_nested('[[][]]') ➞ True\n is_nested('[[]][[') ➞ True\n '''\n","prompt":"この関数は、角括弧だけを含む文字列を入力として受け取ります。括弧が有効な順序で\n 並んでいて、その中に少なくとも1つの括弧が入れ子になっている場合、関数はTrueを\n 返すようにしてください。\n\n is_nested('[[]]') ➞ True\n is_nested('[]]]]]]][[[[[]') ➞ False\n is_nested('[][]') ➞ False\n is_nested('[]') ➞ False\n is_nested('[[][]]') ➞ True\n is_nested('[[]][[') ➞ True","entry_point":"is_nested","canonical_solution":" opening_bracket_index = []\n closing_bracket_index = []\n for i in range(len(string)):\n if string[i] == '[':\n opening_bracket_index.append(i)\n else:\n closing_bracket_index.append(i)\n closing_bracket_index.reverse()\n cnt = 0\n i = 0\n l = len(closing_bracket_index)\n for idx in opening_bracket_index:\n if i < l and idx < closing_bracket_index[i]:\n cnt += 1\n i += 1\n return cnt >= 2\n\n \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('[[]]') == True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate('[]]]]]]][[[[[]') == False\n assert candidate('[][]') == False\n assert candidate(('[]')) == False\n assert candidate('[[[[]]]]') == True\n assert candidate('[]]]]]]]]]]') == False\n assert candidate('[][][[]]') == True\n assert candidate('[[]') == False\n assert candidate('[]]') == False\n assert candidate('[[]][[') == True\n assert candidate('[[][]]') == True\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate('') == False, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate('[[[[[[[[') == False\n assert candidate(']]]]]]]]') == False\n\n","code":"def is_nested(string):"} -{"task_id":"JHumanEval\/133","prompt_en":"def sum_squares(lst):\n \"\"\"You are given a list of numbers.\n You need to return the sum of squared numbers in the given list,\n round each element in the list to the upper int(Ceiling) first.\n Examples:\n For lst = [1,2,3] the output should be 14\n For lst = [1,4,9] the output should be 98\n For lst = [1,3,5,7] the output should be 84\n For lst = [1.4,4.2,0] the output should be 29\n For lst = [-2.4,1,1] the output should be 6\n \n\n \"\"\"\n","prompt":"数字のリストが与えられます。\n 与えられたリスト内の各数値をまず切り上げ（天井関数を使って最も近い整数に丸める）、\n その後それぞれの数値を二乗した値の合計を返してください。\n 例:\n lst = [1,2,3] のときの出力は 14\n lst = [1,4,9] のときの出力は 98\n lst = [1,3,5,7] のときの出力は 84\n lst = [1.4,4.2,0] のときの出力は 29\n lst = [-2.4,1,1] のときの出力は 6","entry_point":"sum_squares","canonical_solution":" import math\n squared = 0\n for i in lst:\n squared += math.ceil(i)**2\n return squared\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1,2,3])==14, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1.0,2,3])==14, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1,3,5,7])==84, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1.4,4.2,0])==29, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([-2.4,1,1])==6, \"This prints if this assert fails 1 (good for debugging!)\"\n\n assert candidate([100,1,15,2])==10230, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([10000,10000])==200000000, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([-1.4,4.6,6.3])==75, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([-1.4,17.9,18.9,19.9])==1086, \"This prints if this assert fails 1 (good for debugging!)\"\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([0])==0, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([-1])==1, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate([-1,1,0])==2, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def sum_squares(lst):"} -{"task_id":"JHumanEval\/134","prompt_en":"def check_if_last_char_is_a_letter(txt):\n '''\n Create a function that returns True if the last character\n of a given string is an alphabetical character and is not\n a part of a word, and False otherwise.\n Note: \"word\" is a group of characters separated by space.\n\n Examples:\n check_if_last_char_is_a_letter(\"apple pie\") ➞ False\n check_if_last_char_is_a_letter(\"apple pi e\") ➞ True\n check_if_last_char_is_a_letter(\"apple pi e \") ➞ False\n check_if_last_char_is_a_letter(\"\") ➞ False \n '''\n","prompt":"与えられた文字列の最後の文字がアルファベットであり、かつ単語の一部でなければTrueを、\n そうでなければFalseを返す関数を作成せよ。\n 注意：単語とはスペースで区切られた文字の並びである。\n 例:\n check_if_last_char_is_a_letter(\"apple pie\") ➞ False\n check_if_last_char_is_a_letter(\"apple pi e\") ➞ True\n check_if_last_char_is_a_letter(\"apple pi e \") ➞ False\n check_if_last_char_is_a_letter(\"\") ➞ False","entry_point":"check_if_last_char_is_a_letter","canonical_solution":" \n check = txt.split(' ')[-1]\n return True if len(check) == 1 and (97 <= ord(check.lower()) <= 122) else False\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"apple\") == False\n assert candidate(\"apple pi e\") == True\n assert candidate(\"eeeee\") == False\n assert candidate(\"A\") == True\n assert candidate(\"Pumpkin pie \") == False\n assert candidate(\"Pumpkin pie 1\") == False\n assert candidate(\"\") == False\n assert candidate(\"eeeee e \") == False\n assert candidate(\"apple pie\") == False\n assert candidate(\"apple pi e \") == False\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def check_if_last_char_is_a_letter(txt):"} -{"task_id":"JHumanEval\/135","prompt_en":"def can_arrange(arr):\n \"\"\"Create a function which returns the largest index of an element which\n is not greater than or equal to the element immediately preceding it. If\n no such element exists then return -1. The given array will not contain\n duplicate values.\n\n Examples:\n can_arrange([1,2,4,3,5]) = 3\n can_arrange([1,2,3]) = -1\n \"\"\"\n","prompt":"直前の要素よりも大きくない要素の中で、最も大きなインデックスを持つ要素を探して\n そのインデックスを返す関数を作成してください。そのような要素が存在しない場合は、\n -1を返してください。与えられる配列には重複する値は含まれません。\n 例:\n can_arrange([1,2,4,3,5]) = 3\n can_arrange([1,2,3]) = -1","entry_point":"can_arrange","canonical_solution":" ind=-1\n i=1\n while i 0, lst))\n return (max(smallest) if smallest else None, min(largest) if largest else None)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([2, 4, 1, 3, 5, 7]) == (None, 1)\n assert candidate([2, 4, 1, 3, 5, 7, 0]) == (None, 1)\n assert candidate([1, 3, 2, 4, 5, 6, -2]) == (-2, 1)\n assert candidate([4, 5, 3, 6, 2, 7, -7]) == (-7, 2)\n assert candidate([7, 3, 8, 4, 9, 2, 5, -9]) == (-9, 2)\n assert candidate([]) == (None, None)\n assert candidate([0]) == (None, None)\n assert candidate([-1, -3, -5, -6]) == (-1, None)\n assert candidate([-1, -3, -5, -6, 0]) == (-1, None)\n assert candidate([-6, -4, -4, -3, 1]) == (-3, 1)\n assert candidate([-6, -4, -4, -3, -100, 1]) == (-3, 1)\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n","code":"def largest_smallest_integers(lst):"} -{"task_id":"JHumanEval\/137","prompt_en":"def compare_one(a, b):\n \"\"\"\n Create a function that takes integers, floats, or strings representing\n real numbers, and returns the larger variable in its given variable type.