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Here's hoping somebody can shed some light on this question because it has me stumped. I have a string that looks like this:

s = "abcdef [[xxxx xxx|ghijk]] lmnop [[qrs]] tuv [[xx xxxx|wxyz]] 0123456789"

I want this result:

abcdef ghijk lmnop qrs tuv wxyz 0123456789

Having reviewed numerous questions and answers here, the closest I have come to a solution is:

s = "abcdef [[xxxx xxx|ghijk]] lmnop [[qrs]] tuv [[xx xxxx|wxyz]] 0123456789"
s = re.sub('\[\[.*?\|', '', s)
s = re.sub('[\]\]]', '', s)
--> abcdef ghijk lmnop wxyz 0123456789

Since not every substring within double brackets contains a pipe, the re.sub removes everything from '[[' to next '|' instead of checking within each set of double brackets.

Any assistance would be most appreciated.

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3
  • Can you please describe the discriminating factor, i.e., why those particular substrings? Commented Jun 30, 2015 at 21:35
  • Try using raw strings for your regular expressions, like r'\[\[.*?\|'. Commented Jun 30, 2015 at 21:36
  • @TigerhawkT3 in my text whenever a pipe occurs between a set of double brackets, everything before the pipe is a description I don't need in the final result. Commented Jun 30, 2015 at 22:05

4 Answers 4

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What about this:

In [187]: re.sub(r'([\[|\]])|((?<=\[)\w+\s+\w+(?=|))', '', s)
Out[187]: 'abcdef ghijk lmnop qrs tuv wxyz 0123456789'
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1 Comment

Should have specified that the x's between '[[' and the '|' represent a substring unknown to me. How can I alter your suggested regex to include any character?
1

I purpose you a contrary method, instead of remove it you can just catch patterns you want. I think this way can make your code more semantics.

There are two patterns you wish to catch:

  1. Case: words outside [[...]]

    Pattern: Any words are either leaded by ']] ' or trailed by ' [['.

    Regex: (?<=\]\]\s)\w+|\w+(?=\s\[\[)

  2. Case: words inside [[...]]

    Pattern: Any words are trailed by ']]'

    Regex: \w+(?=\]\])

Example code

1 #!/usr/bin/env python
2 import re
3
4 s = "abcdef [[xxxx xxx|ghijk]] lmnop [[qrs]] tuv [[xx xxxx|wxyz]] 0123456789    "
5
6 p = re.compile('(?<=\]\]\s)\w+|\w+(?=\s\[\[)|\w+(?=\]\])')
7 print p.findall(s)

Result:

['abcdef', 'ghijk', 'lmnop', 'qrs', 'tuv', 'wxyz', '0123456789']
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0 
>>> import re
>>> s = "abcdef [[xxxx xxx|ghijk]] lmnop [[qrs]] tuv [[xx xxxx|wxyz]] 0123456789"
>>> re.sub(r'(\[\[[^]]+?\|)|([\[\]])', '', s)
'abcdef ghijk lmnop qrs tuv wxyz 0123456789'

This searches for and removes the following two items:

  1. Two opening brackets followed by a bunch of stuff that isn't a closing bracket followed by a pipe.
  2. Opening or closing brackets.
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just ran your solution through a sample of my file and it worked with every entry. Comes after a lot of frustration on the issue. Many thanks!
0

As a general regex using built-in re module you can use follwing regex that used look-around:

(?<!\[\[)\b([\w]+)\b(?!\|)|\[\[([^|]*)\]\]

you can use re.finditer to get the desire result :

>>> g=re.finditer(r'(?<!\[\[)\b([\w]+)\b(?!\|)|(?<=\[\[)[^|]*(?=\]\])',s)
>>> [j.group() for j in g]
['abcdef', 'ghijk', 'lmnop', 'qrs', 'tuv', 'wxyz', '0123456789']

The preceding regex contains from 2 part one is :

(?<=\[\[)[^|]*(?=\]\])

which match any combinations of word characters that not followed by | and not precede by [[.

the second part is :

\[\[([^|]*)\]\]

that will match any thing between 2 brackets except |.

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