Python regular expressions - re.search() vs re.findall()

For school I'm supposed to write a Python RE script that extracts IP addresses. The regular expression I'm using seems to work with re.search() but not with re.findall().

exp = "(\d{1,3}\.){3}\d{1,3}"
ip = "blah blah 192.168.0.185 blah blah"
match = re.search(exp, ip)
print match.group()

The match for that is always 192.168.0.185, but its different when I do re.findall()

exp = "(\d{1,3}\.){3}\d{1,3}"
ip = "blah blah 192.168.0.185 blah blah"
matches = re.findall(exp, ip)
print matches[0]

0.

I'm wondering why re.findall() yields 0. when re.search() yields 192.168.0.185, since I'm using the same expression for both functions.

And what can I do to make it so re.findall() will actually follow the expression correctly? Or am I making some kind of mistake?

python regex

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1 Answer

You're only capturing the 0 in that regex, as it'll be the last one that's caught.

Change the expression to capture the entire IP, and the repeated part to be a non-capturing group:

In [2]: ip = "blah blah 192.168.0.185 blah blah"

In [3]: exp = "((?:\d{1,3}\.){3}\d{1,3})"

In [4]: m = re.findall(exp, ip)

In [5]: m
Out[5]: ['192.168.0.185']

In [6]:

And if it helps to explain the regex:

In [6]: re.compile(exp, re.DEBUG)
subpattern 1
  max_repeat 3 3
    subpattern None
      max_repeat 1 3
        in
          category category_digit
      literal 46
  max_repeat 1 3
    in
      category category_digit

This explains the subpatterns. Subpattern 1 is what gets captured by findall.

answered 2012-01-25T10:24:59.120

Python regular expressions - re.search() vs re.findall()

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