\n Return None if the values are equal.\n Note: If a real number is represented as a string, the floating point might be . or ,\n\n compare_one(1, 2.5) ➞ 2.5\n compare_one(1, \"2,3\") ➞ \"2,3\"\n compare_one(\"5,1\", \"6\") ➞ \"6\"\n compare_one(\"1\", 1) ➞ None\n \"\"\"\n","prompt":"整数、浮動小数点数、または実数を表す文字列を引数として受け取り、その中で最も大きい値を\n その元の型で返す関数を作成してください。もし値が同じであれば、Noneを返してください。\n 注意：実数が文字列として表されている場合、小数点はピリオド（.）またはカンマ（,）の可能性があります。\n\n compare_one(1, 2.5) ➞ 2.5\n compare_one(1, \"2,3\") ➞ \"2,3\"\n compare_one(\"5,1\", \"6\") ➞ \"6\"\n compare_one(\"1\", 1) ➞ None","entry_point":"compare_one","canonical_solution":" temp_a, temp_b = a, b\n if isinstance(temp_a, str): temp_a = temp_a.replace(',','.')\n if isinstance(temp_b, str): temp_b = temp_b.replace(',','.')\n if float(temp_a) == float(temp_b): return None\n return a if float(temp_a) > float(temp_b) else b \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(1, 2) == 2\n assert candidate(1, 2.5) == 2.5\n assert candidate(2, 3) == 3\n assert candidate(5, 6) == 6\n assert candidate(1, \"2,3\") == \"2,3\"\n assert candidate(\"5,1\", \"6\") == \"6\"\n assert candidate(\"1\", \"2\") == \"2\"\n assert candidate(\"1\", 1) == None\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def compare_one(a, b):"} -{"task_id":"JHumanEval\/138","prompt_en":"def is_equal_to_sum_even(n):\n \"\"\"Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers\n Example\n is_equal_to_sum_even(4) == False\n is_equal_to_sum_even(6) == False\n is_equal_to_sum_even(8) == True\n \"\"\"\n","prompt":"与えられた数値nが、ちょうど4つの正の偶数の合計として表現できるかどうかを評価してください。\n 例\n is_equal_to_sum_even(4) == False\n is_equal_to_sum_even(6) == False\n is_equal_to_sum_even(8) == True","entry_point":"is_equal_to_sum_even","canonical_solution":" return n%2 == 0 and n >= 8\n","test":"def check(candidate):\n assert candidate(4) == False\n assert candidate(6) == False\n assert candidate(8) == True\n assert candidate(10) == True\n assert candidate(11) == False\n assert candidate(12) == True\n assert candidate(13) == False\n assert candidate(16) == True\n","code":"def is_equal_to_sum_even(n):"} -{"task_id":"JHumanEval\/139","prompt_en":"def special_factorial(n):\n \"\"\"The Brazilian factorial is defined as:\n brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!\n where n > 0\n\n For example:\n >>> special_factorial(4)\n 288\n\n The function will receive an integer as input and should return the special\n factorial of this integer.\n \"\"\"\n","prompt":"ブラジリアン階乗は次のように定義される：\n brazilian_factorial(n) = n！* (n-1)! * (n-2)! * ... * 1!\n ただし n > 0\n\n 例えば:\n >>> special_factorial(4)\n 288\n\n この関数は入力として整数を受け取り、整数の特殊な階乗を返す。","entry_point":"special_factorial","canonical_solution":" fact_i = 1\n special_fact = 1\n for i in range(1, n+1):\n fact_i *= i\n special_fact *= fact_i\n return special_fact\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(4) == 288, \"Test 4\"\n assert candidate(5) == 34560, \"Test 5\"\n assert candidate(7) == 125411328000, \"Test 7\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1) == 1, \"Test 1\"\n\n","code":"def special_factorial(n):"} -{"task_id":"JHumanEval\/140","prompt_en":"def fix_spaces(text):\n \"\"\"\n Given a string text, replace all spaces in it with underscores, \n and if a string has more than 2 consecutive spaces, \n then replace all consecutive spaces with - \n \n fix_spaces(\"Example\") == \"Example\"\n fix_spaces(\"Example 1\") == \"Example_1\"\n fix_spaces(\" Example 2\") == \"_Example_2\"\n fix_spaces(\" Example 3\") == \"_Example-3\"\n \"\"\"\n","prompt":"文字列テキストが与えられた場合、その中のすべての空白をアンダースコアに置換し、\n 文字列が2つ以上の連続した空白を持つ場合、すべての連続した空白を - に置換する\n\n fix_spaces(\"Example\") == \"Example\"\n fix_spaces(\"Example 1\") == \"Example_1\"\n fix_spaces(\" Example 2\") == \"_Example_2\"\n fix_spaces(\" Example 3\") == \"_Example-3\"","entry_point":"fix_spaces","canonical_solution":" new_text = \"\"\n i = 0\n start, end = 0, 0\n while i < len(text):\n if text[i] == \" \":\n end += 1\n else:\n if end - start > 2:\n new_text += \"-\"+text[i]\n elif end - start > 0:\n new_text += \"_\"*(end - start)+text[i]\n else:\n new_text += text[i]\n start, end = i+1, i+1\n i+=1\n if end - start > 2:\n new_text += \"-\"\n elif end - start > 0:\n new_text += \"_\"\n return new_text\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"Example\") == \"Example\", \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(\"Mudasir Hanif \") == \"Mudasir_Hanif_\", \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate(\"Yellow Yellow Dirty Fellow\") == \"Yellow_Yellow__Dirty__Fellow\", \"This prints if this assert fails 3 (good for debugging!)\"\n \n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"Exa mple\") == \"Exa-mple\", \"This prints if this assert fails 4 (good for debugging!)\"\n assert candidate(\" Exa 1 2 2 mple\") == \"-Exa_1_2_2_mple\", \"This prints if this assert fails 4 (good for debugging!)\"\n\n","code":"def fix_spaces(text):"} -{"task_id":"JHumanEval\/141","prompt_en":"def file_name_check(file_name):\n \"\"\"Create a function which takes a string representing a file's name, and returns\n 'Yes' if the the file's name is valid, and returns 'No' otherwise.\n A file's name is considered to be valid if and only if all the following conditions \n are met:\n - There should not be more than three digits ('0'-'9') in the file's name.\n - The file's name contains exactly one dot '.'\n - The substring before the dot should not be empty, and it starts with a letter from \n the latin alphapet ('a'-'z' and 'A'-'Z').\n - The substring after the dot should be one of these: ['txt', 'exe', 'dll']\n Examples:\n file_name_check(\"example.txt\") # => 'Yes'\n file_name_check(\"1example.dll\") # => 'No' (the name should start with a latin alphapet letter)\n \"\"\"\n","prompt":"ファイル名を表す文字列を受け取り、そのファイル名が有効であれば'Yes'を返し、そうでなければ'No'\n を返す関数を作成する。ファイル名が有効であるとみなされるのは、\n 以下の条件をすべて満たす場合のみである：\n - ファイル名に3桁以上の数字（'0'-'9'）があってはならない。\n - ファイル名に含まれるドット '.' はひとつのみ。