Python regex group() works, but findall() results are not as I expected

Question

I'm parsing some logs and want to extract all parameter names of a certain type. For simplicity, I'll just include a small substring.

log = 'WT.tz=-8&WT.bh=23&WT.ul=en-US'

#I want to find all strings that start with WT and get WT and all the following characters until I find an & or the end of the string. I tested this on an online regex page and it seems to work great.
regex = r'(?s)(?=WT).+?(?=(=))'

# if I try to find the first I get what I expected
re.search(regex,log).group()
>> 'WT.tz'

#when I try to find all I do not get what I thought I was going to get.
re.findall(regex,log)
>> ['=','=','=']

vks · Accepted Answer · 2015-10-08 17:05:36Z

2

findall returns all the groups.You have a group (=).So remove it.

regex = r'(?s)WT.+?(?==)'

                   ^^^^^

Also there is no need for lookahead.

Output: ['WT.tz', 'WT.bh', 'WT.ul']

edited Oct 8, 2015 at 17:05

answered Oct 8, 2015 at 17:00

vks

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LetzerWille · Accepted Answer · 2015-10-08 17:17:37Z

0

log = 'WT.tz=-8&WT.bh=23&WT.ul=en-US'

print(re.findall(r'WT\.[^&]*\b',log))

['WT.tz=-8', 'WT.bh=23', 'WT.ul=en-US']

answered Oct 8, 2015 at 17:17

LetzerWille

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