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up vote 1 down vote favorite

Now I have a list of pattern:

patterns = ['php', 'java', 'c++']

and I want to match it in another string, say, r'c++ primer'. I want to use python re module to do it, but the problem is, if I use:

for pattern in patterns:
    re.findall(pattern, r'php php java java c++ c++')

I will get an error because '+' has special meanning in regular expression.

So how can I fix something like c++ or c* in this situation?

Notice that I have a lot of patterns to match so I don't want to convert everything like c++ to c\+\+ manually.

Thanks for your attention.

share|improve this question
    
Made an edit to the question that clarified \+\+ was not what the asker is looking for. But you SHOULD do a Ctrl+H search with "c++" and replace with "c\+\+". – leewangzhong Dec 8 '13 at 7:51
    
Sorry what do you mean by Ctrl+H search? – Warbean Dec 8 '13 at 7:54
    
@Warbean He means, Search & Replace in the editor. – thefourtheye Dec 8 '13 at 7:54
    
Can someone with the relevant power please accept my edit to the question? He means he doesn't want to change to "c\+\+". – leewangzhong Dec 8 '13 at 7:56
    
Depending on what you are doing, the regular string methods might be enough, so no worrying about making valid regex patterns, e.g: for p in patterns: print "php php java java c++ c++".count(p) will show how many times the string occurs, or for p in patterns: print p in "php php java ..." will show if the string contains the pattern at all – dbr Dec 8 '13 at 8:36

2 Answers 2

up vote 4 down vote

Use a character class. Outside of a character class + a special meaning so it is not going to work as is, you need to escape it first: r'c\+\+'.

>>> import re
>>> re.findall(r'[+]{2}', r'c++ primer') 
['++']

Update 1:

If you've predefined regexes then use re.escape on those patterns:

>>> patterns = ['php', 'java', 'c++']
>>> for pattern in patterns:
        print re.findall(re.escape(pattern), r'php php java java c++ c++')
...     
['php', 'php']
['java', 'java']
['c++', 'c++']

Update 2:

>>> to_be_escaped = ('c++',)  #patterns that need to be escaped
>>> new_patterns = [re.escape(p) if p in to_be_escaped else p for p in patterns]
>>> for pattern in new_patterns:
        print re.findall(pattern, r'php php java java c++ c++ .net')
...     
['php', 'php']
['java', 'java']
['c++', 'c++']
['.net']
share|improve this answer
    
Sorry that is not my situation. I have edit my question to make myself understood. Can you please review it to help me? Thank you. – Warbean Dec 8 '13 at 7:52
    
@Warbean I've updated my solution. – Ashwini Chaudhary Dec 8 '13 at 7:53
    
+1 Neat solution :) – thefourtheye Dec 8 '13 at 7:55
    
That's what I want! Thank you so much! – Warbean Dec 8 '13 at 7:55
    
Sorry I have a new problem. I just want to convert 'c++', but not someting like '.net', because '.' is not special in regx. But re.escape will convert '.net' to '\.net', too. Can I choose which to escape and which not? Thanks so much! – Warbean Dec 8 '13 at 8:06
up vote 3 down vote

Escape + with \ like this

pattern = r'c\+\+'
import re
print re.findall(pattern, r'c++ primer')

Output

['c++']

Edit:

import re
patterns = ['php', 'java', 'c\+\+']
for pattern in patterns:
    print re.findall(pattern, r'php php java java c++ c++')

Output

['php', 'php']
['java', 'java']
['c++', 'c++']
share|improve this answer
    
Sorry that is not my situation. I have edit my question to make myself understood, can you please review it to help me? Thank you. – Warbean Dec 8 '13 at 7:51
    
@Warbean Just change the patterns like this patterns = ['php', 'java', 'c\+\+'] – thefourtheye Dec 8 '13 at 7:53
    
Can' solve my problem sill. @Ashwini Chaudhary has solved it. Thank you all the same.^_^ – Warbean Dec 8 '13 at 7:56
    
@Warbean You mean to say that the sample which I have shown will not work? – thefourtheye Dec 8 '13 at 7:59
    
Not meaning that. I just don't want to convert 'c++' to 'c\+\+' manually because I have a list of something like 'c++' in hand now, all I want is auto converting. So re.escape('c++') help me. – Warbean Dec 8 '13 at 8:02

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