\n - ドットの前の部分文字列は空であってはならず、英文字（'a'-'z'および'A'-'Z'）から始まる文字でなければならない。\n - ドットの後の部分文字列は、以下のいずれかでなければならない： ['txt'、'exe'、'dll']。\n 例:\n file_name_check(\"example.txt\") # => 'Yes'\n file_name_check(\"1example.dll\") # => 'No' (名前は英文字で始まらないといけない）","entry_point":"file_name_check","canonical_solution":" suf = ['txt', 'exe', 'dll']\n lst = file_name.split(sep='.')\n if len(lst) != 2:\n return 'No'\n if not lst[1] in suf:\n return 'No'\n if len(lst[0]) == 0:\n return 'No'\n if not lst[0][0].isalpha():\n return 'No'\n t = len([x for x in lst[0] if x.isdigit()])\n if t > 3:\n return 'No'\n return 'Yes'\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"example.txt\") == 'Yes'\n assert candidate(\"1example.dll\") == 'No'\n assert candidate('s1sdf3.asd') == 'No'\n assert candidate('K.dll') == 'Yes'\n assert candidate('MY16FILE3.exe') == 'Yes'\n assert candidate('His12FILE94.exe') == 'No'\n assert candidate('_Y.txt') == 'No'\n assert candidate('?aREYA.exe') == 'No'\n assert candidate('\/this_is_valid.dll') == 'No'\n assert candidate('this_is_valid.wow') == 'No'\n assert candidate('this_is_valid.txt') == 'Yes'\n assert candidate('this_is_valid.txtexe') == 'No'\n assert candidate('#this2_i4s_5valid.ten') == 'No'\n assert candidate('@this1_is6_valid.exe') == 'No'\n assert candidate('this_is_12valid.6exe4.txt') == 'No'\n assert candidate('all.exe.txt') == 'No'\n assert candidate('I563_No.exe') == 'Yes'\n assert candidate('Is3youfault.txt') == 'Yes'\n assert candidate('no_one#knows.dll') == 'Yes'\n assert candidate('1I563_Yes3.exe') == 'No'\n assert candidate('I563_Yes3.txtt') == 'No'\n assert candidate('final..txt') == 'No'\n assert candidate('final132') == 'No'\n assert candidate('_f4indsartal132.') == 'No'\n \n \n\n # Check some edge cases that are easy to work out by hand.\n assert candidate('.txt') == 'No'\n assert candidate('s.') == 'No'\n\n","code":"def file_name_check(file_name):"} -{"task_id":"JHumanEval\/142","prompt_en":"def sum_squares(lst):\n \"\"\"\"\n This function will take a list of integers. For all entries in the list, the function shall square the integer entry if its index is a \n multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not \n change the entries in the list whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries. \n \n Examples:\n For lst = [1,2,3] the output should be 6\n For lst = [] the output should be 0\n For lst = [-1,-5,2,-1,-5] the output should be -126\n \"\"\"\n","prompt":"\"\n この関数は整数のリストを受け取ります。リスト内の各要素に対して、そのインデックスが3の倍数で\n あればその整数を二乗し、インデックスが4の倍数でかつ3の倍数でない場合はその整数を三乗します。\n インデックスが3または4の倍数でない要素については、何も変更しません。最後に、すべての要素の\n 合計値を返します。\n \n 例:\n lst = [1,2,3] の時、返り値は 6\n lst = [] の時、返り値は 0\n lst = [-1,-5,2,-1,-5] の時、返り値は -126","entry_point":"sum_squares","canonical_solution":" result =[]\n for i in range(len(lst)):\n if i %3 == 0:\n result.append(lst[i]**2)\n elif i % 4 == 0 and i%3 != 0:\n result.append(lst[i]**3)\n else:\n result.append(lst[i])\n return sum(result)\n","test":"def check(candidate):\n\n # Check some simple cases\n \n assert candidate([1,2,3]) == 6\n assert candidate([1,4,9]) == 14\n assert candidate([]) == 0\n assert candidate([1,1,1,1,1,1,1,1,1]) == 9\n assert candidate([-1,-1,-1,-1,-1,-1,-1,-1,-1]) == -3\n assert candidate([0]) == 0\n assert candidate([-1,-5,2,-1,-5]) == -126\n assert candidate([-56,-99,1,0,-2]) == 3030\n assert candidate([-1,0,0,0,0,0,0,0,-1]) == 0\n assert candidate([-16, -9, -2, 36, 36, 26, -20, 25, -40, 20, -4, 12, -26, 35, 37]) == -14196\n assert candidate([-1, -3, 17, -1, -15, 13, -1, 14, -14, -12, -5, 14, -14, 6, 13, 11, 16, 16, 4, 10]) == -1448\n \n \n # Don't remove this line:\n","code":"def sum_squares(lst):"} -{"task_id":"JHumanEval\/143","prompt_en":"def words_in_sentence(sentence):\n \"\"\"\n You are given a string representing a sentence,\n the sentence contains some words separated by a space,\n and you have to return a string that contains the words from the original sentence,\n whose lengths are prime numbers,\n the order of the words in the new string should be the same as the original one.\n\n Example 1:\n Input: sentence = \"This is a test\"\n Output: \"is\"\n\n Example 2:\n Input: sentence = \"lets go for swimming\"\n Output: \"go for\"\n\n Constraints:\n * 1 <= len(sentence) <= 100\n * sentence contains only letters\n \"\"\"\n","prompt":"文を表す文字列が与えられ、その文には空白で区切られたいくつかの単語が含まれている。\n 元の文の単語を含みその長さが素数である文字列を返す必要がある。\n 新しい文字列の単語の順序は元の文字列と同じでなければならない。\n\n 例 1:\n 入力: sentence = \"This is a test\"\n 出力: \"is\"\n\n 例 2:\n 入力: sentence = \"lets go for swimming\"\n 出力: \"go for\"\n\n 制約:\n * 1 <= len(sentence) <= 100\n * sentence contains only letters","entry_point":"words_in_sentence","canonical_solution":" new_lst = []\n for word in sentence.split():\n flg = 0\n if len(word) == 1:\n flg = 1\n for i in range(2, len(word)):\n if len(word)%i == 0:\n flg = 1\n if flg == 0 or len(word) == 2:\n new_lst.append(word)\n return \" \".join(new_lst)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"This is a test\") == \"is\"\n assert candidate(\"lets go for swimming\") == \"go for\"\n assert candidate(\"there is no place available here\") == \"there is no place\"\n assert candidate(\"Hi I am Hussein\") == \"Hi am Hussein\"\n assert candidate(\"go for it\") == \"go for it\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"here\") == \"\"\n assert candidate(\"here is\") == \"is\"\n\n","code":"def words_in_sentence(sentence):"} -{"task_id":"JHumanEval\/144","prompt_en":"def simplify(x, n):\n \"\"\"Your task is to implement a function that will simplify the expression\n x * n. The function returns True if x * n evaluates to a whole number and False\n otherwise. Both x and n, are string representation of a fraction, and have the following format,\n \/ where both numerator and denominator are positive whole numbers.\n\n You can assume that x, and n are valid fractions, and do not have zero as denominator.\n\n simplify(\"1\/5\", \"5\/1\") = True\n simplify(\"1\/6\", \"2\/1\") = False\n simplify(\"7\/10\", \"10\/2\") = False\n \"\"\"\n","prompt":"あなたの仕事は、式 x * n を簡単にする関数を実装することです。\n この関数は、x * n が整数になる場合はTrueを、そうでない場合はFalseを\n 返します。xとnはともに分数の文字列表現であり、<分子>\/<分母>という形式で、\n 分子と分母はともに正の整数です。\n \n xとnが有効な分数であり、分母がゼロでないことは仮定してかまいません。