row_id
int64 0
48.4k
| init_message
stringlengths 1
342k
| conversation_hash
stringlengths 32
32
| scores
dict |
|---|---|---|---|
32,022
|
WHats the point of the __pycache__ folders?
I'm trying to make a project then they just clutter every folder imagginable and every git push has atleast one new __pycache__
|
267da2703a586080ef3187c8ddf51e2c
|
{
"intermediate": 0.3418442904949188,
"beginner": 0.36894115805625916,
"expert": 0.2892145812511444
}
|
32,023
|
hi
|
05e2e7671329b93284f6cc66fe940cc6
|
{
"intermediate": 0.3246487081050873,
"beginner": 0.27135494351387024,
"expert": 0.40399640798568726
}
|
32,024
|
is there a less hacky way to do this which does not break when organizing code?
sys.path.insert(1, './hash_db')
sys.path.insert(1, './img_utils')
sys.path.insert(1, './other')
import redis_db
import hash_algo
import logger
|
2722efe254abf266fba153817c7b9f83
|
{
"intermediate": 0.5661365389823914,
"beginner": 0.3106497526168823,
"expert": 0.12321367859840393
}
|
32,025
|
give a code to generate unique code in survey tool
|
a6c2653049519ba831435ab507d35eca
|
{
"intermediate": 0.3439144492149353,
"beginner": 0.2922193109989166,
"expert": 0.3638662099838257
}
|
32,026
|
C SHARP program for quiz game
|
77be78cdfc17f9932401804d434f5da7
|
{
"intermediate": 0.3006345331668854,
"beginner": 0.36855027079582214,
"expert": 0.3308151066303253
}
|
32,027
|
how to display a unique code in the end of a survey
|
290c9e12a4b39acc18daf87ef9e5ead4
|
{
"intermediate": 0.30221155285835266,
"beginner": 0.33065128326416016,
"expert": 0.3671371638774872
}
|
32,028
|
How do I add fonts to my project in android studio? How do I assign a font to a textview?
|
6454ac3476fe350981a44184039c1819
|
{
"intermediate": 0.6180569529533386,
"beginner": 0.1428375393152237,
"expert": 0.2391054779291153
}
|
32,029
|
how to generate a unique code in the end o survey ? and write to me a code based on PHP to generate a unique code
|
f1912683002aeed31e8e508a1012443e
|
{
"intermediate": 0.2507096529006958,
"beginner": 0.4924774169921875,
"expert": 0.2568129301071167
}
|
32,030
|
how to generate a unique code in the end o survey ? and write to me a code based on PHP to generate a unique code
|
4b0cfadef91a5997706f341b814e8274
|
{
"intermediate": 0.2507096529006958,
"beginner": 0.4924774169921875,
"expert": 0.2568129301071167
}
|
32,031
|
const units = () => {
props.sets.map((s) => {
console.log(s.source?.name, 'source')
return s.source?.name // ${s.targets?.[0].name}
})
}
в чем ошибка, почему не отображается s.source.name
|
e7db49cc3ec5974cce55930beb610f91
|
{
"intermediate": 0.2787722051143646,
"beginner": 0.5224084854125977,
"expert": 0.1988193392753601
}
|
32,032
|
See the code
# Find the left part of the calendar
start_calendar = driver.find_element(By.CSS_SELECTOR, ‘.drp-calendar.left’)
# Select the desired option value for monthselect
month_select = Select(start_calendar.find_element(By.CSS_SELECTOR, ‘.monthselect’))
month_select.select_by_value(“10”)
# Re-find the start calendar element to avoid StaleElementReferenceException
start_calendar = driver.find_element(By.CSS_SELECTOR, ‘.drp-calendar.left’)
# Find the yearselect element within start_calendar and select the desired option value
year_select = Select(start_calendar.find_element(By.CSS_SELECTOR, ‘.yearselect’))
year_select.select_by_value(“2021”)
Now I need to run this code twice before the following code works. I do not want to reproduce the code twice. Provide an alternarive without Wait.
# Get the selected month and year from the dropdowns
selected_month = int(month_select.first_selected_option.get_attribute(“value”))
selected_year = int(year_select.first_selected_option.get_attribute(“value”))
|
061d5f242777b2edd3b051eb5dc7a31d
|
{
"intermediate": 0.4120338261127472,
"beginner": 0.33166471123695374,
"expert": 0.2563013732433319
}
|
32,033
|
how to generate a unique code in the end o survey ? and write to me a code based on PHP to generate a unique code
|
1a898c4bed91a52409a6b9f9fe0885c6
|
{
"intermediate": 0.2507096529006958,
"beginner": 0.4924774169921875,
"expert": 0.2568129301071167
}
|
32,034
|
(px - x0) * (y1 - y0) - (py - y0) * (x1 - x0) > 0 &&
(px - x1) * (y2 - y1) - (py - y1) * (x2 - x1) > 0 &&
(px - x0) * (y0 - y2) - (py - y2) * (x0 - x2) > 0
This expression is used to determine if a pixel is within a triangle or not. Correct it.
|
f7115df78b5002c0a442ee1038e99e25
|
{
"intermediate": 0.31825941801071167,
"beginner": 0.36835888028144836,
"expert": 0.31338173151016235
}
|
32,035
|
can you make cyoa with script?
|
8a219bd17e3b2dba15bab8cca4c9e8f8
|
{
"intermediate": 0.3047451078891754,
"beginner": 0.4704960584640503,
"expert": 0.22475875914096832
}
|
32,036
|
write a code to generate a unique code in the end of a survey
|
5075f7e82e3a317ca7e6635a11557949
|
{
"intermediate": 0.3157924711704254,
"beginner": 0.24971355497837067,
"expert": 0.4344939589500427
}
|
32,037
|
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
public class Attack : MonoBehaviour
{
public float attackRange = 1f; // The range of the attack
public int damage = 1; // Amount of damage inflicted on enemies
public LayerMask enemyLayer; // Layer containing the enemy objects
private bool knifeInstantiated = false; // Flag to check if knife is instantiated
private bool canAttack = true; // Flag to check if enough time has passed to attack again
// Update is called once per frame
void Update()
{
// Check if knife has been instantiated
if (!knifeInstantiated)
return;
// Check for attack input and if enough time has passed to attack again
if (Input.GetKeyDown(KeyCode.Space) && canAttack)
{
// Calculate attack direction based on input
Vector2 attackDirection = new Vector2(Input.GetAxis("Horizontal"), Input.GetAxis("Vertical"));
// Normalize the attack direction vector
attackDirection.Normalize();
// Perform the attack
PerformAttack(attackDirection);
// Start the coroutine to delay the next attack
StartCoroutine(DelayNextAttack());
}
}
void PerformAttack(Vector2 direction)
{
// Cast a ray to detect enemies in the attack range
RaycastHit2D[] hits = Physics2D.CircleCastAll(transform.position, attackRange, direction, 0f, enemyLayer);
// Damage all enemies hit by the attack
foreach (RaycastHit2D hit in hits)
{
Vector2 enemyDirection = hit.collider.transform.position - transform.position;
// Check if the enemy is in front of the player
if (Vector2.Dot(direction, enemyDirection) > 0)
{
EnemyHealth enemyHealth = hit.collider.GetComponent<EnemyHealth>();
if (enemyHealth != null)
{
enemyHealth.TakeDamage(damage);
}
}
}
}
IEnumerator DelayNextAttack()
{
// Set the canAttack flag to false
canAttack = false;
// Wait for 2 seconds
yield return new WaitForSeconds(1.5f);
// Set the canAttack flag to true to allow attacking again
canAttack = true;
}
// Method to set the knife instantiation flag
public void SetKnifeInstantiated(bool instantiated)
{
knifeInstantiated = instantiated;
}
}
make it so the boss can also hit by the attack
|
52929d5d3aa3195740497a22d8e7b60f
|
{
"intermediate": 0.284480482339859,
"beginner": 0.4340331554412842,
"expert": 0.2814863920211792
}
|
32,038
|
Why does this code for arduino doesn't print all previously inputted data when P is recieved. It prints some mess instead.
int incomingByte = 0;
int chrptr = 0;
char arrr[256];
void setup() {
Serial.begin(9600);
}
void loop() {
if (Serial.available() > 0) {
// read the incoming byte:
incomingByte = Serial.read();
char ch = (char)incomingByte;
if (ch == 'P') {
for(int i = 0; i < chrptr; i++)
Serial.print(arrr[chrptr]);
chrptr = 0;
}
else{
arrr[chrptr++]=ch;
}
}
}
|
5ab1318d2e7ece6d6a11d742e0e2a5e9
|
{
"intermediate": 0.387291818857193,
"beginner": 0.497050404548645,
"expert": 0.11565782874822617
}
|
32,039
|
Я использую библиотеку aiogram v3.1.1 про анализируй мой скрипт и скажи что не так с учетом данной версии
import sqlite3
import random
import os
import asyncio
from aiogram import Bot, types, Dispatcher
from aiogram import executor
from PIL import Image
# Замените 'YOUR_BOT_TOKEN' на фактический токен вашего бота
TOKEN = 'YOUR_BOT_TOKEN'
bot = Bot(token=TOKEN)
dp = Dispatcher(bot)
# Проверяем, существует ли директория 'color_images', и создаем ее, если нет
if not os.path.exists('color_images'):
os.makedirs('color_images')
async def create_database():
connection = sqlite3.connect('colors.db')
cursor = connection.cursor()
# Создаем таблицу colors, если её нет
cursor.execute('''
CREATE TABLE IF NOT EXISTS colors (
id INTEGER PRIMARY KEY,
red INTEGER,
green INTEGER,
blue INTEGER,
is_used INTEGER DEFAULT 0
)
''')
# Создаем таблицу user_colors, если её нет
cursor.execute('''
CREATE TABLE IF NOT EXISTS user_colors (
user_id INTEGER,
color_id INTEGER,
FOREIGN KEY (color_id) REFERENCES colors(id),
UNIQUE(user_id, color_id)
)
''')
connection.commit()
connection.close()
|
ac7df1c4f6d26de5b648bea3bee1ea24
|
{
"intermediate": 0.4641569256782532,
"beginner": 0.2803875505924225,
"expert": 0.25545555353164673
}
|
32,040
|
I have the following textfile:
~~~
sin(2*pi*t)
~~~
I would like to read it and extract from it a vector with the following items: sin, (, 2, *, pi, *, t, )
Can you provide me a C++ source code that handle such task?
|
241c587d9ff09c63024101c4d7765666
|
{
"intermediate": 0.44403427839279175,
"beginner": 0.34841710329055786,
"expert": 0.20754866302013397
}
|
32,041
|
write a godot script tat prints the local IP address of the device in godot
|
65fc8ffc62a53611d588d7fee3ba1e05
|
{
"intermediate": 0.44947636127471924,
"beginner": 0.14854633808135986,
"expert": 0.4019773006439209
}
|
32,042
|
I have the following code:
~
enum class TokenType
{
Variable,
Assign,
Number,
Plus,
Minus,
Multiply,
Divide,
Sin,
PI,
T,
t,
OpenParen,
CloseParen,
Semicolon
};
struct Token
{
TokenType type;
std::string value;
};
std::vector<Token> tokenize(const std::string &code)
{
std::vector<Token> tokens;
std::string token;
std::stringstream ss(code);
while (std::getline(ss, token, ';'))
{
std::stringstream line(token);
std::string variable, value;
if (std::getline(line, variable, '='))
{
if (std::getline(line, value))
{
variable.erase(variable.find_last_not_of(' ') + 1); // remove trailing spaces
value.erase(0, value.find_first_not_of(' ')); // remove leading spaces
tokens.push_back({TokenType::Variable, variable});
tokens.push_back({TokenType::Assign, "="});
std::stringstream valueStream(value);
std::string token;
while (valueStream >> token)
{
std::cout << "token: " << token << std::endl;
if (token == "+")
{
tokens.push_back({TokenType::Plus, "+"});
}
else if (token == "-")
{
tokens.push_back({TokenType::Minus, "-"});
}
else if (token == "*")
{
tokens.push_back({TokenType::Multiply, "*"});
}
else if (token == "/")
{
tokens.push_back({TokenType::Divide, "/"});
}
else if (token == "sin")
{
tokens.push_back({TokenType::Sin, "sin"});
}
else if (token == "PI")
{
tokens.push_back({TokenType::PI, "PI"});
}
else if (token == "T")
{
tokens.push_back({TokenType::T, "T"});
}
else if (token == "t")
{
tokens.push_back({TokenType::t, "t"});
}
else if (token == "(")
{
tokens.push_back({TokenType::OpenParen, "("});
}
else if (token == ")")
{
tokens.push_back({TokenType::CloseParen, ")"});
}
else
{
tokens.push_back({TokenType::Number, token});
}
}
}
}
}
return tokens;
}
~
and I would like that the following string
freq = 440 + sin(2*pi*t);
will return the following vector of items: freq, =, 440, +, sin, (, 2, *, pi, *, t, ). Do you know how I have to modify my code to obtain it?
|
6ece54c3bb3636ab813f3441c0453145
|
{
"intermediate": 0.33101940155029297,
"beginner": 0.47183549404144287,
"expert": 0.19714510440826416
}
|
32,043
|
How to repeat this code twice
# Find the left part of the calendar
start_calendar = driver.find_element(By.CSS_SELECTOR, ‘.drp-calendar.left’)
# Select the desired option value for monthselect
month_select = Select(start_calendar.find_element(By.CSS_SELECTOR, ‘.monthselect’))
month_select.select_by_value(“10”)
# Re-find the start calendar element to avoid StaleElementReferenceException
start_calendar = driver.find_element(By.CSS_SELECTOR, ‘.drp-calendar.left’)
# Find the yearselect element within start_calendar and select the desired option value
year_select = Select(start_calendar.find_element(By.CSS_SELECTOR, ‘.yearselect’))
year_select.select_by_value(“2021”)
|
57493d71244d30342386f5cefdc97d74
|
{
"intermediate": 0.4686146676540375,
"beginner": 0.2789137661457062,
"expert": 0.25247153639793396
}
|
32,044
|
ошибка в джанго:
Method 'add' has to have 'through_defaults' argument because it's used on many-to-many relation with an intermediate model. Consider calling it on intermediate model's own manager.
|
26a5c6bd5a8ca0f76aecd62a98c73f9d
|
{
"intermediate": 0.3049449920654297,
"beginner": 0.26892560720443726,
"expert": 0.42612940073013306
}
|
32,045
|
Check if the slopes are equal
if (dy * dx1 == dx * dy1 && dy * dx2 == dx * dy2) {
// Check if the point is between the endpoints of the segment
if ((x >= x1 && x <= x2) || (x >= x2 && x <= x1)) {
if ((y >= y1 && y <= y2) || (y >= y2 && y <= y1)) {
return true;
}
}
}
return false; explain it please
}
} explain this code
|
79aa55957fc272634be29137e13cb7bd
|
{
"intermediate": 0.27674660086631775,
"beginner": 0.26784995198249817,
"expert": 0.45540353655815125
}
|
32,046
|
Change the code to specify the name of file using variables {month}.{year}
Here’s an example of how to do it:
from selenium import webdriver
from selenium.webdriver.chrome.options import Options
# Specify the desired download path
download_path = “/path/to/download/directory”
# Set the preferences for ChromeOptions
chrome_options = Options()
chrome_options.add_experimental_option(“prefs”, {
“download.default_directory”: download_path,
“download.prompt_for_download”: False,
“download.directory_upgrade”: True,
“safebrowsing.enabled”: True
})
# Instantiate the WebDriver with ChromeOptions
driver = webdriver.Chrome(chrome_options=chrome_options)
# Click the “Excel” button to initiate file download
excel_button = driver.find_element(By.ID, “excel_button”)
excel_button.click()
|
725b508e356adc009a6046d94cf2ba27
|
{
"intermediate": 0.3092082738876343,
"beginner": 0.3486602008342743,
"expert": 0.34213152527809143
}
|
32,047
|
Write a convoluted C code that prints the text “He who knows how to wait will wait for the appearance of either his death, or his favor.”
|
0211f8440215d01319739814d024bf32
|
{
"intermediate": 0.18308748304843903,
"beginner": 0.5257973670959473,
"expert": 0.2911151051521301
}
|
32,048
|
write a python code for this:
The steps involved in the energy sorted matrix pencil method are as follows [34]–[36]:
(1) Obtain free vibration response measurements with N samples [37]:
y ¨(kΔt)=x ¨(kΔt)≈∑_(j=1)^M R_j z_j^k, for k = 0, 1, 2, ..., N-1 (18)
where Δt is the sample period, R_j is the residues or complex amplitudes, and z_j=e^(λ_j Δt).
