如何找到所有出现的子字符串?

问题描述

Python 有 string.find() 和 string.rfind() 来获取字符串中子串的索引.

Python has string.find() and string.rfind() to get the index of a substring in a string.

我想知道是否有类似 string.find_all() 的东西可以返回所有找到的索引(不仅是从头开始的第一个索引，也不是从最后的第一个索引).

I'm wondering whether there is something like string.find_all() which can return all found indexes (not only the first from the beginning or the first from the end).

例如:

string = "test test test test"

print string.find('test') # 0
print string.rfind('test') # 15

#this is the goal
print string.find_all('test') # [0,5,10,15]

推荐答案

没有简单的内置字符串函数可以满足您的需求，但您可以使用更强大的 正则表达式:

There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:

import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]

如果您想查找重叠匹配，lookahead 会这样做:

If you want to find overlapping matches, lookahead will do that:

[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]

如果你想要一个没有重叠的反向查找，你可以将正负前瞻组合成这样的表达式:

If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:

search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]

re.finditer 返回一个generator，所以你可以把上面的 [] 改成 () 来获得一个生成器而不是一个列表，如果你只迭代一次结果，这将更有效.

re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.

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