查找字符串中第 n 次出现的子字符串 [英] Find the nth occurrence of substring in a string

问题描述

这看起来应该很简单，但我是 Python 新手，想用最 Python 化的方式来做.

This seems like it should be pretty trivial, but I am new at Python and want to do it the most Pythonic way.

我想找到对应于字符串中子字符串第 n 次出现的索引.

I want to find the index corresponding to the n'th occurrence of a substring within a string.

必须有一些与我想做的事情相同的事情

There's got to be something equivalent to what I WANT to do which is

mystring.find("substring", 2nd)

如何在 Python 中实现这一点?

How can you achieve this in Python?

推荐答案

Mark 的迭代方法将是通常的方法，我认为.

Mark's iterative approach would be the usual way, I think.

这里有一个字符串分割的替代方法，它通常对查找相关的过程很有用:

Here's an alternative with string-splitting, which can often be useful for finding-related processes:

def findnth(haystack, needle, n):
    parts= haystack.split(needle, n+1)
    if len(parts)<=n+1:
        return -1
    return len(haystack)-len(parts[-1])-len(needle)

这是一个快速(有点脏，因为你必须选择一些与针头不匹配的箔条)单线:

And here's a quick (and somewhat dirty, in that you have to choose some chaff that can't match the needle) one-liner:

'foo bar bar bar'.replace('bar', 'XXX', 1).find('bar')

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