在列表中没有分隔符的多个分隔符处拆分

时间:2013-06-11 16:53:15

标签: python regex split

使用re库这应该是一项非常简单的任务。但是,我似乎无法在分隔符][上分割我的字符串。

我已阅读Splitting a string with multiple delimiters in PythonPython: Split string with multiple delimitersPython: How to get multiple elements inside square brackets

我的字符串:

data = "This is a string spanning over multiple lines.
        At somepoint there will be square brackets.

        [like this]

        And then maybe some more text.

        [And another text in square brackets]"

它应该返回:

['This is a string spanning over multiple lines.\nAt somepoint there will be square brackets.','like this', 'And then maybe some more text.', 'And another text in square brackets']

尝试的简短示例:

data2 = 'A new string. [with brackets] another line [and a bracket]'

我试过了:

re.split(r'(\[|\])', data2)
re.split(r'([|])', data2)

但是这些会导致我的结果列表中的分隔符或完全错误的列表:

['A new string. ', '[', 'with brackets', ']', ' another line ', '[', 'and a bracket', ']', '']

结果应该是:

['A new string.', 'with brackets', 'another line', 'and a bracket']

作为一项特殊要求,应删除分隔符前后的所有换行符和空格,并且不应包含在列表中。

4 个答案:

答案 0 :(得分:7)

>>> re.split(r'\[|\]', data2)
['A new string. ', 'with brackets', ' another line ', 'and a bracket', '']

答案 1 :(得分:4)

正如arshajii指出的那样,对于这个特殊的正则表达式,你根本不需要任何组。

如果确实需要组来表示更复杂的正则表达式,则可以使用非捕获组进行拆分而不捕获分隔符。它可能对其他情况很有用,但在语法上凌乱有点过分。

  

(?:...)

A non-capturing version of regular parentheses. Matches whatever regular expression is inside the parentheses, but the substring matched by the group cannot be retrieved after performing a match or referenced later in the pattern.

http://docs.python.org/2/library/re.html

因此,这里不必要的复杂但具有示范性的例子将是:

re.split(r'(?:\[|\])', data2)

答案 2 :(得分:2)

改为使用它(没有捕获组):

re.split(r'\s*\[|]\s*', data)

或更短:

re.split(r'\s*[][]\s*', data)

答案 3 :(得分:0)

Couuld要么拆分要么全部找到,例如:

data2 = 'A new string. [with brackets] another line [and a bracket]'

使用拆分和过滤掉前导/尾随空格:

import re
print filter(None, re.split(r'\s*[\[\]]\s*', data2))
# ['A new string.', 'with brackets', 'another line', 'and a bracket']

或者可能采用findall方法:

print re.findall(r'[^\b\[\]]+', data2)
# ['A new string. ', 'with brackets', ' another line ', 'and a bracket'] # needs a little work on leading/trailing stuff...