regex - Python re.findall() is not working as expected

Question

I have the code:

import re
sequence="aabbaa"
rexp=re.compile("(aa|bb)+")
rexp.findall(sequence)

This returns ['aa']

If we have

import re
sequence="aabbaa"
rexp=re.compile("(aa|cc)+")
rexp.findall(sequence)

we get ['aa','aa']

Why is there a difference and why (for the first) do we not get ['aa','bb','aa']?

Thanks!

Answer 1

let me explain what you are doing:

regex = re.compile("(aa|bb)+")

you are creating a regex which will look for aa or bb and then will try to find if there are more aa or bb after that, and it will keep looking for aa or bb until it doesnt find. since you want your capturing group to return only the aa or bb then you only get the last captured/found group.

however, if you have a string like this: aaxaabbxaa you will get aa,bb,aa because you first look at the string and find aa, then you look for more, and find only an x, so you have 1 group. then you find another aa, but then you find a bb, and then an x so you stop and you have your second group which is bb. then you find another aa. and so your final result is aa,bb,aa

i hope this explains what you are DOING. and it is as expected. to get ANY group of aa or bb you need to remove the + which is telling the regex to seek multiple groups before returning a match. and just have regex return each match of aa or bb...

so your regex should be:

regex = re.compile("(aa|bb)")

cheers.