Python matching regex multiple times in a row (not the findall way) -

this question not asking finding 'a' multiple times in string etc.

what match:

[ a-za-z0-9]{1,3}\. 

regexp multiple times, 1 way of doing using |

'[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.|[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.|[ a-za-z0-9]{1,3}\.[ a-za-z0-9]{1,3}\.' 

so matches regexp 4 or 3 or 2 times. matches stuff like:

a. v. b. m a.b. 

is there way make more coding like?

i tried doing

([ a-za-z0-9]{1,3}\.){2,4}  

but functionality not same expected. 1 matches:

regex.findall(string) [u' b.', u'b.'] 

string is:

a. v. b. split them a.b. split somethinf words. more words, ten 

is there way this? goal match possible english abbreviations , names mary j. e. things sentence tokenizer recognizes sentence punctuation not.

i want match of this:

u.s. , c.v.a.b. , a. v. p.  

first of regex work expect :

>>> s="aa2.jhf.jev.d23.llo." >>> import re >>> re.search(r'([ a-za-z0-9]{1,3}\.){2,4}',s).group(0) 'aa2.jhf.jev.d23.' 

but if want match sub strings u.s. , c.v.a.b. , a. v. p. need put whole of regex in capture group :

>>> s= 'a. v. b. split them a.b. split somethinf words. say' more  >>> re.findall(r'(([ a-za-z0-9]{1,3}\.){2,4})',s) [('a. v. b.', ' b.'), ('m a.b.', 'b.')] 

then use list comprehension first matches :

>>> [i[0] in re.findall(r'(([ a-za-z0-9]{1,3}\.){2,4})',s)] ['a. v. b.', 'm a.b.']