python regex using re.compile and match

Question

eag = 'linux'
rpat = re.compile("^\s*%s\s*=\s*('.*\'')" % eag)

trying to grab r'^LIN' in a line in a text file, linux = r'^LIN',

lines = [line.strip() for line in open (myfile, 'r')]
for line in lines:
    if re.match(rpat, line)
        matched = re.match(rpat,line)
        got_it = matched.group(1)
        # do something here

Not quite sure if my rpat is correct
There is some space in the front of linux then some space until = then some space r'^LIN',

Not sure what you are trying to achieve, but when I am debugging regex I like to use this website. gskinner.com/RegExr — jluzwick
– jluzwick, Commented Aug 28, 2013 at 17:30

Martijn Pieters · Accepted Answer · 2013-08-28 17:27:53Z

2

I'd use:

re.compile("^\s*%s\s*=\s*(r'[^']+')" % re.escape(eag))

This matches the r as well, which you omitted.

Demo:

>>> import re
>>> sample = "linux = r'^LIN',"
>>> eag = 'linux'
>>> rpat = re.compile("^\s*%s\s*=\s*(r'[^']+')" % re.escape(eag))
>>> rpat.match(sample).group(1)
"r'^LIN'"

answered Aug 28, 2013 at 17:27

Martijn Pieters

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Comments

Leonid Shvechikov · Accepted Answer · 2013-08-28 17:29:12Z

1

Your regex is quite messy.

I have tried to fix it:

rpat = re.compile(r"^\s*%s\s*=\s*(r'.*')\s*" % eag)

Your forgot r to match r'LIN' and forgot about trailing spaces.

answered Aug 28, 2013 at 17:29

Leonid Shvechikov

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Comments

demongolem · Accepted Answer · 2013-08-28 17:58:43Z

0

rpat = re.compile("^\s*%s\s*=\s*(.*\')" % eag) works too. Just had to get rid of ''

edited Aug 28, 2013 at 17:58

demongolem

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answered Aug 28, 2013 at 17:37

ealeon

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