Python中字符串中子字符串的重叠计数 [英] Overlapping count of substring in a string in Python

问题描述

我想在一个字符串中找到一个子字符串的所有计数(重叠和非重叠).我找到了两个答案，其中一个使用正则表达式，这不是我的意图，另一个比我需要的效率低得多.我需要类似的东西:

I want to find all the counts (overlapping and non-overlapping) of a sub-string in a string. I found two answers one of which is using regex which is not my intention and the other was much more in-efficient than I need. I need something like:

'ababaa'.count('aba') == 2

str.count() 只计算简单的子串.我该怎么办?

str.count() just counts simple substrings. What should I do?

推荐答案

def sliding(a, n):
    return (a[i:i+n] for i in xrange(len(a) - n + 1))

def substring_count(a, b):
    return sum(s == b for s in sliding(a, len(b)))

assert list(sliding('abcde', 3)) == ['abc', 'bcd', 'cde']    
assert substring_count('ababaa', 'aba') == 2

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