How To Limit Regex's Findall() Method

Is there a regex equivalent of BeautifulSoup's limit=X argument for the findall method? I mean, how to find the first X words in question and then break the code execution? thank y

Solution 1:

Use re.finditer and itertools.islice:

from itertools import islice
import re

limit = 2

for x in islice(re.finditer(r'\d+', '1 2 33'), limit):
    print(x.group())

As a function:

def findall_limiter(pattern, string, flags=0):
    return islice(re.finditer(pattern, string, flags), limit)

eg.

for match in findall_limiter(r'\d+', '1 2 33', 2):
    # do stuff

Solution 2:

You can use re.finditer as it returns an iterator instead of generating all the values at once:

In [21]: strs="12345678"

In [22]: it=re.finditer("\d",strs)

In [23]: [next(it).group(0) for _ in xrange(4)] #returns only 4 mathces
Out[23]: ['1', '2', '3', '4']

Though this might raise StopIteration error when the limit is greater than the number of matches. A simple workaround is to use exception handling or use itertools.isclice :

In [26]: def limiter(strs,pattern,limit):
    it=re.finditer(pattern,strs)
    try:
        for _ in xrange(limit):
            yield next(it).group(0)
    except StopIteration:        
        pass
   ....:     

In [27]: list(limiter("12345","\d",3))
Out[27]: ['1', '2', '3']

In [28]: list(limiter("12345","\d",6))
Out[28]: ['1', '2', '3', '4', '5']

In [29]: list(limiter("12345","\d",10))
Out[29]: ['1', '2', '3', '4', '5']

help on re.finditer:

In [24]: re.finditer?
Type:       function
String Form:<function finditer at 0xb74114c4>
File:       /usr/lib/python2.7/re.py
Definition: re.finditer(pattern, string, flags=0)
Docstring:
Return an iterator over all non-overlapping matches in the
string.  For each match, the iterator returns a match object.

Empty matches are included in the result.