\n \n simplify(\"1\/5\", \"5\/1\") = True\n simplify(\"1\/6\", \"2\/1\") = False\n simplify(\"7\/10\", \"10\/2\") = False","entry_point":"simplify","canonical_solution":" a, b = x.split(\"\/\")\n c, d = n.split(\"\/\")\n numerator = int(a) * int(c)\n denom = int(b) * int(d)\n if (numerator\/denom == int(numerator\/denom)):\n return True\n return False\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"1\/5\", \"5\/1\") == True, 'test1'\n assert candidate(\"1\/6\", \"2\/1\") == False, 'test2'\n assert candidate(\"5\/1\", \"3\/1\") == True, 'test3'\n assert candidate(\"7\/10\", \"10\/2\") == False, 'test4'\n assert candidate(\"2\/10\", \"50\/10\") == True, 'test5'\n assert candidate(\"7\/2\", \"4\/2\") == True, 'test6'\n assert candidate(\"11\/6\", \"6\/1\") == True, 'test7'\n assert candidate(\"2\/3\", \"5\/2\") == False, 'test8'\n assert candidate(\"5\/2\", \"3\/5\") == False, 'test9'\n assert candidate(\"2\/4\", \"8\/4\") == True, 'test10'\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"2\/4\", \"4\/2\") == True, 'test11'\n assert candidate(\"1\/5\", \"5\/1\") == True, 'test12'\n assert candidate(\"1\/5\", \"1\/5\") == False, 'test13'\n\n","code":"def simplify(x, n):"} -{"task_id":"JHumanEval\/145","prompt_en":"def order_by_points(nums):\n \"\"\"\n Write a function which sorts the given list of integers\n in ascending order according to the sum of their digits.\n Note: if there are several items with similar sum of their digits,\n order them based on their index in original list.\n\n For example:\n >>> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]\n >>> order_by_points([]) == []\n \"\"\"\n","prompt":"各数字の桁の合計に基づいて、与えられた整数のリストを昇順に並べる\n 関数を作成してください。\n 注意：もし桁の合計が同じである複数の項目がある場合は、\n 元のリストでの位置に基づいて並べてください。\n 例えば\n >> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11].\n >> order_by_points([]) == [].","entry_point":"order_by_points","canonical_solution":" def digits_sum(n):\n neg = 1\n if n < 0: n, neg = -1 * n, -1 \n n = [int(i) for i in str(n)]\n n[0] = n[0] * neg\n return sum(n)\n return sorted(nums, key=digits_sum)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]\n assert candidate([1234,423,463,145,2,423,423,53,6,37,3457,3,56,0,46]) == [0, 2, 3, 6, 53, 423, 423, 423, 1234, 145, 37, 46, 56, 463, 3457]\n assert candidate([]) == []\n assert candidate([1, -11, -32, 43, 54, -98, 2, -3]) == [-3, -32, -98, -11, 1, 2, 43, 54]\n assert candidate([1,2,3,4,5,6,7,8,9,10,11]) == [1, 10, 2, 11, 3, 4, 5, 6, 7, 8, 9]\n assert candidate([0,6,6,-76,-21,23,4]) == [-76, -21, 0, 4, 23, 6, 6]\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def order_by_points(nums):"} -{"task_id":"JHumanEval\/146","prompt_en":"def specialFilter(nums):\n \"\"\"Write a function that takes an array of numbers as input and returns \n the number of elements in the array that are greater than 10 and both \n first and last digits of a number are odd (1, 3, 5, 7, 9).\n For example:\n specialFilter([15, -73, 14, -15]) => 1 \n specialFilter([33, -2, -3, 45, 21, 109]) => 2\n \"\"\"\n","prompt":"数値の配列を入力とし、配列中の要素のうち、10より大きく、\n かつ数値の最初と最後の桁の両方が奇数（1, 3, 5, 7, 9）である要素の数を返す関数を書く。\n 例えば\n specialFilter([15, -73, 14, -15]) => 1 \n specialFilter([33, -2, -3, 45, 21, 109]) => 2","entry_point":"specialFilter","canonical_solution":" \n count = 0\n for num in nums:\n if num > 10:\n odd_digits = (1, 3, 5, 7, 9)\n number_as_string = str(num)\n if int(number_as_string[0]) in odd_digits and int(number_as_string[-1]) in odd_digits:\n count += 1\n \n return count \n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([5, -2, 1, -5]) == 0 \n assert candidate([15, -73, 14, -15]) == 1\n assert candidate([33, -2, -3, 45, 21, 109]) == 2\n assert candidate([43, -12, 93, 125, 121, 109]) == 4\n assert candidate([71, -2, -33, 75, 21, 19]) == 3\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([1]) == 0 \n assert candidate([]) == 0 \n\n","code":"def specialFilter(nums):"} -{"task_id":"JHumanEval\/147","prompt_en":"def get_max_triples(n):\n \"\"\"\n You are given a positive integer n. You have to create an integer array a of length n.\n For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.\n Return the number of triples (a[i], a[j], a[k]) of a where i < j < k, \n and a[i] + a[j] + a[k] is a multiple of 3.\n\n Example :\n Input: n = 5\n Output: 1\n Explanation: \n a = [1, 3, 7, 13, 21]\n The only valid triple is (1, 7, 13).\n \"\"\"\n","prompt":"正の整数 n が与えられるので、長さ n の整数配列 a を作成せよ。\n 各 i (1 ≤ i ≤ n) に対して、 a[i] = i * i - i + 1 とする。\n i < j < k において、a[i] + a[j] + a[k] が3の倍数となるような三つ組 (a[i], a[j], a[k]) を返す。\n\n 例 :\n 入力: n = 5\n 出力t: 1\n 解説: \n a = [1, 3, 7, 13, 21]\n 唯一の妥当な三つ組は (1, 7, 13)である。","entry_point":"get_max_triples","canonical_solution":" A = [i*i - i + 1 for i in range(1,n+1)]\n ans = []\n for i in range(n):\n for j in range(i+1,n):\n for k in range(j+1,n):\n if (A[i]+A[j]+A[k])%3 == 0:\n ans += [(A[i],A[j],A[k])]\n return len(ans)\n","test":"def check(candidate):\n\n assert candidate(5) == 1\n assert candidate(6) == 4\n assert candidate(10) == 36\n assert candidate(100) == 53361\n","code":"def get_max_triples(n):"} -{"task_id":"JHumanEval\/148","prompt_en":"def bf(planet1, planet2):\n '''\n There are eight planets in our solar system: the closerst to the Sun \n is Mercury, the next one is Venus, then Earth, Mars, Jupiter, Saturn, \n Uranus, Neptune.\n Write a function that takes two planet names as strings planet1 and planet2. \n The function should return a tuple containing all planets whose orbits are \n located between the orbit of planet1 and the orbit of planet2, sorted by \n the proximity to the sun. \n The function should return an empty tuple if planet1 or planet2\n are not correct planet names. \n Examples\n bf(\"Jupiter\", \"Neptune\") ==> (\"Saturn\", \"Uranus\")\n bf(\"Earth\", \"Mercury\") ==> (\"Venus\")\n bf(\"Mercury\", \"Uranus\") ==> (\"Venus\", \"Earth\", \"Mars\", \"Jupiter\", \"Saturn\")\n '''\n","prompt":"私たちの太陽系には8つの惑星があります：太陽に最も近いのはVenus, Earth, Mars, Jupiter, Saturn, \n Uranus, Neptuneです。\n planet1とplanet2という2つの惑星名を文字列として受け取る関数を作成してください。\n この関数は、planet1の軌道とplanet2の軌道の間に位置するすべての惑星を太陽に近い順に並べたタプルを返すべきです。\n planet1またはplanet2が正確な惑星名でない場合、関数は空のタプルを返すべきです。