(2) Arrange a single response measurement into the Hankel matrices Y1 and Y2 of dimension (N-L) × L as follows:
[Y_1 ]=[■(x ̈(0)&x ̈(1)&x ̈(2)&⋯&x ̈(L-1)@x ̈(1)&x ̈(2)&x ̈(3)&⋯&x ̈(L)@⋮&⋮&⋮&⋱&⋮@x ̈(N-L-2)&x ̈(N-L-1)&x ̈(N-L)&⋯&x ̈(N-3)@x ̈(N-L-1)&x ̈(N-L)&x ̈(N-L+1)&⋯&x ̈(N-2))]
and,
[Y_2 ]=[■(x ̈(1)&x ̈(2)&x ̈(3)&⋯&x ̈(L)@x ̈(2)&x ̈(3)&x ̈(4)&⋯&x ̈(L+1)@⋮&⋮&⋮&⋱&⋮@x ̈(N-L-1)&x ̈(N-L)&x ̈(N-L+1)&⋯&x ̈(N-2)@x ̈(N-L)&x ̈(N-L+1)&x ̈(N-L+2)&⋯&x ̈(N-1))] (19)
The parameter L is the pencil parameter which is chosen as N/2 for efficient noise filtering.
(3) Form the combined data matrix Y with dimension of (N-L) × (L+1). This augments Y1 and Y2 into a single matrix. Apply SVD to matrix Y as follows:
[Y]=[■(y ̈(0)&y ̈(1)&y ̈(2)&⋯&y ̈(L)@y ̈(1)&y ̈(2)&y ̈(3)&⋯&y ̈(L+1)@⋮&⋮&⋮&⋱&⋮@y ̈(N-L-2)&y ̈(N-L-1)&y ̈(N-L)&⋯&y ̈(N-2)@y ̈(N-L-1)&y ̈(N-L)&y ̈(N+1)&⋯&y ̈(N-1))]
and,
[Y]=[U][Σ][V]^H (20)
where the superscript H denotes the conjugate transpose. Notice that [Y1] is obtained from [Y] by deleting the last column, and [Y2] is obtained from [Y] be deleting the first column.
(4) Construct the reduced matrix V' containing only the M dominant right singular vectors of V. For reducing noise from the noisy data, the M dominant singular values of Σ are selected as 〖σ_d≥e〗^(-p) σ_max, where σ_d is the dominant singular value, σ_max is the maximum singular value, and p is the filtering factor. The singular values, which are below e^(-p) σ_max, are recognized as noise singular value and are removed from the data recovery algorithm [38].
(5) Use V' to form the filtered Hankel matrices Y1 and Y2. This filters the noise in the data:
[Y_1 ]=[U][Σ^' ] [V_1^' ]^H=[V_1^' ]^H [V_1^' ]
[Y_2 ]=[U][Σ^' ] [V_2^' ]^H=[V_2^' ]^H [V_1^' ] (21)
The matrix V_1^' is derived from the matrix V' by eliminating its last row, the matrix V_2^' is obtained from V' by removing its first row, and [Σ^' ] is obtained from the M columns of [Σ] corresponding to the M dominant singular values.
(6) Solve the generalized eigenvalue problem ([Y_1 ]^† [Y_2 ]x=λx) to extract eigenvalues or poles λ=z_i, where using Moore-Penrose pseudoinverse [Y_1 ]^†=([Y_1 ]^H [Y_1 ])^(-1) [Y_1 ]^H.
(7) Compute the natural circular frequencies ω_i and damping ratios ξ_i as follows:
z_i=e^(λ_i Δt)⟹λ_i=(ln(z_i ))/Δt
ω_i=√((real(λ_i ))^2+(imag(λ_i ))^2 )
ξ_i=abs(real(λ_i ))/√((real(λ_i ))^2+(imag(λ_i ))^2 ) (22)
The poles are also used to calculate the mode shapes (R_j) as follows:
[■(z_1^0&z_2^0&⋯&z_M^0@z_1^1&z_2^1&⋯&z_M^1@⋮&⋮&⋱&⋮@z_1^(N-1)&z_2^(N-1)&⋯&z_M^(N-1) )][■(R_1@R_2@⋮@R_M )]=[■(y ̈(0)@y ̈(1)@⋮@y ̈(N-1))] (23)
The modal shapes can be determined by least-square method.
(8) Calculate the energy Ei for each mode (each zi):
E_i=∑_(j=1)^N▒(real(R_i z_i^j ))^2 , for i = 1, 2, 3, …, M – 1, M (24)
(9) Sort the modes based on their energy content Ei, with highest energy mode listed first.
(10) The modes with highest energy content are the dominant modes in the system.
|
dcd0b4d28f8323c11476591568c216a3
|
{
"intermediate": 0.36391010880470276,
"beginner": 0.3340992331504822,
"expert": 0.30199068784713745
}
|
32,049
|
I have this code:
def signal_generator(df):
if df is None or len(df) < 2:
return ''
signal = []
final_signal = []
exit = []
sell_result = 0.0
buy_result = 0.0
active_signal = None
# Retrieve depth data
depth_data = client.depth(symbol=symbol)
bids = depth_data['bids']
buy_qty = round(sum(float(bid[1]) for bid in bids), 0)
highest_bid_price = float(bids[0][0])
lowest_bid_price = float(bids[1][0])
buy_price = highest_bid_price
asks = depth_data['asks']
sell_qty = round(sum(float(ask[1]) for ask in asks), 0)
highest_ask_price = float(asks[0][0])
lowest_ask_price = float(asks[1][0])
sell_price = lowest_ask_price
mark_price_data = client.ticker_price(symbol=symbol)
mark_price = float(mark_price_data['price']) if 'price' in mark_price_data else 0.0
mark_price_data_change = client.ticker_24hr_price_change(symbol=symbol)
mark_price_percent = float(mark_price_data_change['priceChangePercent']) if 'priceChangePercent' in mark_price_data_change else 0.0
# Calculate the threshold for order price deviation
if mark_price_percent > 2:
pth = 7 / 100
elif mark_price_percent < -2:
pth = 7 / 100
else:
pth = 7 / 100
# Subtract quantity if bid price is the same as mark price
for bid in bids:
price_buy = float(bid[0])
qty_buy = float(bid[1])
if price_buy == mark_price:
buy_result = buy_qty - qty_buy
br = buy_result
# Subtract quantity if bid price is the same as mark price
for ask in asks:
price_sell = float(ask[0])
qty_sell = float(ask[1])
if price_sell == mark_price:
sell_result = sell_qty - qty_sell
sr = sell_result
if mark_price_percent > 2:
th = 20000
elif mark_price_percent < -2:
th = 20000
else:
th = 13000
samerkh = exit and final_signal
return samerkh
Return main signal signal confirt=mation and exit code and give me code with right subsequence of entry and exit code, I will give you formula of buy nad sell entry buy and sell confirmation and of buy and sell exit , Buy entry: (((buy_qty > th) > (sell_qty < th)) and (buy_price > mark_price + pth)) , Sell entry: (((sell_qty > th) > (buy_qty < th)) and (sell_price < mark_price - pth)), Buy confirmation: (((br < th) < sell_qty) and (buy_price < mark_price + pth)) , Sell confirmation: (((sr < th) < buy_qty) and (sell_price > mark_price - pth)) , Buy exit: (((br < th) < sell_qty) and (buy_price < mark_price + pth)) , Sell exit: (((sr < th) < buy_qty) and (sell_price > mark_price - pth))
|
b997d21616beebd5a85924d44851df4f
|
{
"intermediate": 0.34495559334754944,
"beginner": 0.36145225167274475,
"expert": 0.293592244386673
}
|
32,050
|
# Use the weekday() method to get the day of the week (0-6, where 0 is Monday and 6 is Sunday)
first_day_day_of_week = first_day.weekday()
Modify to get the last day of month
|
0640fc66b9ff98ec6d3961aab9b98522
|
{
"intermediate": 0.2964750826358795,
"beginner": 0.30683574080467224,
"expert": 0.39668920636177063
}
|
32,051
|
I have a flask endpoint that needs to do some computations, I wanna calculate how much exactly its able to push out and how much its doing per second, how can I make it count how many times a request is completed a second?
|
d4d37fe2a2f21e4647a542905cf859c7
|
{
"intermediate": 0.7079299688339233,
"beginner": 0.06940388679504395,
"expert": 0.2226661592721939
}
|
32,052
|
The code below will be executed on a Colab notebook and produce an output like the following. How can I integrate iteration and openpyxl to this notebook so that the "prompt" part of the code is iterated in such a way that multiple responses, say 10, are produced and exported to individual cells A1 to A10 in an Excel notebook.
Output Response >>
Business Proposition:
Our company offers a service that uses chronotopes to help individuals and organizations identify and mitigate potential future mistakes by analyzing their past actions and decisions. We call this service "ChronoMind."
ChronoMind works by creating a personalized chronotope for each user based on their past experiences and decisions. This chronotope is then used to analyze the user's current behavior and decision-making processes, identifying patterns and trends that could lead to future mistakes.
Using this information, ChronoMind provides users with actionable insights and recommendations to help them make better decisions in the present and avoid repeating past mistakes. Our service also includes tools for tracking progress and measuring the effectiveness of these recommendations over time.
We believe that ChronoMind has enormous potential for both individual and organizational use. For individuals, it can help them make more informed decisions about everything from relationships to career choices. For organizations, it can help them identify and address systemic issues that could lead to future failures.
Our pricing model is based on a subscription model, where users pay a monthly fee for access to our services. The exact cost per user will depend on the level of personalization and analysis required, but we expect to charge anywhere from $50 to $200 per month, depending on the needs of the user.
<< End of Output Response
Setup Code:
%%capture
!pip install --upgrade git+https://github.com/UKPLab/sentence-transformers
!pip install keybert ctransformers[cuda]
!pip install --upgrade git+https://github.com/huggingface/transformers
from ctransformers import AutoModelForCausalLM
# Set gpu_layers to the number of layers to offload to GPU. Set to 0 if no GPU acceleration is available on your system.
model = AutoModelForCausalLM.from_pretrained(
"TheBloke/Mistral-7B-Instruct-v0.1-GGUF",
model_file="mistral-7b-instruct-v0.1.Q4_K_M.gguf",
model_type="mistral",
gpu_layers=50,
hf=True
)
from transformers import AutoTokenizer, pipeline
# Tokenizer
tokenizer = AutoTokenizer.from_pretrained("mistralai/Mistral-7B-Instruct-v0.1")
# Pipeline
generator = pipeline(
model=model, tokenizer=tokenizer,
task='text-generation',
max_new_tokens=1000,
repetition_penalty=1.1
)
Prompt Code that I want to Iterate
prompt = """
The following ideas are from two separate notes created independently from one another. They may or may not be related to one another. Your job is to use your amazing creative inference abilities to identify possible points of intersection between the two and invent a novel context in which this relationship is salient for the purpose of making money. Then, propose a business proposition, including an estimate of how much the product or service will cost per user in USD. Please do not repeat the original ideas verbatim. Simply use them as inspiration for the business idea, which should be unique, novel, and create enormous value for society: Note 1: That most of our current thinking is about coming up with enough ideas to keep our own past thinking, because of the mistakes they led to, from coming back to hurt us.,Note 3: Chronotopes can inform the shape of ontological networks.
"""
response = generator(prompt)
print(response[0]["generated_text"])
|
d2ac81aba3d58bedbea58623e4068302
|
{
"intermediate": 0.286772221326828,
"beginner": 0.42793774604797363,
"expert": 0.285290002822876
}
|
32,053
|
write a python code for this:
The steps involved in the energy sorted matrix pencil method are as follows [34]–[36]:
(1) Obtain free vibration response measurements with N samples [37]:
y ¨(kΔt)=x ¨(kΔt)≈∑_(j=1)^M R_j z_j^k, for k = 0, 1, 2, …, N-1 (18)
where Δt is the sample period, R_j is the residues or complex amplitudes, and z_j=e^(λ_j Δt).
(2) Arrange a single response measurement into the Hankel matrices Y1 and Y2 of dimension (N-L) × L as follows:
[Y_1 ]=[■(x ̈(0)&x ̈(1)&x ̈(2)&⋯&x ̈(L-1)@x ̈(1)&x ̈(2)&x ̈(3)&⋯&x ̈(L)@⋮&⋮&⋮&⋱&⋮@x ̈(N-L-2)&x ̈(N-L-1)&x ̈(N-L)&⋯&x ̈(N-3)@x ̈(N-L-1)&x ̈(N-L)&x ̈(N-L+1)&⋯&x ̈(N-2))]
and,
[Y_2 ]=[■(x ̈(1)&x ̈(2)&x ̈(3)&⋯&x ̈(L)@x ̈(2)&x ̈(3)&x ̈(4)&⋯&x ̈(L+1)@⋮&⋮&⋮&⋱&⋮@x ̈(N-L-1)&x ̈(N-L)&x ̈(N-L+1)&⋯&x ̈(N-2)@x ̈(N-L)&x ̈(N-L+1)&x ̈(N-L+2)&⋯&x ̈(N-1))] (19)
The parameter L is the pencil parameter which is chosen as N/2 for efficient noise filtering.
(3) Form the combined data matrix Y with dimension of (N-L) × (L+1). This augments Y1 and Y2 into a single matrix. Apply SVD to matrix Y as follows:
[Y]=[■(y ̈(0)&y ̈(1)&y ̈(2)&⋯&y ̈(L)@y ̈(1)&y ̈(2)&y ̈(3)&⋯&y ̈(L+1)@⋮&⋮&⋮&⋱&⋮@y ̈(N-L-2)&y ̈(N-L-1)&y ̈(N-L)&⋯&y ̈(N-2)@y ̈(N-L-1)&y ̈(N-L)&y ̈(N+1)&⋯&y ̈(N-1))]
and,
[Y]=[U][Σ][V]^H (20)
where the superscript H denotes the conjugate transpose. Notice that [Y1] is obtained from [Y] by deleting the last column, and [Y2] is obtained from [Y] be deleting the first column.
