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I'm currently working on python code to extract a specific strings on a text file between ' '. Here's the code I'm working on:

import re
pattern = "(\w+'*',)"
with open('C:\Documents and Settings\Ash\Desktop\strcount.txt', "r") as text:
    for line in text:
        print re.findall(pattern, line)

and the list of strings in the text file

('FLBC8U', 24)
('cHvQuW', 24)
('7FDm@', 24)
('15ii?', 24)
('H!oDe', 24)
('RB6#U', 24)
('uAmer', 24)
('6NmDJ', 24)
('d-MS1', 24)
('Ejf&B', 24)

I only wanted to take the string in the middle of ' ' single quotation mark before the comma , so the number and the bracket is ignored

Brian Tompsett - 汤莱恩
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TheGameDoctor
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  • possible duplicate of [Python Regex to find a string in double quotes within a string](http://stackoverflow.com/questions/9519734/python-regex-to-find-a-string-in-double-quotes-within-a-string) or [Regex for quoted string with escaping quotes](http://stackoverflow.com/questions/249791/regex-for-quoted-string-with-escaping-quotes) – Robin Jun 02 '14 at 09:24

5 Answers5

2
s = "'FLBC8U', 24"

print re.findall("'([^']*)'", s)[0]
FLBC8U
Padraic Cunningham
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0

This should solve it for you:

(?<=\(')((?!').)*(?=')

Explanation here

Dhrubajyoti Gogoi
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  • please add some details on why your proposed solution answer the given question. One-line-like answers are likely to be deleted in the future. – furins Jun 02 '14 at 09:40
0

My regex:

(?<=\').*(?=\',)

You can check the result here.

By the way, this site is quite helpful for playing with regex in python.

Ray
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Try using this pattern:

   \(\'(.*)\',.+\)

example: http://regex101.com/r/aZ6tU6

l'L'l
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If the strings are simply in a line, you can just do this:

>>> "('FLBC8U', 24)"[1:-1].split(',')[0]
"'FLBC8U'"
Burhan Khalid
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