Simple regex in python

Question

I'm trying to simply get everything after the colon in the following:

hfarnsworth:204b319de6f41bbfdbcb28da724dda23

And then everything before the space in the following:

29ca0a80180e9346295920344d64d1ce ::: 25basement

Here's what I have:

for line in f: 
    line = line.rstrip() #to remove \n
    line = re.compile('.* ',line) #everything before space. 
    print line

Any tips to point me in the corrent direction? Thanks!

Also, is re.compile the correct function to use if I want the matched string returned? I'm pretty new at python too. Thanks!!

eumiro · Accepted Answer · 2012-02-03 12:00:58Z

4

string = "hfarnsworth:204b319de6f41bbfdbcb28da724dda23"
print(string.split(":")[1:])
string = "29ca0a80180e9346295920344d64d1ce ::: 25basement"
print(string.split(" ")[0])

edited Feb 3, 2012 at 12:00

eumiro

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answered Feb 3, 2012 at 11:48

xeno

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Comments

Roman Bodnarchuk · Accepted Answer · 2012-02-03 11:48:03Z

3

At first you should probably take a careful look at the doc for re.compile. It doesn't expect for second parameter to be a string to lookup. Try to use re.search or re.findall. E.g.:

>>> s = "29ca0a80180e9346295920344d64d1ce ::: 25basement"
>>> re.findall('(\S*) ', s)[0]
'29ca0a80180e9346295920344d64d1ce'
>>> re.search('(\S*) ', s).groups()
('29ca0a80180e9346295920344d64d1ce',)

BTW, this is not the task for regular expressions. Consider using some simple string operations (like split).

edited Feb 3, 2012 at 11:48

answered Feb 3, 2012 at 11:41

Roman Bodnarchuk

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Comments

ptitpoulpe · Accepted Answer · 2012-02-03 11:44:05Z

0

this regular expression seems to work

r"^(?:[^:]*\:)?([^:]*)(?::::.*)?$"

answered Feb 3, 2012 at 11:44

ptitpoulpe

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