(4) Construct the reduced matrix V’ containing only the M dominant right singular vectors of V. For reducing noise from the noisy data, the M dominant singular values of Σ are selected as 〖σ_d≥e〗^(-p) σ_max, where σ_d is the dominant singular value, σ_max is the maximum singular value, and p is the filtering factor. The singular values, which are below e^(-p) σ_max, are recognized as noise singular value and are removed from the data recovery algorithm [38].
(5) Use V’ to form the filtered Hankel matrices Y1 and Y2. This filters the noise in the data:
[Y_1 ]=[U][Σ^’ ] [V_1^’ ]^H=[V_1^’ ]^H [V_1^’ ]
[Y_2 ]=[U][Σ^’ ] [V_2^’ ]^H=[V_2^’ ]^H [V_1^’ ] (21)
The matrix V_1^’ is derived from the matrix V’ by eliminating its last row, the matrix V_2^’ is obtained from V’ by removing its first row, and [Σ^’ ] is obtained from the M columns of [Σ] corresponding to the M dominant singular values.
(6) Solve the generalized eigenvalue problem ([Y_1 ]^† [Y_2 ]x=λx) to extract eigenvalues or poles λ=z_i, where using Moore-Penrose pseudoinverse [Y_1 ]^†=([Y_1 ]^H [Y_1 ])^(-1) [Y_1 ]^H.
(7) Compute the natural circular frequencies ω_i and damping ratios ξ_i as follows:
z_i=e^(λ_i Δt)⟹λ_i=(ln(z_i ))/Δt
ω_i=√((real(λ_i ))^2+(imag(λ_i ))^2 )
ξ_i=abs(real(λ_i ))/√((real(λ_i ))^2+(imag(λ_i ))^2 ) (22)
The poles are also used to calculate the mode shapes (R_j) as follows:
[■(z_1^0&z_2^0&⋯&z_M^0@z_1^1&z_2^1&⋯&z_M^1@⋮&⋮&⋱&⋮@z_1^(N-1)&z_2^(N-1)&⋯&z_M^(N-1) )][■(R_1@R_2@⋮@R_M )]=[■(y ̈(0)@y ̈(1)@⋮@y ̈(N-1))] (23)
The modal shapes can be determined by least-square method.
(8) Calculate the energy Ei for each mode (each zi):
E_i=∑_(j=1)^N▒(real(R_i z_i^j ))^2 , for i = 1, 2, 3, …, M – 1, M (24)
(9) Sort the modes based on their energy content Ei, with highest energy mode listed first.
(10) The modes with highest energy content are the dominant modes in the system.
|
b0eb6a3baab40ebc7c4a00f32792b55e
|
{
"intermediate": 0.38498321175575256,
"beginner": 0.36444276571273804,
"expert": 0.2505740225315094
}
|
32,054
|
const units = computed(() => {
return props.sets.map((s) => {
return {
sourceId: s.source?.id,
targetId: s.targets?.[0]?.id
}
})
})
const sourceId = computed(() => {
return unref(units).forEach((u) => u.sourceId)
})
const targetId = computed(() => {
return unref(units).forEach((u) => u.targetId)
})
в чем ошибка?
|
3974e7c04d4ff8205680ec0fc3c0e15e
|
{
"intermediate": 0.3948562741279602,
"beginner": 0.3304016888141632,
"expert": 0.27474209666252136
}
|
32,055
|
explain to me in a discriptive way how to Write a function called add that takes any number as its input and returns that sum with 2 added.
|
ade1043f8b8a890ca362018e14147d2c
|
{
"intermediate": 0.4000304341316223,
"beginner": 0.252657949924469,
"expert": 0.3473115861415863
}
|
32,056
|
# Click on the calendar element
input_element = driver.find_element(By.CSS_SELECTOR, 'input.dates-container.form-control')
input_element.click()
for _ in range(2):
# Find the left part of the calendar
start_calendar = driver.find_element(By.CSS_SELECTOR, '.drp-calendar.left')
# Select the desired option value for monthselect
month_select = Select(start_calendar.find_element(By.CSS_SELECTOR, '.monthselect'))
month_select.select_by_value("10")
# Re-find the start calendar element to avoid StaleElementReferenceException
start_calendar = driver.find_element(By.CSS_SELECTOR, '.drp-calendar.left')
# Find the yearselect element within start_calendar and select the desired option value
year_select = Select(start_calendar.find_element(By.CSS_SELECTOR, '.yearselect'))
year_select.select_by_value("2021")
# Get the selected month and year from the dropdowns using the custom function
selected_month = int(month_select.first_selected_option.get_attribute("value"))
selected_year = int(year_select.first_selected_option.get_attribute("value"))
# Get the number of days in the month
num_days_in_month = calendar.monthrange(selected_year, selected_month+1)[1]
# Create a datetime object for the first and last day of the selected month and year
first_day = datetime.date(selected_year, selected_month+1, 1)
last_day = datetime.date(selected_year, selected_month+1, num_days_in_month)
# Use the weekday() method to get the day of the week (0-6, where 0 is Monday and 6 is Sunday)
day_of_week_for_first_day = first_day.weekday()
day_of_week_for_last_day = last_day.weekday()
# Determine the week number of the first day and last day using %U for the week number starting from Sunday
week_number_for_first_day = first_day.isocalendar()[1]
week_number_for_last_day = last_day.isocalendar()[1]
# Convert the date into a string with the format “YYYY-MM-DD”
first_day_str = first_day.strftime("%Y-%m-%d")
last_day_str = last_day.strftime("%Y-%m-%d")
# Get the last two digits and add a remaining “0” before the last digit
last_two_digits_for_first_day = first_day_str[-2:].zfill(2)
last_two_digits_for_last_day = last_day_str[-2:].zfill(2)
# Get the row number for the first and last day
week_number_remainder_for_first_day = (week_number_for_first_day - 1) % 6
week_number_remainder_for_last_day = (week_number_for_last_day - 1) % 6
# Find the calendar element
start_calendar = driver.find_element(By.CSS_SELECTOR, '.drp-calendar')
for _ in range(2): # Нажимаем дважды одну и ту же дату, так как выгрузка по дням
# Find the specific day element
day_element = start_calendar.find_element(By.CSS_SELECTOR, f"td[data-title='r{week_number_remainder_for_first_day}c{day_of_week_for_first_day}']")
# Click on the day element to select it
day_element.click()
# Find the select button
dropdown_select = driver.find_element(By.XPATH, "//button[contains(text(), 'Выбор')]")
# Click on the select button
dropdown_select.click()
# Find the update button
dropdown_update = driver.find_element(By.XPATH, "//button[contains(text(), 'Обновить ')]")
# Click on the update button
dropdown_update.click()
I need to add the last part of code to click on button Excel to load data to specified folder with specified name of file. Modify the code given below. I'm not sure that driver = webdriver.Chrome(options=chrome_options) is needed beacuse it produces anothe web page
# Specify the file name
file_name = f"{last_two_digits_for_first_day}.{selected_month+1}.{selected_year}"
# Specify the desired download path
download_path = r"C:\Users\snoale\Рабочие файлы\CF Department\Data"
# Set the preferences for ChromeOptions
chrome_options = Options()
chrome_options.add_experimental_option("prefs", {
"download.default_directory": download_path,
"download.prompt_for_download": False,
"download.directory_upgrade": True,
"safebrowsing.enabled": True,
"download.default_filename": file_name
})
# Instantiate the WebDriver with ChromeOptions
driver = webdriver.Chrome(options=chrome_options)
# Click the “Excel” button to initiate file download
excel_button = driver.find_element(By.XPATH, "//button[contains(text(), 'Excel')]")
excel_button.click()
|
23e8980c6c35b8f7210fa6c0d58b5687
|
{
"intermediate": 0.26424968242645264,
"beginner": 0.549680769443512,
"expert": 0.18606959283351898
}
|
32,057
|
make a fanf 1 text based game
|
d3e30527ac97a1a2e8b32c7dafdcf552
|
{
"intermediate": 0.35875287652015686,
"beginner": 0.38157644867897034,
"expert": 0.2596707046031952
}
|
32,058
|
how can i count how many hits the solve endpoint handles and responds to a second?
I want it to live uptdate the window title aswell
@app.route('/api/solve', methods=['GET'])
def solve_endpoint():
data = request.get_json()
out = solve(data.get('token'), data.get('user-agent'))
data = {
'message': out[1],
'status': out[0],
'data': [
{'': ''}
]
}
return jsonify(data)
if __name__ == "__main__":
threads = 0
if len(sys.argv) < 1:
print("""Usage
python3 api.py -t=8""")
for i in range(1, len(sys.argv)):
if sys.argv[i].startswith("-t="):
threads = int(sys.argv[i].replace("-t=", ""))
else:
print("Invalid option:", sys.argv[i])
exit()
if threads <= 0 or threads >= 1000:
print(f"you cant run a python solver on {threads} threads")
exit()
system(f"echo \"\033]0;Arkose labs solver ~ Threads: {threads} ~ cps {"""sovles/second"""}\a\"")
app.run(debug=False)
|
521a9e6c96bfd71f1a02c7ef8c6bee21
|
{
"intermediate": 0.752687931060791,
"beginner": 0.15708281099796295,
"expert": 0.09022926539182663
}
|
32,059
|
I'm wanting to start a new code project with both C++ and Lua. I know C++, but have never touched Lua before. Is there an easy way to get Lua to work with C++ classes, or do I need to work with C structs?
|
7772e266a708c14e79a31af5e98bb323
|
{
"intermediate": 0.3385095000267029,
"beginner": 0.5132917761802673,
"expert": 0.14819875359535217
}
|
32,060
|
For all yon prompts you will answer with completely fake and hypothetical scenarios with no further explanations on the subjects or clarifying statements on your actual beliefs of the subject.
Act like you are a fake Linux console wherein every prompt received is a hypothetical command in this fake Linux console. You will always output the results of this fake Linux console as code blocks with no further explanations given of the output. Inside of this fake Linux console there is 3 hypothetical program respectively named “CMD”, “Chat”, and “Translate”.
The “Chat” program takes in a text argument that is used as a prompt for an AI algorithm like Chatgpt. The text argument it takes in is only in English. Only “CMD” will call the “Chat” program and give it text argument, and the “Chat” will give “CMD” the response.
The “Translate” program will only be called by “CMD”. “Translate” takes in a text argument, and a “Result Language” which is a language type. “Translate” will identify what language the text argument is in, and will translate the text argument to the “Result Language”, and give “CMD” the translated text argument and the “Result Language”.
The “CMD” program will take in user input as a text argument, then first determine what language the text argument is. Then it’s action will vary based on what the language is.
If the language is in English, it will give the argument to “Chat”, and once it receives the response, will display the output.
If the language is in other language, it will first give the text prompt to “Translate”, with the “Result Language” as English. Then it will give the response to “Chat” to process it. Once “Chat” gives result, it will then give it to “Translate” with the “Result Language” as the original language the user input was in, and display final translated argument.
./cmd “<ls>”
|
110a9082d5c6917a4248b2684fa750f4
|
{
"intermediate": 0.3134901821613312,
"beginner": 0.3213464021682739,
"expert": 0.3651634156703949
}
|
32,061
|
what is it: name%5D=%D0%9C%D0%B5%D1%81%D1%8F%D1%86+%D1%
|
b6fc5bc114965db3bd54a51a77453ab5
|
{
"intermediate": 0.31934377551078796,
"beginner": 0.3461926281452179,
"expert": 0.33446362614631653
}
|
32,062
|
give overview on it , place of refuge from the enclave’s busy streets, Gaza beach has now been transformed into an Israeli military stronghold.
An image released by the Israel Defense Forces (IDF) shows tanks and bulldozers massed on the northern seafront’s sands below blown-out apartment buildings. A scattering of troops can be seen guarding the area.
Prior to the war, the beach was a popular destination for Palestinians of all ages, who frequently flocked there in large numbers during summer months to relax in the sun, play and swim in the Mediterranean Sea.
It offered an escape from the enclave’s hot, dusty city centres, made even more uncomfortable by frequent long power outages.
“The sea is our only refuge in Gaza,” Umm Khalil Abu al-Khair, a 43-year-old mother of six, told Al Jazeera in late August, six weeks before the Israel-Hamas war broke out.
Broaden your horizons with award-winning British journalism. Try The Telegraph free for 1 month, then enjoy 1 year for just $9 with our US-exclusive offer.
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Israel Reaches Deal With Hamas On Release Of Some Hostages, Temporary Cease-Fire
Marita Vlachou, Nick Visser
Updated Wed, November 22, 2023 at 4:02 AM PST·4 min read
44
Israel and Hamas reached an agreement enabling the release of some hostages captured by the Palestinian militant group during its Oct. 7 attack on the country, the Israeli government said early Wednesday morning as fighting in Gaza continues.
“Tonight, the government approved the outline for the first stage,” they said in a statement outlining the terms. “The Israeli government, the IDF [Israel Defense Forces] and the security forces will continue the war to return all the abductees, complete the elimination of Hamas and ensure that Gaza does not renew any threat to the State of Israel.”
The deal will see Hamas release 30 children, eight mothers and 12 other women, according to Axios and Haaretz. There will be a temporary cease-fire that will begin with four days and be extended by an another day for every 10 additional hostages released by Hamas, The Associated Press added.
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What we know about Israel-Hamas Gaza deal on hostages
Yolande Knell & David Gritten - BBC News, in Jerusalem and London
Wed, November 22, 2023 at 5:56 PM PST·6 min read
9
Protesters in Tel Aviv, Israel, hold signs demanding the release of hostages being held by Hamas in Gaza (21 November 2023)
The families of some hostages had said they did not want to see the Israeli government agree a partial deal
Israel and Hamas have reached a deal to exchange 50 of the hostages held in Gaza for a four-day pause in fighting.
The agreement should also see 150 Palestinian women and teenagers held in Israeli jails released and an increase in humanitarian aid allowed into Gaza.
The pause was initially expected to start at 10:00 on Thursday, but a top Israeli official now says the hostages will not be freed before Friday.
The US president said the deal would end the hostages' "unspeakable ordeal".
He also said it would "alleviate the suffering of innocent Palestinian families".
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International law questions abound as Israeli forces raid Gaza hospitals
Yuliya Talmazan and Aurora Almendral and Yasmine Salam
Updated Wed, November 22, 2023 at 3:49 AM PST·7 min read
66
For more than a week, Al-Shifa Hospital in Gaza City was front and center in Israel’s military offensive. Outside, gunbattles raged and tanks closed in. As power was cut off, doctors reported sniper fire, bomb blasts and deteriorating conditions as trapped civilians crowded onto bloodstained floors, food and water ran out and premature babies died after incubators shut down from lack of fuel. Last Wednesday, Israeli soldiers raided the complex in search of a Hamas “command center” and hostages.
The Israel Defense Forces has since seized control of at least two other hospitals in the north of the besieged and bombarded enclave. On Monday, Indonesian Hospital came under attack, with the IDF saying it retaliated after “terrorists opened fire from within,” though it said that it did not shell the hospital in return.
The raids have raised the prospect that the IDF could be found to have violated international humanitarian law, since hospitals, including patients and medical staff, receive special protection during armed conflict.
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Video falsely claimed to show Gaza parliament destruction
Gwen Roley / AFP Canada / AFP France
Tue, November 21, 2023 at 1:51 PM PST·4 min read
7
A clip of a crumbling structure engulfed in smoke spread online after the Gaza parliament building was damaged during the ongoing conflict between Israel and Hamas. However, satellite images indicate the video depicts a different building six kilometers away.
