i have regex matches 3 characters words in string:

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

when use string:

and tiger attacked you.

this result:

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

as can see matches word of 3 characters, want expression take "you." "." 4 chars word.

i have same problem ",", ";", ":", etc.

i'm pretty new regex guess happens because characters treated word boundaries.

is there way of doing this?

thanks in advance,

edit

thaks answers of @brenbarn , @kendall frey managed regex looking for:

(?<!\w)[^\s]{3}(?=$|\s)

if want make sure word preceded , followed space (and not period happening in case), use lookaround.

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

if need match punctuation part of words (such 'in.') \w won't adequate, , can use \s (anything space)

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

Brande