Python regular expression findall

Question

I want to find all 2 word strings in python. I created this:

#!/usr/bin/python
import re

string='a1 a2 a3 a5 a6'
search=re.findall('.. ..',string)
print len(search)
for nk in search:
        print nk

I am getting: a1 a2 a3 a5 While I wanted:a1 a2,a2 a3,a3 a5,... etc The findall should search for all possible patterns? And why returns a1 a2,a3 a5? Thank you.

Answer 1

It returns ['a1 a2', 'a3 a5'] , because these are the only patterns which can be found: after applying the first one, the 'a1 a2' part is gone and ' a3 a5 a6' is left. The next possible pattern is 'a3 a5' , and ' a6' is left over and cannot be matched further.

'a1 a3' , 'a1 a5' etc. cannot be found because this combinations don't occur. Remember, you search for two arbitrary characters, followed by a space character, followed by 2 arbitrary characters.

With

r=re.compile(r"(\S{2})(?:\s|$)")
pairs =r.findall("a1 a2 a3 a5 a6")

or

pairs = re.findall(r"(\S{2})(?:\s|$)", "a1 a2 a3 a5 a6")

you find all 2-character combination which are wither followed by a space or by the end of the string: ['a1', 'a2', 'a3', 'a5', 'a6'] . If you combine these, you will find all possible combinations:

for ifirst in range(len(pairs) - 1):
    for second in pairs[ifirst + 1:]:
        print " ".join((pairs[ifirst], second))