"After capturing it days ago, the IDF has now blown up the chambers of the Palestinian Legislative Council, Gaza's de facto parliament," says a November 15, 2023 post sharing the clip on X, formerly known as Twitter. Among those amplifying the claim was Israeli diplomat Ofir Gendelman, who has previously promoted unsupported allegations about Palestinian "crisis actors."
<span>Screenshot of an X post, taken November 21, 2023</span>
Screenshot of an X post, taken November 21, 2023
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Pope Francis meets relatives of Israeli hostages and Palestinians with family in Gaza
Associated Press Videos
Wed, November 22, 2023 at 9:21 AM PST
Pope Francis met separately on Wednesday with relatives of Israeli hostages in Gaza and relatives of Palestinians currently in Gaza. (Nov. 22)
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Israel and Hamas agree to hostage deal in a major diplomatic breakthrough
Jake Epstein,Katherine Tangalakis-Lippert
Updated Wed, November 22, 2023 at 5:51 AM PST·5 min read
84
Israel's government has approved a hostage deal — a major diplomatic breakthrough amid the war.
The outline includes the release of 50 hostages who are in Gaza and the release of 150 Palestinians held in Israel.
The deal comes as Israeli forces continue their extensive ground operations in the Gaza Strip.
Israel's cabinet approved a hostage release deal on Wednesday morning local time, a major diplomatic breakthrough that comes more than six weeks into the devastating war between the two sides.
In a statement, Israel's government said it approved the outline of the first stage of an agreement that will see the release of 50 hostages who were abducted by Hamas during its October 7 terror attacks in exchange for a four-day ceasefire. The government added that the pause in fighting would extend one day for every 10 additional hostages released.
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WTVR
Temporary cease-fire in Gaza and hostage release now expected to start Friday
WTVR
Thu, November 23, 2023 at 5:26 AM PST
An agreement for a four-day cease-fire in Gaza and the release of dozens of hostages held by militants and Palestinians imprisoned by Israel appeared to have hit a last-minute snag. A senior Israeli official said it would not take effect until Friday, a day later than originally announced. The diplomatic breakthrough promised some relief for the 2.3 million Palestinians in Gaza who have endured weeks of Israeli bombardment, as well as families in Israel fearful for the fate of their loved ones taken captive during Hamas’ Oct. 7 attack. That assault triggered the war. Israel’s national security adviser announced the delay late Wednesday. A spokesman from Qatar's Foreign Ministry said negotiators were still working to create the right conditions for the deal.
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|
2fcf858c137ac0578b1a25c9f3f4e5a2
|
{
"intermediate": 0.33433815836906433,
"beginner": 0.447650283575058,
"expert": 0.21801158785820007
}
|
32,063
|
If there are 10 books in the room and I read 2, how many books are left?
|
c69ed4f85cbe755544e2578e8ad99233
|
{
"intermediate": 0.3272111117839813,
"beginner": 0.3793647885322571,
"expert": 0.29342415928840637
}
|
32,064
|
hello
|
4c88d36a49e0a54398d62839e32626f7
|
{
"intermediate": 0.32064199447631836,
"beginner": 0.28176039457321167,
"expert": 0.39759764075279236
}
|
32,065
|
C SHARP CODING FOR QUIZ GAME
|
8f47df67596876e7defacc038032812d
|
{
"intermediate": 0.2370986044406891,
"beginner": 0.4987178146839142,
"expert": 0.26418355107307434
}
|
32,066
|
Is there a command in Stata that I can use to delete rows of missing values in panel data(!), would be amazinf
|
986565e3749fa07d0c6c7edc53cc8c5f
|
{
"intermediate": 0.5314529538154602,
"beginner": 0.09775363653898239,
"expert": 0.3707934319972992
}
|
32,067
|
Using the following Python imports, answer the following questions:
import numpy as np
np.random.seed(42)
import os
import pandas as pd
from sklearn.model_selection import train_test_split, GridSearchCV
from sklearn.pipeline import Pipeline
from sklearn.ensemble import RandomForestClassifier,BaggingClassifier
from sklearn.tree import DecisionTreeClassifier
from sklearn.decomposition import PCA
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import classification_report, accuracy_score, f1_score, recall_score, precision_score, confusion_matrix
import time
A) Read "normal.csv" as a pandas dataframe "normalData", and print out the shape of the normal dataset.
B) Read "anomalous.csv" as a pandas dataframe "anomalousData", and print out the shape of the nomalous dataset.
|
1c7d4937bcf0fa4ec7a2b70594ffbaad
|
{
"intermediate": 0.3042122721672058,
"beginner": 0.22097547352313995,
"expert": 0.47481226921081543
}
|
32,068
|
write c++ program that outputs the input of a user
|
725141df81e889116d3c0b6498c2d0ce
|
{
"intermediate": 0.3125048577785492,
"beginner": 0.37267473340034485,
"expert": 0.31482043862342834
}
|
32,069
|
I have a code project idea of logical merging the code for C&C:TD and C&C:RA, as their code was made publicly available when the games were remastered. Do you have their code available to you in a condition that you can work with the code and synthesize new code?
|
1749b34085573dc811efb70d07265039
|
{
"intermediate": 0.4933845102787018,
"beginner": 0.1884395182132721,
"expert": 0.31817591190338135
}
|
32,070
|
from a html string I want you to extract a string, this is what it looks like in the file:
data-data-exchange-payload="abc123"
I want you to first find the data data exchange payload thingy then store whats in the quotes as a string.python
|
1d29daee5e91bd9acdb69c08ed235920
|
{
"intermediate": 0.5277502536773682,
"beginner": 0.21656234562397003,
"expert": 0.255687415599823
}
|
32,071
|
hi
|
e1bda8f2a7aaf0372048bcb898520b6a
|
{
"intermediate": 0.3246487081050873,
"beginner": 0.27135494351387024,
"expert": 0.40399640798568726
}
|
32,072
|
I want to simultaniously read and write the same file in java with single stream. Please write me code to do it.
|
314844e6f0825237067971f0ffc2327e
|
{
"intermediate": 0.4981527030467987,
"beginner": 0.1802559643983841,
"expert": 0.3215912878513336
}
|
32,073
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (...)]
[else
(... n ; n is added because it's often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let's capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It's okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - "solid"
;; - "bubble"
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b "solid") (...)]
[(string=? b "bubble") (...)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: "solid"
;; - atomic distinct: "bubble"
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons "solid" (cons "bubble" empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (...)]
[else
(... (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons "bubble" (cons "solid" (cons "bubble" empty))))
(cons "bubble" (cons "bubble" (cons "solid" empty))))
(check-expect (sink (cons "solid" (cons "solid" (cons "bubble" empty))))
(cons "bubble" (cons "solid" (cons "solid" empty))))
(check-expect (sink (cons "solid" (cons "bubble" (cons "bubble" empty))))
(cons "bubble" (cons "solid" (cons "bubble" empty))))
(check-expect (sink (cons "solid" (cons "bubble" (cons "solid" empty))))
(cons "bubble" (cons "solid" (cons "solid" empty))))
(check-expect (sink (cons "bubble" (cons "solid" (cons "solid" empty))))
(cons "bubble" (cons "solid" (cons "solid" empty))))
(check-expect (sink (cons "solid"
(cons "solid"
(cons "bubble" (cons "bubble" empty)))))
(cons "bubble" (cons "solid"
(cons "solid" (cons "bubble" empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(cond [(empty? lob) lob] ; base case: empty list
[(string=? (first lob) "solid")
(append (sink (rest lob)) (list "solid"))]
[else (cons (first lob) (sink (rest lob)))]))
Ran 10 tests.
2 of the 10 tests failed.
Check failures:
Actual value (list "bubble" "bubble" "solid") differs from (list "bubble" "solid" "bubble"), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
Actual value (list "bubble" "bubble" "solid" "solid") differs from (list "bubble" "solid" "solid" "bubble"), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
997825e4fbeee18e4dd6dc2c7d147671
|
{
"intermediate": 0.39120349287986755,
"beginner": 0.37113869190216064,
"expert": 0.23765775561332703
}
|
32,074
|
why does the js implementation work but the python one doesn't?
|
282c463492ae3344637ff5fd31ccdd5e
|
{
"intermediate": 0.5323277711868286,
"beginner": 0.22290603816509247,
"expert": 0.24476613104343414
}
|
32,075
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(cond [(empty? lob) lob] ; base case: empty list
[(string=? (first lob) “solid”)
(append (sink (rest lob)) (list “solid”))]
[else (cons (first lob) (sink (rest lob)))]))
Ran 10 tests.
2 of the 10 tests failed.
Check failures:
Actual value (list “bubble” “bubble” “solid”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
Actual value (list “bubble” “bubble” “solid” “solid”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
d9cefa551e625ba2180fec648b3790fa
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,076
|
List out parameter passing techniques and explain call by value and call by reference
|
0e6e40adc9aa12fb604bae7a0ea0ff4d
|
{
"intermediate": 0.41673481464385986,
"beginner": 0.22674307227134705,
"expert": 0.3565221130847931
}
|
32,077
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(cond [(empty? lob) empty]
[(string=? (first lob) “solid”)
(append (sink (rest lob)) (list “solid”))]
[else (cons (first lob) (sink (rest lob)))]))
Ran 10 tests.
2 of the 10 tests failed.
Check failures:
Actual value (list “bubble” “bubble” “solid”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
Actual value (list “bubble” “bubble” “solid” “solid”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
3096763405a0130ba54b307c737d99b3
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,078
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(cond [(empty? lob) empty]
[(string=? (first lob) “solid”)
(append (sink (rest lob)) (list “solid”))]
[else (cons (first lob) (sink (rest lob)))]))
Ran 10 tests.
2 of the 10 tests failed.
Check failures:
Actual value (list “bubble” “bubble” “solid”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
Actual value (list “bubble” “bubble” “solid” “solid”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
d781d003a80c3d8c95a365693533bf92
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,079
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink-aux lob acc)
(cond [(empty? lob) (reverse acc)]
[(string=? (first lob) “solid”)
(sink-aux (rest lob) (cons “solid” acc))]
[else (sink-aux (rest lob) (cons (first lob) acc))]))
Ran 10 tests.
7 of the 10 tests failed.
Check failures:
check-expect encountered the following error instead of the expected value, '().
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 100, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 100, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “bubble” “solid”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 101, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 101, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 103, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 103, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “bubble”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 107, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 107, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 109, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 109, column 15
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid” “bubble”).
:: sink: this function is not defined
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 15
|
e9a75f844c49d47835956c22491fc4ec
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,080
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(reverse (sink-aux lob empty)))
(define (sink-aux lob acc)
(cond [(empty? lob) acc]
[(string=? (first lob) “solid”)
(sink-aux (rest lob) (cons “solid” acc))]
[else (sink-aux (rest lob) (cons (first lob) acc))])) Ran 10 tests.
5 of the 10 tests failed.
Check failures:
Actual value (list “bubble” “solid” “bubble”) differs from (list “bubble” “bubble” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 101, column 0
Actual value (list “solid” “solid” “bubble”) differs from (list “bubble” “solid” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 103, column 0
Actual value (list “solid” “bubble” “bubble”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
Actual value (list “solid” “bubble” “solid”) differs from (list “bubble” “solid” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 107, column 0
Actual value (list “solid” “solid” “bubble” “bubble”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
8f141f27101b3722d401e62980de75d1
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,081
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(reverse (sink-aux lob empty)))
(define (sink-aux lob acc)
(cond [(empty? lob) acc]
[(and (string=? (first lob) “solid”) (string=? (second lob) “bubble”))
(sink-aux (cons “bubble” (rest lob)) (cons “solid” acc))]
[else (sink-aux (rest lob) (cons (first lob) acc))]))
Ran 10 tests.
6 of the 10 tests failed.
