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Python Regular Expression用于查找特定模式中最后一次出现的空格

[英]Python Regular Expression to find the last occurrence of whitespace in a certain pattern

I am using Regular expression in Python. 我在Python中使用正则表达式。 I want to find the string before last occurrence of whitespace in a certain pattern. 我想在特定模式中最后一次出现空格之前找到字符串。 For example In the following text, I want to find "Street". 例如,在下面的文字中,我想找到“街道”。 "On Monday , a worker at a [LOC Te Rapa Tika Street ]". “星期一,[LOC Te Rapa Tika Street]的一名工人”。

Can anyone help me to find the string using regular expression? 任何人都可以帮我找到使用正则表达式的字符串吗?

Thanks 谢谢

Split the string on spaces and get the second last element: 在空格上拆分字符串并获取第二个最后一个元素:

>>> strs = "On Monday , a worker at a [LOC Te Rapa Tika Street ]"
>>> strs.split()[-2]
'Street'

re.split by \\s+ and take the second last token in the returned list (eg using index -2 ). re.split \\s+ re.split by \\s+并获取返回列表中的第二个最后一个标记(例如,使用索引-2 )。

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

>>>  import re
>>>  match = re.search('\[\s?LOC.+\s(\w+)\s?\]', "[LOC Te Rapa Tika Street ]")
>>>  match.group(1)
'Street'

This should work regardless of the spacing on the brackets. 无论支架上的间距如何,这都应该有效。

Edit: After reading your comment, this would work better 编辑:阅读您的评论后,这将更好

   >>>  import re
   >>>  sentence = "A man strolling through the [LOC Pullman Hotel ] in [LOC Waterloo Quadrant ] on Sunday with the bag across his shoulder"
   >>>  match = re.findall('\[\s?LOC[^\]]+\s(\w+)\s?\]', sentence)
   >>>  match
   ['Hotel', 'Quadrant']

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