\n 例\n bf(\"Jupiter\", \"Neptune\") ==> (\"Saturn\", \"Uranus\")\n bf(\"Earth\", \"Mercury\") ==> (\"Venus\")\n bf(\"Mercury\", \"Uranus\") ==> (\"Venus\", \"Earth\", \"Mars\", \"Jupiter\", \"Saturn\")","entry_point":"bf","canonical_solution":" planet_names = (\"Mercury\", \"Venus\", \"Earth\", \"Mars\", \"Jupiter\", \"Saturn\", \"Uranus\", \"Neptune\")\n if planet1 not in planet_names or planet2 not in planet_names or planet1 == planet2:\n return ()\n planet1_index = planet_names.index(planet1)\n planet2_index = planet_names.index(planet2)\n if planet1_index < planet2_index:\n return (planet_names[planet1_index + 1: planet2_index])\n else:\n return (planet_names[planet2_index + 1 : planet1_index])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"Jupiter\", \"Neptune\") == (\"Saturn\", \"Uranus\"), \"First test error: \" + str(len(candidate(\"Jupiter\", \"Neptune\"))) \n assert candidate(\"Earth\", \"Mercury\") == (\"Venus\",), \"Second test error: \" + str(candidate(\"Earth\", \"Mercury\")) \n assert candidate(\"Mercury\", \"Uranus\") == (\"Venus\", \"Earth\", \"Mars\", \"Jupiter\", \"Saturn\"), \"Third test error: \" + str(candidate(\"Mercury\", \"Uranus\")) \n assert candidate(\"Neptune\", \"Venus\") == (\"Earth\", \"Mars\", \"Jupiter\", \"Saturn\", \"Uranus\"), \"Fourth test error: \" + str(candidate(\"Neptune\", \"Venus\")) \n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"Earth\", \"Earth\") == ()\n assert candidate(\"Mars\", \"Earth\") == ()\n assert candidate(\"Jupiter\", \"Makemake\") == ()\n\n","code":"def bf(planet1, planet2):"} -{"task_id":"JHumanEval\/149","prompt_en":"def sorted_list_sum(lst):\n \"\"\"Write a function that accepts a list of strings as a parameter,\n deletes the strings that have odd lengths from it,\n and returns the resulted list with a sorted order,\n The list is always a list of strings and never an array of numbers,\n and it may contain duplicates.\n The order of the list should be ascending by length of each word, and you\n should return the list sorted by that rule.\n If two words have the same length, sort the list alphabetically.\n The function should return a list of strings in sorted order.\n You may assume that all words will have the same length.\n For example:\n assert list_sort([\"aa\", \"a\", \"aaa\"]) => [\"aa\"]\n assert list_sort([\"ab\", \"a\", \"aaa\", \"cd\"]) => [\"ab\", \"cd\"]\n \"\"\"\n","prompt":"文字列のリストを引数として受け取る関数を作成してください。\n この関数は、リストから奇数の長さを持つ文字列を削除し、\n 結果として得られるリストを長さで昇順に並べ替えて返します。\n リストは常に文字列のリストであり、数字の配列ではありません。\n また、重複する文字列が含まれる可能性があります。\n リストは各単語の長さで昇順に並べられるべきで、そのルールに従ってソートされたリストを返してください。\n もし二つの単語が同じ長さであれば、リストをアルファベット順に並べ替えてください。\n 関数はソートされた順序で文字列のリストを返す���きです。\n すべての単語が同じ長さを持つと仮定しても構いません。\n 例えば:\n assert list_sort([\"aa\", \"a\", \"aaa\"]) => [\"aa\"]\n assert list_sort([\"ab\", \"a\", \"aaa\", \"cd\"]) => [\"ab\", \"cd\"]","entry_point":"sorted_list_sum","canonical_solution":" lst.sort()\n new_lst = []\n for i in lst:\n if len(i)%2 == 0:\n new_lst.append(i)\n return sorted(new_lst, key=len)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([\"aa\", \"a\", \"aaa\"]) == [\"aa\"]\n assert candidate([\"school\", \"AI\", \"asdf\", \"b\"]) == [\"AI\", \"asdf\", \"school\"]\n assert candidate([\"d\", \"b\", \"c\", \"a\"]) == []\n assert candidate([\"d\", \"dcba\", \"abcd\", \"a\"]) == [\"abcd\", \"dcba\"]\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([\"AI\", \"ai\", \"au\"]) == [\"AI\", \"ai\", \"au\"]\n assert candidate([\"a\", \"b\", \"b\", \"c\", \"c\", \"a\"]) == []\n assert candidate(['aaaa', 'bbbb', 'dd', 'cc']) == [\"cc\", \"dd\", \"aaaa\", \"bbbb\"]\n\n","code":"def sorted_list_sum(lst):"} -{"task_id":"JHumanEval\/150","prompt_en":"def x_or_y(n, x, y):\n \"\"\"A simple program which should return the value of x if n is \n a prime number and should return the value of y otherwise.\n\n Examples:\n for x_or_y(7, 34, 12) == 34\n for x_or_y(15, 8, 5) == 5\n \n \"\"\"\n","prompt":"素数である場合はxの値を返し、それ以外の場合はyの値を返す簡単なプログラム。\n\n 例:\n for x_or_y(7, 34, 12) == 34\n for x_or_y(15, 8, 5) == 5","entry_point":"x_or_y","canonical_solution":" if n == 1:\n return y\n for i in range(2, n):\n if n % i == 0:\n return y\n break\n else:\n return x\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(7, 34, 12) == 34\n assert candidate(15, 8, 5) == 5\n assert candidate(3, 33, 5212) == 33\n assert candidate(1259, 3, 52) == 3\n assert candidate(7919, -1, 12) == -1\n assert candidate(3609, 1245, 583) == 583\n assert candidate(91, 56, 129) == 129\n assert candidate(6, 34, 1234) == 1234\n \n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1, 2, 0) == 0\n assert candidate(2, 2, 0) == 2\n\n","code":"def x_or_y(n, x, y):"} -{"task_id":"JHumanEval\/151","prompt_en":"def double_the_difference(lst):\n '''\n Given a list of numbers, return the sum of squares of the numbers\n in the list that are odd. Ignore numbers that are negative or not integers.\n \n double_the_difference([1, 3, 2, 0]) == 1 + 9 + 0 + 0 = 10\n double_the_difference([-1, -2, 0]) == 0\n double_the_difference([9, -2]) == 81\n double_the_difference([0]) == 0 \n \n If the input list is empty, return 0.\n '''\n","prompt":"数字のリストが与えられた場合、そのリスト内の奇数の数値の二乗の合計を返してください。\n 負の数や整数でない数は無視してください。\n \n double_the_difference([1, 3, 2, 0]) == 1 + 9 + 0 + 0 = 10\n double_the_difference([-1, -2, 0]) == 0\n double_the_difference([9, -2]) == 81\n double_the_difference([0]) == 0 \n \n 入力リストが空の場合は0を返すようにしてください。","entry_point":"double_the_difference","canonical_solution":" return sum([i**2 for i in lst if i > 0 and i%2!=0 and \".\" not in str(i)])\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([]) == 0 , \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([5, 4]) == 25 , \"This prints if this assert fails 2 (good for debugging!)\"\n assert candidate([0.1, 0.2, 0.3]) == 0 , \"This prints if this assert fails 3 (good for debugging!)\"\n assert candidate([-10, -20, -30]) == 0 , \"This prints if this assert fails 4 (good for debugging!)\"\n\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate([-1, -2, 8]) == 0, \"This prints if this assert fails 5 (also good for debugging!)\"\n assert candidate([0.2, 3, 5]) == 34, \"This prints if this assert fails 6 (also good for debugging!)\"\n lst = list(range(-99, 100, 2))\n odd_sum = sum([i**2 for i in lst if i%2!=0 and i > 0])\n assert candidate(lst) == odd_sum , \"This prints if this assert fails 7 (good for debugging!)\"\n\n","code":"def double_the_difference(lst):"} -{"task_id":"JHumanEval\/152","prompt_en":"def compare(game,guess):\n \"\"\"I think we all remember that feeling when the result of some long-awaited\n event is finally known. The feelings and thoughts you have at that moment are\n definitely worth noting down and comparing.