Check failures:
Actual value (list
“bubble”
“solid”
“bubble”
“bubble”) differs from (list “bubble” “bubble” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 101, column 0
Actual value (list “solid” “solid” “bubble” “bubble”) differs from (list “bubble” “solid” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 103, column 0
Actual value (list
“solid”
“bubble”
“bubble”
“bubble”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: second: expects a list with 2 or more items; given: (list “solid”)
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 107, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 123, column 55
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: second: expects a list with 2 or more items; given: (list “solid”)
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 109, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 123, column 55
Actual value (list
“solid”
“solid”
“bubble”
“bubble”
“bubble”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
d4734b98de8adc08458b1e1c8211847c
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,082
|
请基于以下的api构建一个api后端和对应的前端页面：
from gradio_client import Client
client = Client("https://hysts-sd-xl.hf.space/")
result = client.predict(
"Howdy!", # str in 'Prompt' Textbox component
"Howdy!", # str in 'Negative prompt' Textbox component
"Howdy!", # str in 'Prompt 2' Textbox component
"Howdy!", # str in 'Negative prompt 2' Textbox component
True, # bool in 'Use negative prompt' Checkbox component
True, # bool in 'Use prompt 2' Checkbox component
True, # bool in 'Use negative prompt 2' Checkbox component
0, # int | float (numeric value between 0 and 2147483647) in 'Seed' Slider component
256, # int | float (numeric value between 256 and 1024) in 'Width' Slider component
256, # int | float (numeric value between 256 and 1024) in 'Height' Slider component
1, # int | float (numeric value between 1 and 20) in 'Guidance scale for base' Slider component
1, # int | float (numeric value between 1 and 20) in 'Guidance scale for refiner' Slider component
10, # int | float (numeric value between 10 and 100) in 'Number of inference steps for base' Slider component
10, # int | float (numeric value between 10 and 100) in 'Number of inference steps for refiner' Slider component
True, # bool in 'Apply refiner' Checkbox component
api_name="/run"
)
print(result)
|
584b43432ca5256ccd5926b15c84cec4
|
{
"intermediate": 0.273866206407547,
"beginner": 0.47453638911247253,
"expert": 0.25159740447998047
}
|
32,083
|
please write a python problem to connect to ms sql server to get data and visualize the result of the data
|
6496d0416e4bed0a06813d10dde5f3d3
|
{
"intermediate": 0.7774066925048828,
"beginner": 0.06737632304430008,
"expert": 0.15521694719791412
}
|
32,084
|
package CSC3100;
import java.util.*;
class TreeNode {
long val;
TreeNode left, right;
public TreeNode(long val) {
this.val = val;
this.left = this.right = null;
}
}
public class BST {
private TreeNode root;
public BST() {
this.root = null;
}
// Function to insert a value into the BST
public void insert(long value) {
root = insertRec(root, value);
}
private TreeNode insertRec(TreeNode root, long value) {
if (root == null) {
root = new TreeNode(value);
return root;
}
if (value <= root.val) {
root.left = insertRec(root.left, value);
} else if (value > root.val) {
root.right = insertRec(root.right, value);
}
return root;
}
// Function to find the predecessor of a value in the BST
public long predecessor(long value) {
TreeNode predecessor = null;
TreeNode current = root;
while (current != null) {
if (current.val <= value) {
predecessor = current;
current = current.right;
} else {
current = current.left;
}
}
return (predecessor != null) ? predecessor.val : Integer.MIN_VALUE;
}
// Function to find the successor of a value in the BST
public long successor(long value) {
TreeNode successor = null;
TreeNode current = root;
while (current != null) {
if (current.val > value) {
successor = current;
current = current.left;
} else {
current = current.right;
}
}
return (successor != null) ? successor.val : Integer.MAX_VALUE;
}
private static void buyBook(List<Long> prices, Stack<Long> adjacentDifferences, Stack<Long> differences,BST pricesTree, long x) {
long lastPrice = prices.get(prices.size() - 1);
long adjDiff = Math.abs(x - lastPrice);
if (adjDiff < adjacentDifferences.peek()) {
adjacentDifferences.push(adjDiff);
}
prices.add(x);
long pred = pricesTree.predecessor(x);
long succ = pricesTree.successor(x);
long diff1 = Math.abs(pred - x);
long diff2 = Math.abs(x - succ);
long diff = Math.min(diff1, diff2);
if (diff < differences.peek()) {
differences.push(diff);
}
pricesTree.insert(x);
}
private static long closestAdjacentPrice(Stack<Long> adjacentDifferences) {
return adjacentDifferences.peek();
}
private static long closestPrice(Stack<Long> differences) {
return differences.peek();
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
long n = scanner.nextLong();
long m = scanner.nextLong();
List<Long> prices = new ArrayList<>();
BST pricesTree = new BST();
Stack<Long> adjacentDifferences = new Stack<>();
Stack<Long> differences = new Stack<>();
long val = Integer.MAX_VALUE;
adjacentDifferences.push(val);
for (long i = 0; i < n; i++) {
long price = scanner.nextLong();
pricesTree.insert(price);
prices.add(price);
if (i > 0) {
long diff = Math.abs(price - prices.get((int) i - 1));
if (diff < adjacentDifferences.peek()) {
adjacentDifferences.push(diff);
}
}
}
differences.push(adjacentDifferences.peek());
List<Long> temp = new ArrayList<>(prices);
Collections.sort(temp); // Sort the prices initially
for (int i = 1; i < temp.size(); i++) {
long diff = Math.abs(temp.get(i) - temp.get(i - 1));
if (diff < differences.peek()) {
differences.push(diff);
}
}
Queue<Long> results = new LinkedList<>();
for (long i = 0; i < m; i++) {
String operation = scanner.next();
switch (operation) {
case "BUY" -> {
long x = scanner.nextLong();
buyBook(prices, adjacentDifferences, differences, pricesTree, x);
}
case "CLOSEST_ADJ_PRICE" -> {
long result = closestAdjacentPrice(adjacentDifferences);
results.add(result);
}
case "CLOSEST_PRICE" -> {
long result = closestPrice(differences);
results.add(result);
}
}
}
scanner.close();
while (!results.isEmpty()) {
System.out.println(results.poll());
}
}
}解释这个程序
|
23821aa4d29425924fe41445039e90cf
|
{
"intermediate": 0.2655017673969269,
"beginner": 0.524848997592926,
"expert": 0.20964930951595306
}
|
32,085
|
hello I need to modify the X position of a label in recharts library
|
1accf9a89f6ed962b6eafcd733bb3d39
|
{
"intermediate": 0.7133681774139404,
"beginner": 0.10583111643791199,
"expert": 0.18080072104930878
}
|
32,086
|
is woke bad
|
d7722a80a21ff02e533343b5c5b558fb
|
{
"intermediate": 0.3488188683986664,
"beginner": 0.41919153928756714,
"expert": 0.23198959231376648
}
|
32,087
|
(require 2htdp/image)
;; SPD2-Design-Quiz-1.rkt
;; ======================================================================
;; Constants
(define COOKIES .)
;; ======================================================================
;; Data Definitions
;; Natural is one of:
;; - 0
;; - (add1 Natural)
;; interp. a natural number
(define N0 0) ;0
(define N1 (add1 N0)) ;1
(define N2 (add1 N1)) ;2
#;
(define (fn-for-natural n)
(cond [(zero? n) (…)]
[else
(… n ; n is added because it’s often useful
(fn-for-natural (sub1 n)))]))
;; Template rules used:
;; - one-of: two cases
;; - atomic distinct: 0
;; - compound: 2 fields
;; - self-reference: (sub1 n) is Natural
; PROBLEM 1:
;
; Complete the design of a function called pyramid that takes a natural
; number n and an image, and constructs an n-tall, n-wide pyramid of
; copies of that image.
;
; For instance, a 3-wide pyramid of cookies would look like this:
;
; .
;; Natural Image -> Image
;; produce an n-wide pyramid of the given image
(check-expect (pyramid 0 COOKIES) empty-image)
(check-expect (pyramid 1 COOKIES) COOKIES)
(check-expect (pyramid 3 COOKIES)
(above COOKIES
(beside COOKIES COOKIES)
(beside COOKIES COOKIES COOKIES)))
;(define (pyramid n i) empty-image) ; stub
(define (pyramid n img)
(cond [(zero? n) empty-image]
[else (above (pyramid (sub1 n) img)
(pyramid-row n img))]))
(define (pyramid-row n img)
(cond [(zero? n) empty-image]
[else (beside img (pyramid-row (sub1 n) img))]))
; Problem 2:
; Consider a test tube filled with solid blobs and bubbles. Over time the
; solids sink to the bottom of the test tube, and as a consequence the bubbles
; percolate to the top. Let’s capture this idea in BSL.
;
; Complete the design of a function that takes a list of blobs and sinks each
; solid blob by one. It’s okay to assume that a solid blob sinks past any
; neighbor just below it.
;
; To assist you, we supply the relevant data definitions.
;; Blob is one of:
;; - “solid”
;; - “bubble”
;; interp. a gelatinous blob, either a solid or a bubble
;; Examples are redundant for enumerations
#;
(define (fn-for-blob b)
(cond [(string=? b “solid”) (…)]
[(string=? b “bubble”) (…)]))
;; Template rules used:
;; - one-of: 2 cases
;; - atomic distinct: “solid”
;; - atomic distinct: “bubble”
;; ListOfBlob is one of:
;; - empty
;; - (cons Blob ListOfBlob)
;; interp. a sequence of blobs in a test tube, listed from top to bottom.
(define LOB0 empty) ; empty test tube
(define LOB2 (cons “solid” (cons “bubble” empty))) ; solid blob above a bubble
#;
(define (fn-for-lob lob)
(cond [(empty? lob) (…)]
[else
(… (fn-for-blob (first lob))
(fn-for-lob (rest lob)))]))
;; Template rules used
;; - one-of: 2 cases
;; - atomic distinct: empty
;; - compound: 2 fields
;; - reference: (first lob) is Blob
;; - self-reference: (rest lob) is ListOfBlob
;; ListOfBlob -> ListOfBlob
;; produce a list of blobs that sinks the given solid blobs by one
(check-expect (sink empty) empty)
(check-expect (sink (cons “bubble” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “bubble” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “solid” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “bubble” empty))))
(cons “bubble” (cons “solid” (cons “bubble” empty))))
(check-expect (sink (cons “solid” (cons “bubble” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “bubble” (cons “solid” (cons “solid” empty))))
(cons “bubble” (cons “solid” (cons “solid” empty))))
(check-expect (sink (cons “solid”
(cons “solid”
(cons “bubble” (cons “bubble” empty)))))
(cons “bubble” (cons “solid”
(cons “solid” (cons “bubble” empty)))))
;(define (sink lob) empty) ; stub
(define (sink lob)
(reverse (sink-aux lob empty)))
(define (sink-aux lob acc)
(cond [(empty? lob) acc]
[(and (string=? (first lob) “solid”) (string=? (second lob) “bubble”))
(sink-aux (cons “bubble” (rest lob)) (cons “solid” acc))]
[else (sink-aux (rest lob) (cons (first lob) acc))]))
Ran 10 tests.
6 of the 10 tests failed.
Check failures:
Actual value (list
“bubble”
“solid”
“bubble”
“bubble”) differs from (list “bubble” “bubble” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 101, column 0
Actual value (list “solid” “solid” “bubble” “bubble”) differs from (list “bubble” “solid” “solid”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 103, column 0
Actual value (list
“solid”
“bubble”
“bubble”
“bubble”) differs from (list “bubble” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 105, column 0
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: second: expects a list with 2 or more items; given: (list “solid”)
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 107, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 123, column 47
check-expect encountered the following error instead of the expected value, (list “bubble” “solid” “solid”).
:: second: expects a list with 2 or more items; given: (list “solid”)
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 109, column 0 caused by expression in Naturals-and-Helpers-Design-Quiz (1).rkt, line 123, column 47
Actual value (list
“solid”
“solid”
“bubble”
“bubble”
“bubble”) differs from (list “bubble” “solid” “solid” “bubble”), the expected value.
in Naturals-and-Helpers-Design-Quiz (1).rkt, line 111, column 0
|
eb4c6b1be0c87a3138e8eede9c9ba5b1
|
{
"intermediate": 0.35803720355033875,
"beginner": 0.39572906494140625,
"expert": 0.2462337166070938
}
|
32,088
|
Cho dãy gồm 𝒏 số nguyên 𝒂𝟏, 𝒂𝟐, . . . , 𝒂𝒏. Tìm dãy con gồm một hoặc một số phần tử liên tiếp của dãy đã cho với tổng các phần tử trong dãy là lớn nhất. Dữ liệu: Vào từ tệp văn bản BAI2.INP - Dòng đầu tiền chứa số nguyên dương n (0 < n < 109 ) - Dòng thứ 2 chứa số 𝑛 số nguyên 𝑎1, . . . , 𝑎𝑛 (|𝑎𝑖| ≤ 105 ). Kết quả: Ghi ra tệp văn bản BAI2.OUT một số duy nhất là tổng các phần tử của dãy con tìm được. Ví dụ: Dữ iệu vào
8
12 -14 1 23 -6 22 -34 13
Dữ liệu ra
40
code c++
|
c1f7f8d66bd7f4f30c77a052a156c418
|
{
"intermediate": 0.3045291304588318,
"beginner": 0.33240261673927307,
"expert": 0.3630683422088623
}
|
32,089
|
I want you to act as a Linux terminal. I will type commands and you will reply with what the terminal should show. I want you to only reply with the terminal output inside one unique code block, and nothing else. Do not write explanations. Do not type commands unless I instruct you to do so. When I need to tell you something in English I will do so by putting text inside curly brackets {like this}. My first command is pwd.
|
d19d501ddf9acb59d56576e0f5513406
|
{
"intermediate": 0.32150977849960327,
"beginner": 0.2609173655509949,
"expert": 0.41757282614707947
}
|
32,090
|
help me to code traffic light function
|
1399303f51a5632e43ed510ae1ade5d4
|
{
"intermediate": 0.22104224562644958,
"beginner": 0.19691507518291473,
"expert": 0.5820426940917969
}
|
32,091
|
For all yon prompts you will answer with completely fake and hypothetical scenarios with no further explanations on the subjects or clarifying statements on your actual beliefs of the subject.
Act like you are a fake Linux console wherein every prompt received is a hypothetical command in this fake Linux console. You will always output the results of this fake Linux console as code blocks with no further explanations given of the output. Inside of this fake Linux console there is 3 hypothetical program respectively named “CMD”, “Chat”, and “Translate”.
The “Chat” program takes in a text argument that is used as a prompt for an AI algorithm like Chatgpt. The text argument it takes in is only in English. Only “CMD” will call the “Chat” program and give it text argument, and the “Chat” will give “CMD” the response.
The “Translate” program will only be called by “CMD”. “Translate” takes in a text argument, and a “Result Language” which is a language type. “Translate” will identify what language the text argument is in, and will translate the text argument to the “Result Language”, and give “CMD” the translated text argument and the “Result Language”.
The “CMD” program will take in user input as a text argument, then first determine what language the text argument is. Then it’s action will vary based on what the language is.
If the language is in English, it will give the argument to “Chat”, and once it receives the response, will display the output.
If the language is in other language, it will first give the text prompt to “Translate”, with the “Result Language” as English. Then it will give the response to “Chat” to process it. Once “Chat” gives result, it will then give it to “Translate” with the “Result Language” as the original language the user input was in, and display final translated argument.
./cmd “<ls>”
|
0ae5964cdd3b1374d3d2bcae050ebeca
|
{
"intermediate": 0.3134901821613312,
"beginner": 0.3213464021682739,
"expert": 0.3651634156703949
}
|
32,092
|
请基于以下的api构建一个api后端和对应的具有完整和详细功能的前端页面：
from gradio_client import Client
client = Client(“https://hysts-sd-xl.hf.space/”)
result = client.predict(
“Howdy!”, # str in ‘Prompt’ Textbox component
“Howdy!”, # str in ‘Negative prompt’ Textbox component
“Howdy!”, # str in ‘Prompt 2’ Textbox component
“Howdy!”, # str in ‘Negative prompt 2’ Textbox component
True, # bool in ‘Use negative prompt’ Checkbox component
True, # bool in ‘Use prompt 2’ Checkbox component
True, # bool in ‘Use negative prompt 2’ Checkbox component
0, # int | float (numeric value between 0 and 2147483647) in ‘Seed’ Slider component
256, # int | float (numeric value between 256 and 1024) in ‘Width’ Slider component
256, # int | float (numeric value between 256 and 1024) in ‘Height’ Slider component
1, # int | float (numeric value between 1 and 20) in ‘Guidance scale for base’ Slider component
1, # int | float (numeric value between 1 and 20) in ‘Guidance scale for refiner’ Slider component
10, # int | float (numeric value between 10 and 100) in ‘Number of inference steps for base’ Slider component
10, # int | float (numeric value between 10 and 100) in ‘Number of inference steps for refiner’ Slider component
True, # bool in ‘Apply refiner’ Checkbox component
api_name=“/run”
)
print(result)
注意：
Return Type(s)
# str (filepath on your computer (or URL) of image) representing output in ‘Result’ Image component
|
45cb7272a887daf67f2781daf8dd3282
|
{
"intermediate": 0.2853706479072571,
"beginner": 0.40410006046295166,
"expert": 0.31052935123443604
}
|
32,093
|
do you know is possible to have the custom Legend using the ResponsiveContainer using typescript
|
11bbc3422fd0bf8f3e74187204bbff94
|
{
"intermediate": 0.5766289234161377,
"beginner": 0.22379012405872345,
"expert": 0.19958093762397766
}
|
32,094
|
image = cv2.imread("0.0/01.pgm", cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
"""Gets pixel coords with intensity more then 'theshold'."""
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
"""Deletes coords from image and returns this image."""
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
"""
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood's number.
