Bisect

This module provides support for maintaining a list in sorted order without having to sort the list after each insertion. For long lists of items with expensive comparison operations, this can be an improvement over the more common approach.

bisect.bisect_left(a, x, lo=0, hi=len(a))

Locate the insertion point for x in a to maintain sorted order. The parameters lo and hi may be used to specify a subset of the list which should be considered; by default the entire list is used. If x is already present in a, the insertion point will be before (to the left of) any existing entries. The return value is suitable for use as the first parameter to list.insert() assuming that a is already sorted.

The returned insertion point i partitions the array a into two halves so that all(val < x for val in a[lo:i]) for the left side and all(val >= x for val in a[i:hi]) for the right side.
This is can also be used to find the index of first apperence

>>> data = [('red', 5), ('blue', 1), ('yellow', 8), ('black', 0)]
>>> data.sort(key=lambda r: r[1])
>>> keys = [r[1] for r in data]         # precomputed list of keys
#
>>> data[bisect_left(keys, 0)]
('black', 0)
>>> data[bisect_left(keys, 1)]
('blue', 1)
>>> data[bisect_left(keys, 5)]
('red', 5)
>>> data[bisect_left(keys, 8)]
('yellow', 8)

bisect.bisect_right(a, x, lo=0, hi=len(a))bisect.bisect(a, x, lo=0, hi=len(a))

Similar to bisect_left(), but returns an insertion point which comes after (to the right of) any existing entries of x in a.

The returned insertion point i partitions the array a into two halves so that all(val <= x for val in a[lo:i]) for the left side and all(val > x for val in a[i:hi]) for the right side.

>>> def grade(score, breakpoints=[60, 70, 80, 90], grades='FDCBA'):
...     i = bisect(breakpoints, score)
...     return grades[i]
...
>>> [grade(score) for score in [33, 99, 77, 70, 89, 90, 100]]
['F', 'A', 'C', 'C', 'B', 'A', 'A']

bisect.insort_left(a, x, lo=0, hi=len(a))

Insert x in a in sorted order. This is equivalent to a.insert(bisect.bisect_left(a, x, lo, hi), x) assuming that a is already sorted. Keep in mind that the O(log n) search is dominated by the slow O(n) insertion step.

bisect.insort_right(a, x, lo=0, hi=len(a))bisect.insort(a, x, lo=0, hi=len(a))

Similar to insort_left(), but inserting x in a after any existing entries of x.

li1 = [1, 3, 4, 4, 4, 6, 7]
# inserts at 6th position
bisect.insort(li1, 5)
#1 3 4 4 4 5 6 
li2 = [1, 3, 4, 4, 4, 6, 7]
bisect.insort_right(li2, 5, 0, 4)
# 1 3 4 4 5 4 6 

The following five functions show how to transform them into the standard lookups for sorted lists:

def index(a, x):
    'Locate the leftmost value exactly equal to x'
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    raise ValueError

def find_lt(a, x):
    'Find rightmost value less than x'
    i = bisect_left(a, x)
    if i:
        return a[i-1]
    raise ValueError

def find_le(a, x):
    'Find rightmost value less than or equal to x'
    i = bisect_right(a, x)
    if i:
        return a[i-1]
    raise ValueError

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError

def find_ge(a, x):
    'Find leftmost item greater than or equal to x'
    i = bisect_left(a, x)
    if i != len(a):
        return a[i]
    raise ValueError