重叠出现的字符串计数

String count with overlapping occurrences(重叠出现的字符串计数)

问题描述

What's the best way to count the number of occurrences of a given string, including overlap in Python? This is one way:

def function(string, str_to_search_for):
      count = 0
      for x in xrange(len(string) - len(str_to_search_for) + 1):
           if string[x:x+len(str_to_search_for)] == str_to_search_for:
                count += 1
      return count


function('1011101111','11')

This method returns 5.

Is there a better way in Python?

Well, this might be faster since it does the comparing in C:

def occurrences(string, sub):
    count = start = 0
    while True:
        start = string.find(sub, start) + 1
        if start > 0:
            count+=1
        else:
            return count

这篇关于重叠出现的字符串计数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！