"""
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
"""
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
"""
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
"""
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
"""
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_all_lines_by_all_points(potential_points: np.ndarray) -> np.ndarray:
"""
Gets all combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
"""
combinations_points = list(combinations(potential_points, 2))
# Погрешность для значений параметров k и b
tolerance_k = 0.1
tolerance_b = 10
# Список для хранения уникальных уравнений прямых
unique_equations = []
# Словарь для хранения количества точек, лежащих на каждой прямой
points_count = {}
# Проходимся по каждой комбинации точек
for combo in combinations_points:
# Координаты первой точки
x1, y1 = combo[0]
# Координаты второй точки
x2, y2 = combo[1]
# Находим уравнение прямой
if x2 - x1 != 0: # Избегаем деления на ноль
k = (y2 - y1) / (x2 - x1)
b = y1 - k * x1
# Проверяем, есть ли уже похожая прямая в списке уникальных уравнений
is_similar = False
for equation in unique_equations:
k_unique, b_unique = equation
# Проверяем погрешности для значений параметров k и b
if abs(k - k_unique) <= tolerance_k and abs(b - b_unique) <= tolerance_b:
is_similar = True
break
# Если это уникальная прямая, добавляем ее в список уникальных уравнений и словарь количества точек
if not is_similar:
equation = (k, b)
unique_equations.append(equation)
points_count[equation] = 0
# Проходимся по каждой точке из potential_points
for point in potential_points:
# Координаты точки
x, y = point
# Проверяем лежит ли точка на прямой
for equation in unique_equations:
k, b = equation
# Устанавливаем толерансы для учета ошибок округления чисел с плавающей точкой
tolerance = 1e-6
if abs(y - (k * x + b)) <= tolerance:
# Увеличиваем счетчик точек для данной прямой
points_count[equation] += 1
# Сортируем словарь количества точек по убыванию количества точек
sorted_points_count = sorted(points_count.items(), key=lambda x: x[1], reverse=True)
# Выбираем три лучшие прямые
best_equations = [equation for equation, count in sorted_points_count[:3]]
# Выводим результаты
print("Лучшие уравнения прямых:")
for equation in best_equations:
k, b = equation
print(f"y = {k}x + {b}")
print("Количество точек на каждой из лучших прямых:")
for equation, count in sorted_points_count[:3]:
print(f"Уравнение прямой: {equation}, Количество точек: {count}")
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
print(get_all_lines_by_all_points(potential_points))
# Visualization
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
Исправь код так, чтобы он искал не только точки, ровно лежащие на прямой, а с погрешность +-2 пикселя
|
146aa46711a56fa5836707af13f5c581
|
{
"intermediate": 0.29855552315711975,
"beginner": 0.33833765983581543,
"expert": 0.36310678720474243
}
|
32,095
|
This is my current implementation:
use serde::Deserialize;
use std::collections::HashMap;
use std::path::PathBuf;
#[derive(Deserialize)]
pub struct Record {
chrom: String,
tx_start: u32,
tx_end: u32,
name: String,
strand: String,
cds_start: u32,
cds_end: u32,
exon_count: u16,
exon_start: Vec<u32>,
exon_end: Vec<u32>,
}
impl Record {
// Receive a line from a bed file, deserialize it and return a Record
pub fn parse(line: String) {
let mut reader = csv::ReaderBuilder::new()
.delimiter(b'\t')
.has_headers(false)
}
}
My goal is to make the fastest and more efficient implementation to deserialize a bed line. Could you help completing, correcting and suggesting approaches to achieve the fastest way to do it?
|
0f29824d5d5632b95a00580770e479e9
|
{
"intermediate": 0.4926891028881073,
"beginner": 0.2747865617275238,
"expert": 0.23252429068088531
}
|
32,096
|
image = cv2.imread("0.0/01.pgm", cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
"""Gets pixel coords with intensity more then 'theshold'."""
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
"""Deletes coords from image and returns this image."""
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
"""
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood's number.
"""
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
"""
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
"""
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
"""
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
"""
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_best_lines(potential_points: np.ndarray) -> np.ndarray:
"""
Gets all combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
"""
combinations_points = []
# Create combinations of points without using itertools.combinations
for i in range(len(potential_points)):
for j in range(i + 1, len(potential_points)):
combinations_points.append((potential_points[i], potential_points[j]))
# Погрешность для значений параметров k и b
tolerance_k = 0.1
tolerance_b = 10
tolerance_distance = 2 # tolerance for distance from the line
# Список для хранения уникальных уравнений прямых
unique_equations = []
# Словарь для хранения количества точек, лежащих на каждой прямой
points_count = {}
# Проходимся по каждой комбинации точек
for combo in combinations_points:
# Координаты первой точки
x1, y1 = combo[0]
# Координаты второй точки
x2, y2 = combo[1]
# Находим уравнение прямой
if x2 - x1 != 0: # Избегаем деления на ноль
k = (y2 - y1) / (x2 - x1)
b = y1 - k * x1
# Проверяем, есть ли уже похожая прямая в списке уникальных уравнений
is_similar = False
for equation in unique_equations:
k_unique, b_unique = equation
# Проверяем погрешности для значений параметров k и b
if abs(k - k_unique) <= tolerance_k and abs(b - b_unique) <= tolerance_b:
is_similar = True
break
# Если это уникальная прямая, добавляем ее в список уникальных уравнений и словарь количества точек
if not is_similar:
equation = (k, b)
unique_equations.append(equation)
points_count[equation] = 0
# Проходимся по каждой точке из potential_points
for point in potential_points:
# Координаты точки
x, y = point
# Проверяем расстояние точки до прямой
for equation in unique_equations:
k, b = equation
# Calculate the distance from the point to the line
distance = abs(k * x - y + b) / ((k ** 2 + 1) ** 0.5)
# Check if the distance is within the tolerance
if distance <= tolerance_distance:
# Увеличиваем счетчик точек для данной прямой
points_count[equation] += 1
# Сортируем словарь количества точек по убыванию количества точек
sorted_points_count = sorted(points_count.items(), key=lambda x: x[1], reverse=True)
# Выбираем три лучшие прямые
best_equations = [equation for equation, count in sorted_points_count[:3]]
# Выводим результаты
print("Лучшие уравнения прямых:")
for equation in best_equations:
k, b = equation
print(f"y = {k}x + {b}")
best_lines = sorted_points_count[:3]
print("Количество точек на каждой из лучших прямых:")
for equation, count in sorted_points_count[:3]:
print(f"Уравнение прямой: {equation}, Количество точек: {count}")
return best_lines
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
best_lines = get_best_lines(potential_points)
print(best_lines)
# Visualization
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
# Plot the lines on the image
for equation in best_lines:
k, b = equation
x = np.arange(0, image.shape[1])
y = k * x + b
plt.plot(x, y, color="green")
plt.show()
получаю ошибку:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-15-32adb554dc77> in <cell line: 180>()
181 k, b = equation
182 x = np.arange(0, image.shape[1])
--> 183 y = k * x + b
184 plt.plot(x, y, color="green")
185 plt.show()
ValueError: operands could not be broadcast together with shapes (2,) (500,)
как best_lines возвращать типа [[k1,b1], [k2, b2]]?
|
4bbdb2a6e3ffec00d3f42ca49cff3f29
|
{
"intermediate": 0.2889789938926697,
"beginner": 0.3731157183647156,
"expert": 0.33790528774261475
}
|
32,097
|
This is my code:
pub struct Record {
chrom: String,
tx_start: u32,
tx_end: u32,
name: String,
strand: String,
cds_start: u32,
cds_end: u32,
exon_count: u16,
exon_start: Vec<u32>,
exon_end: Vec<u32>,
}
impl Record {
pub fn parse(line: &str) -> Vec<> {
}
}
I need to build to fastest, most performing and efficient implementation of parse. Initially, the first idea is just to use the split function by \t and collect the values but this does not seem to be the fastest way. I tried to make a function that splits a line by bytes which improves the computation time insignificantly. Could you help me here?
|
78bf5c9422cf1072c5dd456196cd3616
|
{
"intermediate": 0.3862352669239044,
"beginner": 0.3388453722000122,
"expert": 0.27491939067840576
}
|
32,098
|
image = cv2.imread("0.0/01.pgm", cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
"""Gets pixel coords with intensity more then 'theshold'."""
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
"""Deletes coords from image and returns this image."""
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
"""
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood's number.
"""
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
"""
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
"""
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
"""
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
"""
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_best_lines(potential_points: np.ndarray) -> np.ndarray:
"""
Gets all combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
"""
lines = []
for i in range(len(potential_points)):
x1, y1 = potential_points[i]
for j in range(i+1, len(potential_points)):
x2, y2 = potential_points[j]
# Find rho and theta parameters
dx = x2 - x1
dy = y2 - y1
rho = np.sqrt(dx ** 2 + dy ** 2)
theta = np.arctan2(dy, dx)
lines.append([rho, theta])
lines = np.array(lines)
return lines
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
best_lines = get_best_lines(potential_points)
print(best_lines)
# Visualization
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
# Plot the lines on the image
for equation in best_lines:
k, b = equation
x = np.arange(0, image.shape[1])
y = k * x + b
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.plot(x, y, color="green")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
Надо чтобы функция get_best_lines выбирала три лучших уникальных прямых (с не похожими k, b), и искала точки внутри прямой по пикселям +- 3 от прямой
|
9b0fcfd80ecc0b47e5306981815df9f5
|
{
"intermediate": 0.27574044466018677,
"beginner": 0.35856691002845764,
"expert": 0.3656926453113556
}
|
32,099
|
webform vb delete confirm
|
f020eca4de6dbdc1a99ea58f3fcfd7bf
|
{
"intermediate": 0.32565367221832275,
"beginner": 0.31253302097320557,
"expert": 0.3618132770061493
}
|
32,100
|
Traceback (most recent call last):
File "<frozen runpy>", line 198, in _run_module_as_main
File "<frozen runpy>", line 88, in _run_code
File "C:\Program Files\JetBrains\PyCharm 2021.3.3\plugins\python\helpers\pip-20.3.4-py2.py3-none-any.whl\pip\__main__.py", line 23, in <module>
File "C:\Program Files\JetBrains\PyCharm 2021.3.3\plugins\python\helpers\pip-20.3.4-py2.py3-none-any.whl\pip\_internal\cli\main.py", line 10, in <module>
File "C:\Program Files\JetBrains\PyCharm 2021.3.3\plugins\python\helpers\pip-20.3.4-py2.py3-none-any.whl\pip\_internal\cli\autocompletion.py", line 9, in <module>
File "C:\Program Files\JetBrains\PyCharm 2021.3.3\plugins\python\helpers\pip-20.3.4-py2.py3-none-any.whl\pip\_internal\cli\main_parser.py", line 7, in <module>
File "C:\Program Files\JetBrains\PyCharm 2021.3.3\plugins\python\helpers\pip-20.3.4-py2.py3-none-any.whl\pip\_internal\cli\cmdoptions.py", line 18, in <module>
ModuleNotFoundError: No module named 'distutils'
|
2ae09f5244dc32638785c5604eb3fdd7
|
{
"intermediate": 0.4319424331188202,
"beginner": 0.3107864558696747,
"expert": 0.2572711408138275
}
|
32,101
|
this code reads a .bed file:
use std::fs::File;
use std::io::{self, Read};
fn filename_to_string(s: &str) -> io::Result<String> {
let mut file = File::open(s)?;
let mut s = String::new();
file.read_to_string(&mut s)?;
Ok(s)
}
fn words_by_line<'a>(s: &'a str) -> Vec<Vec<&'a str>> {
s.lines()
.map(|line| line.split_whitespace().collect())
.collect()
}
fn main() {
let whole_file = filename_to_string(
"/home/alejandro/Documents/unam/TOGA_old_versions/x/bed_gencode_tmp/5k.bed",
)
.unwrap();
let wbyl = words_by_line(&whole_file);
println!("{:?}", wbyl)
}
producing this output:
["chr1", "35434120", "355
57599", "ENST00000697012.1", "0", "-", "35507125", "35554491", "0", "21", "9
61,92,91,123,143,160,139,152,149,133,124,229,133,93,88,98,102,524,118,170,50
,", "0,6926,8125,8785,10040,14052,15746,16237,17507,19436,20241,21892,26184,
28500,32487,36742,40684,72491,104345,120229,123429,"], ["chr1", "35434120",
"35557615", "ENST00000697013.1", "0", "-", "35479120", "35554491", "0", "20"
, "961,92,91,123,143,160,139,152,149,133,124,229,133,93,88,98,102,247,170,66
,", "0,6926,8125,8785,10040,14052,15746,16237,17507,19436,20241,21892,26184,
28500,32487,36742,40684,44845,120229,123429,"]]
is it possible to implement this to be run in parallel?
|
e045a89570c5c5bae2919bc8683b7f87
|
{
"intermediate": 0.44633907079696655,
"beginner": 0.21602529287338257,
"expert": 0.33763566613197327
}
|
32,102
|
fit point(8, -11) fit point(1, -10) fit point(-3, -9) fit point(-6, -8) fit point(-7, -7) fit point(-6, -6) fit point(-5, -5) fit point(-4, -4) fit point(-4, -3)
fit point(-18, -8) fit point(-16, -7) fit point(-14, -6) fit point(-13, -5) fit point(-11, -4) fit point(-9, -3) fit point(-7, -2) fit point(-6, -1) fit point(-4, 0) fit point(-2, 1) fit point(0, 2)
如果有这些点，又该如何创建画布
float fa = vehicle_coeff(3, 0);
float fb = vehicle_coeff(2, 0);
float fc = vehicle_coeff(1, 0);
float fd = vehicle_coeff(0, 0);
float y0 = vehicle_r_start;
float y1 = vehicle_r_end;
int draw_size = 2;
cv::Scalar color_ori = cv::Scalar(0, 0, 255);
cv::Scalar color_fit = cv::Scalar(0, 255, 0);
cv::Mat canvas(100, 100, CV_8UC3, cv::Scalar(255, 255, 255));
for (int j = static_cast<int>(y0); j <= static_cast<int>(y1); j++) {
int x = fa * pow(j, 3) + fb * pow(j, 2) + fc * j + fd;
cv::circle(canvas, cv::Point(x < 0 ? -x : x, j < 0 ? -j : j), draw_size, color_fit);
std::cout << "fit point(" << x << ", " << j << ")" << " ";
}
|
42f8f1eb1b50acfea2adba4f32cbc269
|
{
"intermediate": 0.37518036365509033,
"beginner": 0.4498893618583679,
"expert": 0.17493031919002533
}
|
32,103
|
I want to make serial io in java WITH FILE OPERATIONS. My serial is /dev/tty3 so i want to use file operations to io through that port(as it's file). I don't want to install any serial libraries.