\n Your task is to determine if a person correctly guessed the results of a number of matches.\n You are given two arrays of scores and guesses of equal length, where each index shows a match. \n Return an array of the same length denoting how far off each guess was. If they have guessed correctly,\n the value is 0, and if not, the value is the absolute difference between the guess and the score.\n \n \n example:\n\n compare([1,2,3,4,5,1],[1,2,3,4,2,-2]) -> [0,0,0,0,3,3]\n compare([0,5,0,0,0,4],[4,1,1,0,0,-2]) -> [4,4,1,0,0,6]\n \"\"\"\n","prompt":"待ち望んでいた出来事の結果がようやく判明したときの感覚は、誰もが覚えていると思う。\n その瞬間に抱いた感情や思考は、間違いなくメモして比較する価値がある。\n あなたの仕事は、人がいくつかの試合の結果を正確に予想したかどうかを判断することです。\n スコアと予想の2つの配列が等しい長さで与えられます。各インデックスは1つの試合を示しています。\n 各予想がどれだけ外れていたかを示す同じ長さの配列を返してください。予想が正確であれば、\n その値は0です。そうでなければ、その値は予想とスコアの絶対的な差です。\n \n \n 例:\n\n compare([1,2,3,4,5,1],[1,2,3,4,2,-2]) -> [0,0,0,0,3,3]\n compare([0,5,0,0,0,4],[4,1,1,0,0,-2]) -> [4,4,1,0,0,6]","entry_point":"compare","canonical_solution":" return [abs(x-y) for x,y in zip(game,guess)]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate([1,2,3,4,5,1],[1,2,3,4,2,-2])==[0,0,0,0,3,3], \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([0,0,0,0,0,0],[0,0,0,0,0,0])==[0,0,0,0,0,0], \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1,2,3],[-1,-2,-3])==[2,4,6], \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate([1,2,3,5],[-1,2,3,4])==[2,0,0,1], \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def compare(game,guess):"} -{"task_id":"JHumanEval\/153","prompt_en":"def Strongest_Extension(class_name, extensions):\n \"\"\"You will be given the name of a class (a string) and a list of extensions.\n The extensions are to be used to load additional classes to the class. The\n strength of the extension is as follows: Let CAP be the number of the uppercase\n letters in the extension's name, and let SM be the number of lowercase letters \n in the extension's name, the strength is given by the fraction CAP - SM. \n You should find the strongest extension and return a string in this \n format: ClassName.StrongestExtensionName.\n If there are two or more extensions with the same strength, you should\n choose the one that comes first in the list.\n For example, if you are given \"Slices\" as the class and a list of the\n extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should\n return 'Slices.SErviNGSliCes' since 'SErviNGSliCes' is the strongest extension \n (its strength is -1).\n Example:\n for Strongest_Extension('my_class', ['AA', 'Be', 'CC']) == 'my_class.AA'\n \"\"\"\n","prompt":"クラスの名前（文字列）と拡張子のリストが与えられます。\n この拡張子は、指定されたクラスに追加のクラスをロードするために使用されます。\n 拡張子の強度は次のように計算されます：CAPは拡張子の名前に含まれる\n 大文字の数、SMは小文字の数です。強度は、CAP - SM で与えられます。\n 最も強い拡張子を見つけて、この形式の文字列を返してください：ClassName.StrongestExtensionName。\n 同じ強度を持つ2つ以上の拡張子がある場合は、リストで最初に来るものを選びます。\n 例えば、\"Slices\"というクラスと、拡張子のリスト['SErviNGSliCes', 'Cheese', 'StuFfed'] が与えられた場合、'\n SErviNGSliCes'が最も強い拡張子（強度は-1）となるため、'Slices.SErviNGSliCes'を返すべきです。\n 例：\n Strongest_Extension('my_class', ['AA', 'Be', 'CC']) == 'my_class.AA'","entry_point":"Strongest_Extension","canonical_solution":" strong = extensions[0]\n my_val = len([x for x in extensions[0] if x.isalpha() and x.isupper()]) - len([x for x in extensions[0] if x.isalpha() and x.islower()])\n for s in extensions:\n val = len([x for x in s if x.isalpha() and x.isupper()]) - len([x for x in s if x.isalpha() and x.islower()])\n if val > my_val:\n strong = s\n my_val = val\n\n ans = class_name + \".\" + strong\n return ans\n\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('Watashi', ['tEN', 'niNE', 'eIGHt8OKe']) == 'Watashi.eIGHt8OKe'\n assert candidate('Boku123', ['nani', 'NazeDa', 'YEs.WeCaNe', '32145tggg']) == 'Boku123.YEs.WeCaNe'\n assert candidate('__YESIMHERE', ['t', 'eMptY', 'nothing', 'zeR00', 'NuLl__', '123NoooneB321']) == '__YESIMHERE.NuLl__'\n assert candidate('K', ['Ta', 'TAR', 't234An', 'cosSo']) == 'K.TAR'\n assert candidate('__HAHA', ['Tab', '123', '781345', '-_-']) == '__HAHA.123'\n assert candidate('YameRore', ['HhAas', 'okIWILL123', 'WorkOut', 'Fails', '-_-']) == 'YameRore.okIWILL123'\n assert candidate('finNNalLLly', ['Die', 'NowW', 'Wow', 'WoW']) == 'finNNalLLly.WoW'\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate('_', ['Bb', '91245']) == '_.Bb'\n assert candidate('Sp', ['671235', 'Bb']) == 'Sp.671235'\n \n","code":"def Strongest_Extension(class_name, extensions):"} -{"task_id":"JHumanEval\/154","prompt_en":"def cycpattern_check(a , b):\n \"\"\"You are given 2 words. You need to return True if the second word or any of its rotations is a substring in the first word\n cycpattern_check(\"abcd\",\"abd\") => False\n cycpattern_check(\"hello\",\"ell\") => True\n cycpattern_check(\"whassup\",\"psus\") => False\n cycpattern_check(\"abab\",\"baa\") => True\n cycpattern_check(\"efef\",\"eeff\") => False\n cycpattern_check(\"himenss\",\"simen\") => True\n\n \"\"\"\n","prompt":"2つの単語が与えられる。2番目の単語またはその回転させた文字列が最初の単語の部分文字列である場合、Trueを返す必要がある。\n cycpattern_check(\"abcd\",\"abd\") => False\n cycpattern_check(\"hello\",\"ell\") => True\n cycpattern_check(\"whassup\",\"psus\") => False\n cycpattern_check(\"abab\",\"baa\") => True\n cycpattern_check(\"efef\",\"eeff\") => False\n cycpattern_check(\"himenss\",\"simen\") => True","entry_point":"cycpattern_check","canonical_solution":" l = len(b)\n pat = b + b\n for i in range(len(a) - l + 1):\n for j in range(l + 1):\n if a[i:i+l] == pat[j:j+l]:\n return True\n return False\n","test":"def check(candidate):\n\n # Check some simple cases\n #assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n #assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(\"xyzw\",\"xyw\") == False , \"test #0\"\n assert candidate(\"yello\",\"ell\") == True , \"test #1\"\n assert candidate(\"whattup\",\"ptut\") == False , \"test #2\"\n assert candidate(\"efef\",\"fee\") == True , \"test #3\"\n assert candidate(\"abab\",\"aabb\") == False , \"test #4\"\n assert candidate(\"winemtt\",\"tinem\") == True , \"test #5\"\n\n","code":"def cycpattern_check(a , b):"} -{"task_id":"JHumanEval\/155","prompt_en":"def even_odd_count(num):\n \"\"\"Given an integer. return a tuple that has the number of even and odd digits respectively.\n\n Example:\n even_odd_count(-12) ==> (1, 1)\n even_odd_count(123) ==> (1, 2)\n \"\"\"\n","prompt":"整数が与えられた場合、偶数桁数と奇数桁数をそれぞれ持つタプルを返す。