python3 bisect() bisect_left()函数_python3 bisect_left

前提：列表有序！！！

bisect()和bisect_right()等同，那下面就介绍bisect_left()和bisec_right()的区别！

用法：

index1 = bisect(ls, x)   #第1个参数是列表，第2个参数是要查找的数，返回值为索引
index2 = bisect_left(ls, x)
index3 = bisec_right(ls, x)

case 1
如果列表中没有元素x，那么bisect_left(ls, x)和bisec_right(ls, x)返回相同的值，该值是x在ls中“合适的插入点索引，使得数组有序”。此时，ls[index2] > x，ls[index3] > x。

import bisect
ls = [1,5,9,13,17]
index1 = bisect.bisect(ls,7)
index2 = bisect.bisect_left(ls,7)
index3 = bisect.bisect_right(ls,7)
print("index1 = {}, index2 = {}, index3 = {}".format(index1, index2, index3))
 

程序运行结果为，index1 = 2, index2 = 2, index3 = 2

case 2
如果列表中只有一个元素等于x，那么bisect_left(ls, x)的值是x在ls中的索引，ls[index2] = x。而bisec_right(ls, x)的值是x在ls中的索引加1，ls[index3] > x。

import bisect
ls = [1,5,9,13,17]
index1 = bisect.bisect(ls,9)
index2 = bisect.bisect_left(ls,9)
index3 = bisect.bisect_right(ls,9)
print("index1 = {}, index2 = {}, index3 = {}".format(index1, index2, index3))

程序运行结果为，index1 = 3, index2 = 2, index3 = 3

case 3
如果列表中存在多个元素等于x，那么bisect_left(ls, x)返回最左边的那个索引，此时ls[index2] = x。bisect_right(ls, x)返回最右边的那个索引加1，此时ls[index3] > x。

import bisect
ls = [1,5,5,5,17]
index1 = bisect.bisect(ls,5)
index2 = bisect.bisect_left(ls,5)
index3 = bisect.bisect_right(ls,5)
print("index1 = {}, index2 = {}, index3 = {}".format(index1, index2, index3))
 

程序运行结果为，index1 = 4, index2 = 1, index3 = 4

引用自：https://blog.csdn.net/YMWM_/article/details/122378152