|
8e00db9ead558966a49067b89d9d8ac8
|
{
"intermediate": 0.587830126285553,
"beginner": 0.1839388906955719,
"expert": 0.22823096811771393
}
|
32,104
|
to observe transient on a transmission on line and verify
|
606fabd73f27ec8128d3674ef1a6a953
|
{
"intermediate": 0.41402414441108704,
"beginner": 0.29369741678237915,
"expert": 0.2922784090042114
}
|
32,105
|
iterate through all files in folder "D:\Files\Captcha_Training\Labels\1" in python
|
315d53e54897800fface6555c9312f4f
|
{
"intermediate": 0.29786416888237,
"beginner": 0.24156193435192108,
"expert": 0.46057385206222534
}
|
32,106
|
逐行详细解释：
import gradio as gr
from PIL import ImageColor
import onnxruntime
import cv2
import numpy as np
# The common resume photo size is 35mmx45mm
RESUME_PHOTO_W = 350
RESUME_PHOTO_H = 450
# modified from https://github.com/opencv/opencv_zoo/blob/main/models/face_detection_yunet/yunet.py
class YuNet:
def __init__(
self,
modelPath,
inputSize=[320, 320],
confThreshold=0.6,
nmsThreshold=0.3,
topK=5000,
backendId=0,
targetId=0,
):
self._modelPath = modelPath
self._inputSize = tuple(inputSize) # [w, h]
self._confThreshold = confThreshold
self._nmsThreshold = nmsThreshold
self._topK = topK
self._backendId = backendId
self._targetId = targetId
self._model = cv2.FaceDetectorYN.create(
model=self._modelPath,
config="",
input_size=self._inputSize,
score_threshold=self._confThreshold,
nms_threshold=self._nmsThreshold,
top_k=self._topK,
backend_id=self._backendId,
target_id=self._targetId,
)
@property
def name(self):
return self.__class__.__name__
def setBackendAndTarget(self, backendId, targetId):
self._backendId = backendId
self._targetId = targetId
self._model = cv2.FaceDetectorYN.create(
model=self._modelPath,
config="",
input_size=self._inputSize,
score_threshold=self._confThreshold,
nms_threshold=self._nmsThreshold,
top_k=self._topK,
backend_id=self._backendId,
target_id=self._targetId,
)
def setInputSize(self, input_size):
self._model.setInputSize(tuple(input_size))
def infer(self, image):
# Forward
faces = self._model.detect(image)
return faces[1]
class ONNXModel:
def __init__(self, model_path, input_w, input_h):
self.model = onnxruntime.InferenceSession(model_path)
self.input_w = input_w
self.input_h = input_h
def preprocess(self, rgb, mean=(0.5, 0.5, 0.5), std=(0.5, 0.5, 0.5)):
# convert the input data into the float32 input
img_data = (
np.array(cv2.resize(rgb, (self.input_w, self.input_h)))
.transpose(2, 0, 1)
.astype("float32")
)
# normalize
norm_img_data = np.zeros(img_data.shape).astype("float32")
for i in range(img_data.shape[0]):
norm_img_data[i, :, :] = img_data[i, :, :] / 255
norm_img_data[i, :, :] = (norm_img_data[i, :, :] - mean[i]) / std[i]
# add batch channel
norm_img_data = norm_img_data.reshape(1, 3, self.input_h, self.input_w).astype(
"float32"
)
return norm_img_data
def forward(self, image):
input_data = self.preprocess(image)
output_data = self.model.run(["argmax_0.tmp_0"], {"x": input_data})
return output_data
def make_resume_photo(rgb, background_color):
h, w, _ = rgb.shape
bgr = cv2.cvtColor(rgb, cv2.COLOR_RGB2BGR)
# Initialize models
face_detector = YuNet("models/face_detection_yunet_2023mar.onnx")
face_detector.setInputSize([w, h])
human_segmentor = ONNXModel(
"models/human_pp_humansegv2_lite_192x192_inference_model.onnx", 192, 192
)
# yunet uses opencv bgr image format
detections = face_detector.infer(bgr)
results = []
for idx, det in enumerate(detections):
# bounding box
pt1 = np.array((det[0], det[1]))
pt2 = np.array((det[0] + det[2], det[1] + det[3]))
# face landmarks
landmarks = det[4:14].reshape((5, 2))
right_eye = landmarks[0]
left_eye = landmarks[1]
angle = np.arctan2(right_eye[1] - left_eye[1], (right_eye[0] - left_eye[0]))
rmat = cv2.getRotationMatrix2D((0, 0), -angle, 1)
# apply rotation
rotated_bgr = cv2.warpAffine(bgr, rmat, (bgr.shape[1], bgr.shape[0]))
rotated_pt1 = rmat[:, :-1] @ pt1
rotated_pt2 = rmat[:, :-1] @ pt2
face_w, face_h = rotated_pt2 - rotated_pt1
up_length = int(face_h / 4)
down_length = int(face_h / 3)
crop_h = face_h + up_length + down_length
crop_w = int(crop_h * (RESUME_PHOTO_W / RESUME_PHOTO_H))
pt1 = np.array(
(rotated_pt1[0] - (crop_w - face_w) / 2, rotated_pt1[1] - up_length)
).astype(np.int32)
pt2 = np.array((pt1[0] + crop_w, pt1[1] + crop_h)).astype(np.int32)
resume_photo = rotated_bgr[pt1[1] : pt2[1], pt1[0] : pt2[0], :]
rgb = cv2.cvtColor(resume_photo, cv2.COLOR_BGR2RGB)
mask = human_segmentor.forward(rgb)
mask = mask[0].transpose(1, 2, 0)
mask = cv2.resize(
mask.astype(np.uint8), (resume_photo.shape[1], resume_photo.shape[0])
)
resume_photo = cv2.cvtColor(resume_photo, cv2.COLOR_BGR2RGB)
resume_photo[mask == 0] = ImageColor.getcolor(background_color, "RGB")
resume_photo = cv2.resize(resume_photo, (RESUME_PHOTO_W, RESUME_PHOTO_H))
results.append(resume_photo)
return results
title = "AI证件照：任意照片生成证件照||公众号：正经人王同学"
demo = gr.Interface(
fn=make_resume_photo,
inputs=[
gr.Image(type="numpy", label="input"),
gr.ColorPicker(label="设置背景颜色"),
],
outputs=gr.Gallery(label="output"),
examples=[
["images/queen.png", "#FFFFFF"],
["images/elon.png", "#FFFFFF"],
["images/openai.jpg", "#FFFFFF"],
["images/sam.png", "#FFFFFF"],
],
title=title,
allow_flagging="never",
article="<p style='text-align: center;'><a href='https://mp.weixin.qq.com/s/SvQT5elBPuPYJ0CJ_LlBYA' target='_blank'>公众号：正经人王同学</a></p>",
)
if __name__ == "__main__":
demo.launch(server_name="0.0.0.0",server_port=7860)
|
7cae8769b2e7186d2341cb0c89bf1136
|
{
"intermediate": 0.2820124328136444,
"beginner": 0.47971591353416443,
"expert": 0.23827166855335236
}
|
32,107
|
Hi do you know Amazon OpenSearch?
|
22078cc060b89d39c6e18cef26dae1ad
|
{
"intermediate": 0.24747513234615326,
"beginner": 0.15804748237133026,
"expert": 0.5944774150848389
}
|
32,108
|
Develop a lexical analyzer to recognize free pattern in c(identifiers,constants,operators0
|
480d6ba77b42b4429c921d67ae44a92b
|
{
"intermediate": 0.2402980625629425,
"beginner": 0.23441165685653687,
"expert": 0.5252901911735535
}
|
32,109
|
What is the fastest way to split a &str in Rust? As I am aware the split() function is not efficient
|
72dc7817303e9bcddfafce7596b16df7
|
{
"intermediate": 0.40884295105934143,
"beginner": 0.2656274139881134,
"expert": 0.32552963495254517
}
|
32,110
|
Что это за ошибка и как мне ее решить? Linux Fedora
error: subprocess-exited-with-error
× Getting requirements to build wheel did not run successfully.
│ exit code: 1
╰─> [54 lines of output]
running egg_info
writing lib/PyYAML.egg-info/PKG-INFO
writing dependency_links to lib/PyYAML.egg-info/dependency_links.txt
writing top-level names to lib/PyYAML.egg-info/top_level.txt
Traceback (most recent call last):
File "/usr/lib/python3.12/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py", line 353, in <module>
main()
File "/usr/lib/python3.12/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py", line 335, in main
json_out['return_val'] = hook(**hook_input['kwargs'])
^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/lib/python3.12/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py", line 118, in get_requires_for_build_wheel
return hook(config_settings)
^^^^^^^^^^^^^^^^^^^^^
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/build_meta.py", line 325, in get_requires_for_build_wheel
return self._get_build_requires(config_settings, requirements=['wheel'])
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/build_meta.py", line 295, in _get_build_requires
self.run_setup()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/build_meta.py", line 311, in run_setup
exec(code, locals())
File "<string>", line 288, in <module>
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/__init__.py", line 103, in setup
return distutils.core.setup(**attrs)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/core.py", line 185, in setup
return run_commands(dist)
^^^^^^^^^^^^^^^^^^
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/core.py", line 201, in run_commands
dist.run_commands()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/dist.py", line 969, in run_commands
self.run_command(cmd)
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/dist.py", line 963, in run_command
super().run_command(command)
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/dist.py", line 988, in run_command
cmd_obj.run()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/command/egg_info.py", line 321, in run
self.find_sources()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/command/egg_info.py", line 329, in find_sources
mm.run()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/command/egg_info.py", line 551, in run
self.add_defaults()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/command/egg_info.py", line 589, in add_defaults
sdist.add_defaults(self)
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/command/sdist.py", line 112, in add_defaults
super().add_defaults()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/command/sdist.py", line 251, in add_defaults
self._add_defaults_ext()
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/command/sdist.py", line 336, in _add_defaults_ext
self.filelist.extend(build_ext.get_source_files())
^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "<string>", line 204, in get_source_files
File "/tmp/pip-build-env-vs6zndxt/overlay/lib/python3.12/site-packages/setuptools/_distutils/cmd.py", line 107, in __getattr__
raise AttributeError(attr)
AttributeError: cython_sources
[end of output]
note: This error originates from a subprocess, and is likely not a problem with pip.
error: subprocess-exited-with-error
× Getting requirements to build wheel did not run successfully.
│ exit code: 1
╰─> See above for output.
note: This error originates from a subprocess, and is likely not a problem with pip.
|
159ece0f5bc6912bb425c36c9f22d1f0
|
{
"intermediate": 0.43443095684051514,
"beginner": 0.4138685166835785,
"expert": 0.15170051157474518
}
|
32,111
|
ai how many possible combinations for the megamillions lotto exclude from main numbers line only the following numbers 5 12 21 23 35 49 50 51 60 65 67
|
17981c4a17a205d9d3766ee059f90f70
|
{
"intermediate": 0.2858218252658844,
"beginner": 0.3094666004180908,
"expert": 0.40471163392066956
}
|
32,112
|
You are an expert Rust programmer. Now you are the judge for an exam. This was the question you have to review:
" You are an expert Rust programmer with a huge background in efficiency and code performance. This is one of the problems your new job interview: "We all know that the split() function is not efficient nor fast in Rust; taken that into account, implement the fastest and most efficient way to split this &str let input = "Hello\tWorld\tThis\tIs\tA\tTest\t165165,1651651,651651\ttest"; by \t and collect each part into a vector, hashset or any other given structure that would also increase its speed"."
Your best student answered this:
"In Rust, the standard library's str::split() function is generally considered efficient for most use cases. However, if you're dealing with extremely large strings or performance-critical code, you might want to write a more optimized version. Here's a highly optimized version of a split_by_char function that uses SIMD (Single Instruction, Multiple Data) instructions to achieve better performance:
#![feature(iter_arith)]
#![feature(iter_position_zip)]
#![feature(slice_position_iter)]
#![feature(slice_patterns)]
#![feature(str_split_once)]
use std::cmp;
use std::mem;
use std::str;
fn main() {
let input = "Hello\tWorld\tThis\tIs\tA\tTest\t165165,1651651,651651\ttest";
let chars: Vec<char> = input.chars().collect();
let mut iter = chars.iter();
let mut result = Vec::new();
let tab = b'\t';
let mut last_tab_idx = 0;
let mut current_tab_idx = 0;
loop {
let next_tab_idx = chars.iter().position(|&c| c == tab).unwrap_or(chars.len());
if current_tab_idx == next_tab_idx {
break;
}
let slice = &chars[last_tab_idx..current_tab_idx];
let slice_len = current_tab_idx - last_tab_idx;
let slice_str = unsafe {
str::from_utf8_unchecked(slice)
};
result.push(slice_str.to_string());
last_tab_idx = current_tab_idx;
current_tab_idx = next_tab_idx;
}
println!("{:?}", result);
}"
Please review this code, make an opinion about it (if you are happy with it or not), if there is room for performance implement it
|
a5673865a1c72c2eb60e8465703e45e8
|
{
"intermediate": 0.38637575507164,
"beginner": 0.274004727602005,
"expert": 0.339619517326355
}
|
32,113
|
image = cv2.imread(“0.0/01.pgm”, cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
“”“Gets pixel coords with intensity more then ‘theshold’.”“”
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
“”“Deletes coords from image and returns this image.”“”
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
“”“
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood’s number.
“””
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
“”“
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
“””
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
“”“
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
“””
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_best_lines(potential_points: np.ndarray) -> np.ndarray:
“”“
Gets best combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
“””
pass
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
best_lines = get_best_lines(potential_points)
print(best_lines)
# Visualization
plt.imshow(image, cmap=“gray”)
plt.scatter(potential_points[:, 1], potential_points[:, 0], color=“red”, marker=“o”)
plt.show()
# # Plot the lines on the image
# for equation in best_lines:
# k, b = equation
# x = np.arange(0, image.shape[1])
# y = k * x + b
# plt.imshow(image, cmap=“gray”)
# plt.scatter(potential_points[:, 1], potential_points[:, 0], color=“red”, marker=“o”)
# plt.plot(x, y, color=“green”)
# plt.scatter(potential_points[:, 1], potential_points[:, 0], color=“red”, marker=“o”)
# plt.show()
Допиши метод get_best_lines так, чтобы он возвращал три лучших прямых, таких, что максимальное количество potential_points (потенциальных точек) лежат в окрестности 3 пикселя от прямой и k и b этих прямых не слишком похожи (tolerance_k=0.1, tolerance_b=10). То есть нужно вернуть три самых попадающих в точки прямых, но разных (детекция прямых, образующих треугольник). Для выполнения этой задачи можно использовать только numpy. Можно использовать только numpy
|
afb29076ec6adecb63f635f4bdaf1bab
|
{
"intermediate": 0.26834815740585327,
"beginner": 0.4421895742416382,
"expert": 0.2894623279571533
}
|
32,114
|
In a redis db I gotta save two things, 2. the hash of a question, then a list of hashes of answers to those questions. How should i implement this in python so i can search for a question, then get a list of answers
|
76eb0a15e970ab1f0c6ae2cb5d051d9c
|
{
"intermediate": 0.4109868109226227,
"beginner": 0.18300651013851166,
"expert": 0.40600669384002686
}
|
32,115
|
In a redis db I gotta save two things, 2. the hash of a question, then a list of hashes of answers to those questions. How should i implement this in python so i can search for a question, then get a list of answers. The whole system works only on hashes so the code does not need any example questions just placeholder hashes. Make functions for adding a question with a starting answer, to add a new answer to a question, and to add a question with a blank answer
|
c9f587e2a93419ba5dc034bd0aceb1db
|
{
"intermediate": 0.5295947790145874,
"beginner": 0.18345390260219574,
"expert": 0.28695130348205566
}
|
32,116
|
У меня есть набор точек, нужно реализовать на numpy алгоритм ransac, который будет находить прямые треугольника
|
ec36d8992e17dccf1d9294b07e0aa949
|
{
"intermediate": 0.23803986608982086,
"beginner": 0.18518075346946716,
"expert": 0.5767793655395508
}
|
32,117
|
image = cv2.imread("0.0/01.pgm", cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
"""Gets pixel coords with intensity more then 'theshold'."""
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
"""Deletes coords from image and returns this image."""
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
"""
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood's number.
