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I have a regex that matches all three characters words in a string:

\b[^\s]{3}\b

When I use it with the string:

And the tiger attacked you.

this is the result:

regex = re.compile("\b[^\s]{3}\b")
regex.findall(string)
[u'And', u'the', u'you']

As you can see it matches you as a word of three characters, but I want the expression to take "you." with the "." as a 4 chars word.

I have the same problem with ",", ";", ":", etc.

I'm pretty new with regex but I guess it happens because those characters are treated like word boundaries.

Is there a way of doing this?

Thanks in advance,

EDIT

Thaks to the answers of @BrenBarn and @Kendall Frey I managed to get to the regex I was looking for:

(?<!\w)[^\s]{3}(?=$|\s)
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  • 3
    It obviously won't match a 4-character anything if you tell it it must match exactly 3 characters. What exactly are the rules you want to use to decide if/when to match a fourth character? Commented May 2, 2013 at 19:18
  • 1
    I don't whant it to match, I just want you. to be treated as 4-char words so it doesn't match the regex Commented May 2, 2013 at 19:23
  • What characters do you want to count as word boundaries? Commented May 2, 2013 at 19:24
  • Just blank spaces and ends of line Commented May 2, 2013 at 19:27
  • Can you please accept an answer? Also, why are you using \Z and not $? I think they will do the same thing in this case, but $ is more recognizable. Commented May 2, 2013 at 21:00

3 Answers 3

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3

If you want to make sure the word is preceded and followed by a space (and not a period like is happening in your case), then use lookaround.

(?<=\s)\w{3}(?=\s)

If you need it to match punctuation as part of words (such as 'in.') then \w won't be adequate, and you can use \S (anything but a space)

(?<=\s)\S{3}(?=\s)
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5 Comments

He clarified in a comment that he doesn't want to match the punctuation; rather, he wants the period to be counted as a word character so it prevents the "word" you. from matching (because it is more than three characters).
Your example still won't work, because \w will not match periods.
Thanks guys!! I found the solution! I didn't know about lookarounds.
This regex requires words to always have whitespace round them, so it won't match your first and last word.
you could probably fix this by using (?<=\s|^) and (?=\s|$) for the lookarounds though
1

As described in the documentation:

A word is defined as a sequence of alphanumeric or underscore characters, so the end of a word is indicated by whitespace or a non-alphanumeric, non-underscore character.

So if you want a period to count as a word character and not a word boundary, you can't use \b to indicate a word boundary. You'll have to use your own character class. For instance, you can use a regex like \s[^\s]{3}\s if you want to match 3 non-space characters surrounded by spaces. If you still want the boundary to be zero-width (i.e., restrict the match but not be included in it), you could use lookaround, something like (?<=\s)[^\s]{3}(?=\s).

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1

This would be my approach. Also matches words that come right after punctuations.

import re

r = r'''
        \b                   # word boundary
        (                    # capturing parentheses
            [^\s]{3}         # anything but whitespace 3 times
            \b               # word boundary
            (?=[^\.,;:]|$)   # dont allow . or , or ; or : after word boundary but allow end of string
        |                    # OR
            [^\s]{2}         # anything but whitespace 2 times
            [\.,;:]          # a . or , or ; or :
        )
    '''
s = 'And the tiger attacked you. on,bla tw; th: fo.tes'

print re.findall(r, s, re.X)

output:

['And', 'the', 'on,', 'bla', 'tw;', 'th:', 'fo.', 'tes']
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