\n 例:\n even_odd_count(-12) ==> (1, 1)\n even_odd_count(123) ==> (1, 2)","entry_point":"even_odd_count","canonical_solution":" even_count = 0\n odd_count = 0\n for i in str(abs(num)):\n if int(i)%2==0:\n even_count +=1\n else:\n odd_count +=1\n return (even_count, odd_count)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(7) == (0, 1)\n assert candidate(-78) == (1, 1)\n assert candidate(3452) == (2, 2)\n assert candidate(346211) == (3, 3)\n assert candidate(-345821) == (3, 3)\n assert candidate(-2) == (1, 0)\n assert candidate(-45347) == (2, 3)\n assert candidate(0) == (1, 0)\n\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def even_odd_count(num):"} -{"task_id":"JHumanEval\/156","prompt_en":"def int_to_mini_roman(number):\n \"\"\"\n Given a positive integer, obtain its roman numeral equivalent as a string,\n and return it in lowercase.\n Restrictions: 1 <= num <= 1000\n\n Examples:\n >>> int_to_mini_roman(19) == 'xix'\n >>> int_to_mini_roman(152) == 'clii'\n >>> int_to_mini_roman(426) == 'cdxxvi'\n \"\"\"\n","prompt":"正の整数が与えられたとき、ローマ数字に相当する文字列を小文字で返す。\n 制限事項1 <= num <= 1000\n 例:\n >>> int_to_mini_roman(19) == 'xix'\n >>> int_to_mini_roman(152) == 'clii'\n >>> int_to_mini_roman(426) == 'cdxxvi'","entry_point":"int_to_mini_roman","canonical_solution":" num = [1, 4, 5, 9, 10, 40, 50, 90, \n 100, 400, 500, 900, 1000] \n sym = [\"I\", \"IV\", \"V\", \"IX\", \"X\", \"XL\", \n \"L\", \"XC\", \"C\", \"CD\", \"D\", \"CM\", \"M\"] \n i = 12\n res = ''\n while number: \n div = number \/\/ num[i] \n number %= num[i] \n while div: \n res += sym[i] \n div -= 1\n i -= 1\n return res.lower()\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(19) == 'xix'\n assert candidate(152) == 'clii'\n assert candidate(251) == 'ccli'\n assert candidate(426) == 'cdxxvi'\n assert candidate(500) == 'd'\n assert candidate(1) == 'i'\n assert candidate(4) == 'iv'\n assert candidate(43) == 'xliii'\n assert candidate(90) == 'xc'\n assert candidate(94) == 'xciv'\n assert candidate(532) == 'dxxxii'\n assert candidate(900) == 'cm'\n assert candidate(994) == 'cmxciv'\n assert candidate(1000) == 'm'\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def int_to_mini_roman(number):"} -{"task_id":"JHumanEval\/157","prompt_en":"def right_angle_triangle(a, b, c):\n '''\n Given the lengths of the three sides of a triangle. Return True if the three\n sides form a right-angled triangle, False otherwise.\n A right-angled triangle is a triangle in which one angle is right angle or \n 90 degree.\n Example:\n right_angle_triangle(3, 4, 5) == True\n right_angle_triangle(1, 2, 3) == False\n '''\n","prompt":"三角形の3辺の長さを与える。三角形が直角三角形ならTrueを、そうでなければFalseを返す。\n 直角三角形とは、1つの角が直角または90度である三角形のことである。\n 例:\n right_angle_triangle(3, 4, 5) == True\n right_angle_triangle(1, 2, 3) == False","entry_point":"right_angle_triangle","canonical_solution":" return a*a == b*b + c*c or b*b == a*a + c*c or c*c == a*a + b*b\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(3, 4, 5) == True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(1, 2, 3) == False\n assert candidate(10, 6, 8) == True\n assert candidate(2, 2, 2) == False\n assert candidate(7, 24, 25) == True\n assert candidate(10, 5, 7) == False\n assert candidate(5, 12, 13) == True\n assert candidate(15, 8, 17) == True\n assert candidate(48, 55, 73) == True\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(1, 1, 1) == False, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(2, 2, 10) == False\n\n","code":"def right_angle_triangle(a, b, c):"} -{"task_id":"JHumanEval\/158","prompt_en":"def find_max(words):\n \"\"\"Write a function that accepts a list of strings.\n The list contains different words. Return the word with maximum number\n of unique characters. If multiple strings have maximum number of unique\n characters, return the one which comes first in lexicographical order.\n\n find_max([\"name\", \"of\", \"string\"]) == \"string\"\n find_max([\"name\", \"enam\", \"game\"]) == \"enam\"\n find_max([\"aaaaaaa\", \"bb\" ,\"cc\"]) == \"\"aaaaaaa\"\n \"\"\"\n","prompt":"文字列のリストを受け取る関数を書きなさい。\n リストは異なる単語を含む。異なる固有の文字数が最も多い単語を返す。\n 複数の文字列が同じ文字数を持つ場合は、辞書順で最初に来るものを返すことにする。\n \n find_max([\"name\", \"of\", \"string\"]) == \"string\"\n find_max([\"name\", \"enam\", \"game\"]) == \"enam\"\n find_max([\"aaaaaaa\", \"bb\" ,\"cc\"]) == \"\"aaaaaaa\"","entry_point":"find_max","canonical_solution":" return sorted(words, key = lambda x: (-len(set(x)), x))[0]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert (candidate([\"name\", \"of\", \"string\"]) == \"string\"), \"t1\"\n assert (candidate([\"name\", \"enam\", \"game\"]) == \"enam\"), 't2'\n assert (candidate([\"aaaaaaa\", \"bb\", \"cc\"]) == \"aaaaaaa\"), 't3'\n assert (candidate([\"abc\", \"cba\"]) == \"abc\"), 't4'\n assert (candidate([\"play\", \"this\", \"game\", \"of\",\"footbott\"]) == \"footbott\"), 't5'\n assert (candidate([\"we\", \"are\", \"gonna\", \"rock\"]) == \"gonna\"), 't6'\n assert (candidate([\"we\", \"are\", \"a\", \"mad\", \"nation\"]) == \"nation\"), 't7'\n assert (candidate([\"this\", \"is\", \"a\", \"prrk\"]) == \"this\"), 't8'\n\n # Check some edge cases that are easy to work out by hand.\n assert (candidate([\"b\"]) == \"b\"), 't9'\n assert (candidate([\"play\", \"play\", \"play\"]) == \"play\"), 't10'\n\n","code":"def find_max(words):"} -{"task_id":"JHumanEval\/159","prompt_en":"def eat(number, need, remaining):\n \"\"\"\n You're a hungry rabbit, and you already have eaten a certain number of carrots,\n but now you need to eat more carrots to complete the day's meals.\n you should return an array of [ total number of eaten carrots after your meals,\n the number of carrots left after your meals ]\n if there are not enough remaining carrots, you will eat all remaining carrots, but will still be hungry.\n \n Example:\n * eat(5, 6, 10) -> [11, 4]\n * eat(4, 8, 9) -> [12, 1]\n * eat(1, 10, 10) -> [11, 0]\n * eat(2, 11, 5) -> [7, 0]\n \n Variables:\n @number : integer\n the number of carrots that you have eaten.\n @need : integer\n the number of carrots that you need to eat.\n @remaining : integer\n the number of remaining carrots thet exist in stock\n \n Constrain:\n * 0 <= number <= 1000\n * 0 <= need <= 1000\n * 0 <= remaining <= 1000\n\n Have fun :)\n \"\"\"\n","prompt":"あなたはお腹を空かせたウサギです。すでに一定数のニンジンを食べました。