找到第 n 个 '|' 之后的子串 [英] Find the sub-string after nth '|'

问题描述

给定一个字符串如下:

1|2||||auc|0|1||0|||76u|
      ^ 

返回第 5 个|"之后的子字符串的最有效方法是什么?例如，给定上面的字符串，结果应该是:

what is the most efficient way to return the substring after the 5th '|'? For example, given the above string, the result should be:

auc|0|1||0|||76u|

推荐答案

使用 str.split:

s = '1|2||||auc|0|1||0|||76u|'
print s.split('|', 5)[-1]
# auc|0|1||0|||76u|

注意，如果没有至少 5 个 |s，这可能会导致不希望的结果，例如，

Note, this will cause possibly undesired results if there's not at least 5 |s, eg,

'1|2'.split('|', 5)[-1]
# returns 2 - which isn't *after* the 5th

存在于字符串中，因此您可能希望将其包装在 try/except 中并强制处理没有足够 | 的情况，以便结果 在第 5 个之后 是空的，因为没有 5 个.

present in the string, so you may wish to wrap it in a try/except and force handle the case where there aren't sufficient |s so that the result after the 5th is empty as there wasn't 5 present.

try:
    rest = s.split('|', 5)[5]
except IndexError:
    rest = ''

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