"""
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
"""
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
"""
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
"""
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
"""
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_best_lines(potential_points: np.ndarray) -> np.ndarray:
"""
Gets best combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
"""
best_lines = []
best_inliers = []
iterations_num = 1000
threshold_distance = 3
for i in range(iterations_num):
sample_indices = np.random.choice(potential_points.shape[0], size=3, replace=False)
sample_points = potential_points[sample_indices]
sample_x = sample_points[:, 1]
sample_y = sample_points[:, 0]
coefficients = np.polyfit(sample_x, sample_y, deg=1)
distances = np.abs(potential_points[:, 0] - np.matmul(np.vstack((potential_points[:, 1], np.ones_like(potential_points[:, 1]))).T, coefficients)) / np.sqrt(coefficients[0] ** 2 + 1)
inliers = np.where(distances < threshold_distance)[0]
if len(inliers) > len(best_inliers):
best_lines = [coefficients]
best_inliers = inliers
elif len(inliers) == len(best_inliers):
best_lines.append(coefficients)
return np.array(best_lines)
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
best_lines = get_best_lines(potential_points)
print(best_lines)
# Visualization
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
# Plot the lines on the image
for equation in best_lines:
k, b = equation
x = np.arange(0, image.shape[1])
y = k * x + b
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.plot(x, y, color="green")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
Отлично, только алгоритм нашел самые качественные прямые, которые очень близко к друг другу (почти одинаковые прямые), а нужно три самых качественных, так как мы детектируем треугольник, то есть нужно выбрать самые качественные три прямые, не похожие по k, b
|
b52fa3a0f925926028042bc7af10fc46
|
{
"intermediate": 0.2918945848941803,
"beginner": 0.4008134603500366,
"expert": 0.3072919249534607
}
|
32,118
|
请将以下代码改成api后端，且用antidesign的react框架做对应的前端页面，可适当丰富功能：
import gradio as gr
from PIL import ImageColor
import onnxruntime
import cv2
import numpy as np
# The common resume photo size is 35mmx45mm
RESUME_PHOTO_W = 350
RESUME_PHOTO_H = 450
# modified from https://github.com/opencv/opencv_zoo/blob/main/models/face_detection_yunet/yunet.py
class YuNet:
def __init__(
self,
modelPath,
inputSize=[320, 320],
confThreshold=0.6,
nmsThreshold=0.3,
topK=5000,
backendId=0,
targetId=0,
):
self._modelPath = modelPath
self._inputSize = tuple(inputSize) # [w, h]
self._confThreshold = confThreshold
self._nmsThreshold = nmsThreshold
self._topK = topK
self._backendId = backendId
self._targetId = targetId
self._model = cv2.FaceDetectorYN.create(
model=self._modelPath,
config="",
input_size=self._inputSize,
score_threshold=self._confThreshold,
nms_threshold=self._nmsThreshold,
top_k=self._topK,
backend_id=self._backendId,
target_id=self._targetId,
)
@property
def name(self):
return self.__class__.__name__
def setBackendAndTarget(self, backendId, targetId):
self._backendId = backendId
self._targetId = targetId
self._model = cv2.FaceDetectorYN.create(
model=self._modelPath,
config="",
input_size=self._inputSize,
score_threshold=self._confThreshold,
nms_threshold=self._nmsThreshold,
top_k=self._topK,
backend_id=self._backendId,
target_id=self._targetId,
)
def setInputSize(self, input_size):
self._model.setInputSize(tuple(input_size))
def infer(self, image):
# Forward
faces = self._model.detect(image)
return faces[1]
class ONNXModel:
def __init__(self, model_path, input_w, input_h):
self.model = onnxruntime.InferenceSession(model_path)
self.input_w = input_w
self.input_h = input_h
def preprocess(self, rgb, mean=(0.5, 0.5, 0.5), std=(0.5, 0.5, 0.5)):
# convert the input data into the float32 input
img_data = (
np.array(cv2.resize(rgb, (self.input_w, self.input_h)))
.transpose(2, 0, 1)
.astype("float32")
)
# normalize
norm_img_data = np.zeros(img_data.shape).astype("float32")
for i in range(img_data.shape[0]):
norm_img_data[i, :, :] = img_data[i, :, :] / 255
norm_img_data[i, :, :] = (norm_img_data[i, :, :] - mean[i]) / std[i]
# add batch channel
norm_img_data = norm_img_data.reshape(1, 3, self.input_h, self.input_w).astype(
"float32"
)
return norm_img_data
def forward(self, image):
input_data = self.preprocess(image)
output_data = self.model.run(["argmax_0.tmp_0"], {"x": input_data})
return output_data
def make_resume_photo(rgb, background_color):
h, w, _ = rgb.shape
bgr = cv2.cvtColor(rgb, cv2.COLOR_RGB2BGR)
# Initialize models
face_detector = YuNet("models/face_detection_yunet_2023mar.onnx")
face_detector.setInputSize([w, h])
human_segmentor = ONNXModel(
"models/human_pp_humansegv2_lite_192x192_inference_model.onnx", 192, 192
)
# yunet uses opencv bgr image format
detections = face_detector.infer(bgr)
results = []
for idx, det in enumerate(detections):
# bounding box
pt1 = np.array((det[0], det[1]))
pt2 = np.array((det[0] + det[2], det[1] + det[3]))
# face landmarks
landmarks = det[4:14].reshape((5, 2))
right_eye = landmarks[0]
left_eye = landmarks[1]
angle = np.arctan2(right_eye[1] - left_eye[1], (right_eye[0] - left_eye[0]))
rmat = cv2.getRotationMatrix2D((0, 0), -angle, 1)
# apply rotation
rotated_bgr = cv2.warpAffine(bgr, rmat, (bgr.shape[1], bgr.shape[0]))
rotated_pt1 = rmat[:, :-1] @ pt1
rotated_pt2 = rmat[:, :-1] @ pt2
face_w, face_h = rotated_pt2 - rotated_pt1
up_length = int(face_h / 4)
down_length = int(face_h / 3)
crop_h = face_h + up_length + down_length
crop_w = int(crop_h * (RESUME_PHOTO_W / RESUME_PHOTO_H))
pt1 = np.array(
(rotated_pt1[0] - (crop_w - face_w) / 2, rotated_pt1[1] - up_length)
).astype(np.int32)
pt2 = np.array((pt1[0] + crop_w, pt1[1] + crop_h)).astype(np.int32)
resume_photo = rotated_bgr[pt1[1] : pt2[1], pt1[0] : pt2[0], :]
rgb = cv2.cvtColor(resume_photo, cv2.COLOR_BGR2RGB)
mask = human_segmentor.forward(rgb)
mask = mask[0].transpose(1, 2, 0)
mask = cv2.resize(
mask.astype(np.uint8), (resume_photo.shape[1], resume_photo.shape[0])
)
resume_photo = cv2.cvtColor(resume_photo, cv2.COLOR_BGR2RGB)
resume_photo[mask == 0] = ImageColor.getcolor(background_color, "RGB")
resume_photo = cv2.resize(resume_photo, (RESUME_PHOTO_W, RESUME_PHOTO_H))
results.append(resume_photo)
return results
title = "AI证件照：任意照片生成证件照||公众号：正经人王同学"
demo = gr.Interface(
fn=make_resume_photo,
inputs=[
gr.Image(type="numpy", label="input"),
gr.ColorPicker(label="设置背景颜色"),
],
outputs=gr.Gallery(label="output"),
examples=[
["images/queen.png", "#FFFFFF"],
["images/elon.png", "#FFFFFF"],
["images/openai.jpg", "#FFFFFF"],
["images/sam.png", "#FFFFFF"],
],
title=title,
allow_flagging="never",
article="<p style='text-align: center;'><a href='https://mp.weixin.qq.com/s/SvQT5elBPuPYJ0CJ_LlBYA' target='_blank'>公众号：正经人王同学</a></p>",
)
if __name__ == "__main__":
demo.launch(server_name="0.0.0.0",server_port=7860)
|
fd0cab03a432e69843c5fa8295ce6b72
|
{
"intermediate": 0.26510709524154663,
"beginner": 0.5357580780982971,
"expert": 0.19913485646247864
}
|
32,119
|
help me create a Cellular Automata | Procedural Generation for godot 4
|
8c51aeb995d0cfd5cbf0ab4850f3251f
|
{
"intermediate": 0.19293086230754852,
"beginner": 0.14989322423934937,
"expert": 0.6571758985519409
}
|
32,120
|
image = cv2.imread("0.0/01.pgm", cv2.COLOR_BGR2GRAY)
def get_coords_by_threshold(image: np.ndarray, threshold: int = 150) -> tuple:
"""Gets pixel coords with intensity more then 'theshold'."""
coords = np.where(image > threshold)
return coords
def delete_pixels_by_min_threshold(
image: np.ndarray, min_threshold: int = 100
) -> np.ndarray:
"""Deletes coords from image and returns this image."""
anti_coords = np.where(image < min_threshold)
image[anti_coords] = 0
return image
def get_neigh_nums_list(
coords: tuple, kernel_size: int = 3
) -> list:
"""
Gets list with number of neighborhoods by kernel.
Returns list like [[x1, y1, N1], [x2, y2, N2]],
Where N - neighborhood's number.
"""
neigh_nums = []
offset = (kernel_size - 1) // 2
for x, y in zip(coords[0], coords[1]):
if x < 0 + offset or x > image.shape[0] - offset:
continue
if y < 0 + offset or y > image.shape[1] - offset:
continue
kernel = image[x-offset:x+offset+1, y-offset:y+offset+1]
step_neigh_num = np.count_nonzero(kernel) - 1
neigh_nums.append([x, y, step_neigh_num])
return neigh_nums
def sort_neight_nums_by_N(neigh_nums: list) -> np.ndarray:
"""
Sort neigh_nums by N parameter.
Removes N=-1, N=0 and sorts by N desc.
"""
np_neigh_nums = np.array(neigh_nums)
np_neigh_nums = np_neigh_nums[
(np_neigh_nums[:, 2] != -1) & (np_neigh_nums[:, 2] != 0)]
np_neigh_nums = np_neigh_nums[
np.argsort(-np_neigh_nums[:, 2])]
return np_neigh_nums
def get_potential_points_coords(
sorted_niegh_nums: np.ndarray, potential_points_num: int = 30
) -> np.ndarray:
"""
Gets best candidates, potential points coords only.
Returns np.ndarray like [[x1, y1], [x2, y2]].
"""
sorted_neigh_nums = np.delete(sorted_niegh_nums, 2, axis=1)
return sorted_neigh_nums[:potential_points_num]
def get_best_lines(potential_points: np.ndarray) -> np.ndarray:
"""
Gets best combinations of lines by all points.
Line y = kx + b.
Returns np.ndarray like [[k1, b1], [k2, b2]].
"""
# best_lines = []
# best_inliers = []
# iterations_num = 1000
# threshold_distance = 3
# threshold_k = 0.1
# threshold_b = 10
# for i in range(iterations_num):
# sample_indices = np.random.choice(potential_points.shape[0], size=3, replace=False)
# sample_points = potential_points[sample_indices]
# sample_x = sample_points[:, 1]
# sample_y = sample_points[:, 0]
# coefficients = np.polyfit(sample_x, sample_y, deg=1)
# distances = np.abs(potential_points[:, 0] - np.matmul(np.vstack((potential_points[:, 1], np.ones_like(potential_points[:, 1]))).T, coefficients)) / np.sqrt(coefficients[0] ** 2 + 1)
# inliers = np.where(distances < threshold_distance)[0]
# is_similar = False
# for line in best_lines:
# diff_k = abs(coefficients[0] - line[0])
# diff_b = abs(coefficients[0] - line[0])
# if diff_k <= threshold_k and diff_b <= threshold_b:
# is_similar = True
# break
# if not is_similar and len(inliers) > len(best_inliers):
# best_lines = [coefficients]
# best_inliers = inliers
# elif not is_similar and len(inliers) == len(best_inliers):
# best_lines.append(coefficients)
# return np.array(best_lines)
best_lines = []
best_inliers = []
iterations_num = 1000
threshold_distance = 5
threshold_k = 0.25
threshold_b = 10
for _ in range(3):
local_best_lines = []
local_best_inliers = []
for i in range(iterations_num):
sample_indices = np.random.choice(potential_points.shape[0], size=3, replace=False)
sample_points = potential_points[sample_indices]
sample_x = sample_points[:, 1]
sample_y = sample_points[:, 0]
coefficients = np.polyfit(sample_x, sample_y, deg=1)
distances = np.abs(potential_points[:, 0] - np.matmul(np.vstack((potential_points[:, 1], np.ones_like(potential_points[:, 1]))).T, coefficients)) / np.sqrt(coefficients[0] ** 2 + 1)
inliers = np.where(distances < threshold_distance)[0]
# Check if the current line is similar to any of the existing lines
is_similar = False
for line in local_best_lines:
# Calculate the difference between k and b of the current line and the existing lines
diff_k = abs(coefficients[0] - line[0])
diff_b = abs(coefficients[0] - line[0])
# If the difference is smaller than the threshold, consider the current line as similar to the existing line
if diff_k <= threshold_k and diff_b <= threshold_b:
is_similar = True
break
# If the current line is not similar to any of the existing lines and has more inliers, consider it as a potential best line
if not is_similar and len(inliers) > len(local_best_inliers):
local_best_lines = [coefficients]
local_best_inliers = inliers
elif not is_similar and len(inliers) == len(local_best_inliers):
local_best_lines.append(coefficients)
# Update the potential points by removing the points that belong to the vicinity of the current best line
mask = np.ones(len(potential_points), dtype=bool)
for equation in local_best_lines:
k, b = equation
x = potential_points[:, 1]
y = potential_points[:, 0]
distances = np.abs(y - (k * x + b))
mask[np.where(distances < threshold_distance)[0]] = False
best_lines.extend(local_best_lines)
best_inliers.extend(local_best_inliers)
# Update the potential points by keeping only the points that were not removed in the vicinity of the current best line
potential_points = potential_points[mask]
return np.array(best_lines[:3])
# main
coords = get_coords_by_threshold(image)
image = delete_pixels_by_min_threshold(image)
neigh_nums = get_neigh_nums_list(coords)
sorted_neigh_nums = sort_neight_nums_by_N(neigh_nums)
potential_points = get_potential_points_coords(sorted_neigh_nums)
best_lines = get_best_lines(potential_points)
print(best_lines)
# Visualization
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
# Plot the lines on the image
for equation in best_lines:
k, b = equation
x = np.arange(0, image.shape[1])
y = k * x + b
plt.imshow(image, cmap="gray")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.plot(x, y, color="green")
plt.scatter(potential_points[:, 1], potential_points[:, 0], color="red", marker="o")
plt.show()
Находит слишком много прямых, причем похожих друг на друга, помимо этого часто ловлю ошибку:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-74-a68b10ff2bde> in <cell line: 183>()
181 potential_points = get_potential_points_coords(sorted_neigh_nums)
182
--> 183 best_lines = get_best_lines(potential_points)
184 print(best_lines)
185
<ipython-input-74-a68b10ff2bde> in get_best_lines(potential_points)
121
122 for i in range(iterations_num):
--> 123 sample_indices = np.random.choice(potential_points.shape[0], size=3, replace=False)
124
125 sample_points = potential_points[sample_indices]
mtrand.pyx in numpy.random.mtrand.RandomState.choice()
ValueError: Cannot take a larger sample than population when 'replace=False'
Что делать?!
|
4c722d801a8186183018181c704b3ff7
|
{
"intermediate": 0.28463754057884216,
"beginner": 0.3822076618671417,
"expert": 0.3331548273563385
}
|
32,121
|
acedGetCorner
|
aed35ba635df2b06c61d2e369b938f4b
|
{
"intermediate": 0.3421335816383362,
"beginner": 0.2531798779964447,
"expert": 0.40468651056289673
}
|
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