\n これからさらにニンジンを食べなければその日の食事は完了しません。\n あなたは [ 食事の後に食べたニンジンの総数, 食事の後に残ったニンジンの数 ] の配列を返してください。\n もし残りのニンジンが十分でなければ、あなたは残りのニンジンをすべて食べますが、まだお腹が空いています。\n\n 例:\n * eat(5, 6, 10) -> [11, 4]\n * eat(4, 8, 9) -> [12, 1]\n * eat(1, 10, 10) -> [11, 0]\n * eat(2, 11, 5) -> [7, 0]\n \n 変数:\n @number : 整数\n 食べたニンジンの数。\n @need : 整数\n にんじんを何本食べるか。\n @remaining : 整数\n 残りのニンジンの在庫数\n \n 制約:\n * 0 <= number <= 1000\n * 0 <= need <= 1000\n * 0 <= remaining <= 1000\n\n 楽しんで :)","entry_point":"eat","canonical_solution":" if(need <= remaining):\n return [ number + need , remaining-need ]\n else:\n return [ number + remaining , 0]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert True, \"This prints if this assert fails 1 (good for debugging!)\"\n assert candidate(5, 6, 10) == [11, 4], \"Error\"\n assert candidate(4, 8, 9) == [12, 1], \"Error\"\n assert candidate(1, 10, 10) == [11, 0], \"Error\"\n assert candidate(2, 11, 5) == [7, 0], \"Error\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n assert candidate(4, 5, 7) == [9, 2], \"Error\"\n assert candidate(4, 5, 1) == [5, 0], \"Error\"\n\n","code":"def eat(number, need, remaining):"} -{"task_id":"JHumanEval\/160","prompt_en":"def do_algebra(operator, operand):\n \"\"\"\n Given two lists operator, and operand. The first list has basic algebra operations, and \n the second list is a list of integers. Use the two given lists to build the algebric \n expression and return the evaluation of this expression.\n\n The basic algebra operations:\n Addition ( + ) \n Subtraction ( - ) \n Multiplication ( * ) \n Floor division ( \/\/ ) \n Exponentiation ( ** ) \n\n Example:\n operator['+', '*', '-']\n array = [2, 3, 4, 5]\n result = 2 + 3 * 4 - 5\n => result = 9\n\n Note:\n The length of operator list is equal to the length of operand list minus one.\n Operand is a list of of non-negative integers.\n Operator list has at least one operator, and operand list has at least two operands.\n\n \"\"\"\n","prompt":"演算子(operator)とオペランド(operand)の2つのリストが与えられる。ひとつ目のリストは\n 基本的な算術演算を持ち、二つ目のリストは整数のリストである。与えられた2つのリストを\n 使って算術式を構築し、その評価結果を返そう。\n\n 基本的な算術演算:\n 加算 ( + ) \n 減算 ( - ) \n 乗算 ( * )\n 階除算 ( \/\/ ) \n 指数化 ( ** )\n \n 例:\n operator['+', '*', '-']\n array = [2, 3, 4, 5]\n result = 2 + 3 * 4 - 5\n => result = 9\n \n 注：演算子のリストの長さは、オペランドのリストの長さから1を引いた長さに等しい。\n オペランドは非負整数のリストである。\n operator は少なくとも1つの演算子を持ち、operand は少なくとも2つのオペランドを持つ。","entry_point":"do_algebra","canonical_solution":" expression = str(operand[0])\n for oprt, oprn in zip(operator, operand[1:]):\n expression+= oprt + str(oprn)\n return eval(expression)\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(['**', '*', '+'], [2, 3, 4, 5]) == 37\n assert candidate(['+', '*', '-'], [2, 3, 4, 5]) == 9\n assert candidate(['\/\/', '*'], [7, 3, 4]) == 8, \"This prints if this assert fails 1 (good for debugging!)\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def do_algebra(operator, operand):"} -{"task_id":"JHumanEval\/161","prompt_en":"def solve(s):\n \"\"\"You are given a string s.\n if s[i] is a letter, reverse its case from lower to upper or vise versa, \n otherwise keep it as it is.\n If the string contains no letters, reverse the string.\n The function should return the resulted string.\n Examples\n solve(\"1234\") = \"4321\"\n solve(\"ab\") = \"AB\"\n solve(\"#a@C\") = \"#A@c\"\n \"\"\"\n","prompt":"文字列sが与えられます。\n もしs[i]がアルファベットなら、その文字の大文字と小文字を反転させる。そうでない場合��、そのままにしておく。\n もし文字列にアルファベットが一つも含まれていない場合は、文字列全体を逆順にする。\n 関数は結果の文字列を返すようにします。\n 例\n solve(\"1234\") = \"4321\"\n solve(\"ab\") = \"AB\"\n solve(\"#a@C\") = \"#A@c\"","entry_point":"solve","canonical_solution":" flg = 0\n idx = 0\n new_str = list(s)\n for i in s:\n if i.isalpha():\n new_str[idx] = i.swapcase()\n flg = 1\n idx += 1\n s = \"\"\n for i in new_str:\n s += i\n if flg == 0:\n return s[len(s)::-1]\n return s\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(\"AsDf\") == \"aSdF\"\n assert candidate(\"1234\") == \"4321\"\n assert candidate(\"ab\") == \"AB\"\n assert candidate(\"#a@C\") == \"#A@c\"\n assert candidate(\"#AsdfW^45\") == \"#aSDFw^45\"\n assert candidate(\"#6@2\") == \"2@6#\"\n\n # Check some edge cases that are easy to work out by hand.\n assert candidate(\"#$a^D\") == \"#$A^d\"\n assert candidate(\"#ccc\") == \"#CCC\"\n\n # Don't remove this line:\n","code":"def solve(s):"} -{"task_id":"JHumanEval\/162","prompt_en":"def string_to_md5(text):\n \"\"\"\n Given a string 'text', return its md5 hash equivalent string.\n If 'text' is an empty string, return None.\n\n >>> string_to_md5('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'\n \"\"\"\n","prompt":"文字列 text が与えられたとき、その md5 ハッシュと等価な文字列を返す。\n text' が空文字列の場合は None を返す。\n \n >>> string_to_md5('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'","entry_point":"string_to_md5","canonical_solution":" import hashlib\n return hashlib.md5(text.encode('ascii')).hexdigest() if text else None\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'\n assert candidate('') == None\n assert candidate('A B C') == '0ef78513b0cb8cef12743f5aeb35f888'\n assert candidate('password') == '5f4dcc3b5aa765d61d8327deb882cf99'\n\n # Check some edge cases that are easy to work out by hand.\n assert True\n\n","code":"def string_to_md5(text):"} -{"task_id":"JHumanEval\/163","prompt_en":"def generate_integers(a, b):\n \"\"\"\n Given two positive integers a and b, return the even digits between a\n and b, in ascending order.\n\n For example:\n generate_integers(2, 8) => [2, 4, 6, 8]\n generate_integers(8, 2) => [2, 4, 6, 8]\n generate_integers(10, 14) => []\n \"\"\"\n","prompt":"正の整数aとbが与えられたとき、aとbの間にある偶数の数字を昇順で返してください。\n\n 例えば:\n generate_integers(2, 8) => [2, 4, 6, 8]\n generate_integers(8, 2) => [2, 4, 6, 8]\n generate_integers(10, 14) => []","entry_point":"generate_integers","canonical_solution":" lower = max(2, min(a, b))\n upper = min(8, max(a, b))\n\n return [i for i in range(lower, upper+1) if i % 2 == 0]\n","test":"def check(candidate):\n\n # Check some simple cases\n assert candidate(2, 10) == [2, 4, 6, 8], \"Test 1\"\n assert candidate(10, 2) == [2, 4, 6, 8], \"Test 2\"\n assert candidate(132, 2) == [2, 4, 6, 8], \"Test 3\"\n assert candidate(17,89) == [], \"Test 4\"\n\n # Check some edge cases that are easy to work out by hand.\n assert True, \"This prints if this assert fails 2 (also good for debugging!)\"\n\n","code":"def generate_integers(a, b):"}