| | """Bisection algorithms.""" |
| |
|
| |
|
| | def insort_right(a, x, lo=0, hi=None, *, key=None): |
| | """Insert item x in list a, and keep it sorted assuming a is sorted. |
| | |
| | If x is already in a, insert it to the right of the rightmost x. |
| | |
| | Optional args lo (default 0) and hi (default len(a)) bound the |
| | slice of a to be searched. |
| | """ |
| | if key is None: |
| | lo = bisect_right(a, x, lo, hi) |
| | else: |
| | lo = bisect_right(a, key(x), lo, hi, key=key) |
| | a.insert(lo, x) |
| |
|
| |
|
| | def bisect_right(a, x, lo=0, hi=None, *, key=None): |
| | """Return the index where to insert item x in list a, assuming a is sorted. |
| | |
| | The return value i is such that all e in a[:i] have e <= x, and all e in |
| | a[i:] have e > x. So if x already appears in the list, a.insert(i, x) will |
| | insert just after the rightmost x already there. |
| | |
| | Optional args lo (default 0) and hi (default len(a)) bound the |
| | slice of a to be searched. |
| | """ |
| |
|
| | if lo < 0: |
| | raise ValueError('lo must be non-negative') |
| | if hi is None: |
| | hi = len(a) |
| | |
| | |
| | if key is None: |
| | while lo < hi: |
| | mid = (lo + hi) // 2 |
| | if x < a[mid]: |
| | hi = mid |
| | else: |
| | lo = mid + 1 |
| | else: |
| | while lo < hi: |
| | mid = (lo + hi) // 2 |
| | if x < key(a[mid]): |
| | hi = mid |
| | else: |
| | lo = mid + 1 |
| | return lo |
| |
|
| |
|
| | def insort_left(a, x, lo=0, hi=None, *, key=None): |
| | """Insert item x in list a, and keep it sorted assuming a is sorted. |
| | |
| | If x is already in a, insert it to the left of the leftmost x. |
| | |
| | Optional args lo (default 0) and hi (default len(a)) bound the |
| | slice of a to be searched. |
| | """ |
| |
|
| | if key is None: |
| | lo = bisect_left(a, x, lo, hi) |
| | else: |
| | lo = bisect_left(a, key(x), lo, hi, key=key) |
| | a.insert(lo, x) |
| |
|
| | def bisect_left(a, x, lo=0, hi=None, *, key=None): |
| | """Return the index where to insert item x in list a, assuming a is sorted. |
| | |
| | The return value i is such that all e in a[:i] have e < x, and all e in |
| | a[i:] have e >= x. So if x already appears in the list, a.insert(i, x) will |
| | insert just before the leftmost x already there. |
| | |
| | Optional args lo (default 0) and hi (default len(a)) bound the |
| | slice of a to be searched. |
| | """ |
| |
|
| | if lo < 0: |
| | raise ValueError('lo must be non-negative') |
| | if hi is None: |
| | hi = len(a) |
| | |
| | |
| | if key is None: |
| | while lo < hi: |
| | mid = (lo + hi) // 2 |
| | if a[mid] < x: |
| | lo = mid + 1 |
| | else: |
| | hi = mid |
| | else: |
| | while lo < hi: |
| | mid = (lo + hi) // 2 |
| | if key(a[mid]) < x: |
| | lo = mid + 1 |
| | else: |
| | hi = mid |
| | return lo |
| |
|
| |
|
| | |
| | try: |
| | from _bisect import * |
| | except ImportError: |
| | pass |
| |
|
| | |
| | bisect = bisect_right |
| | insort = insort_right |
| |
|