Python pattern-matching. Match 'c[any number of consecutive a's, b's, or c's or b's, c's, or a's etc.]t'

Question

Sorry about the title, I couldn't come up with a clean way to ask my question.

In Python I would like to match an expression 'c[some stuff]t', where [some stuff] could be any number of consecutive a's, b's, or c's and in any order.

For example, these work: 'ct', 'cat', 'cbbt', 'caaabbct', 'cbbccaat'

but these don't: 'cbcbbaat', 'caaccbabbt'

Edit: a's, b's, and c's are just an example but I would really like to be able to extend this to more letters. I'm interested in regex and non-regex solutions.

Answer 1

Not thoroughly tested, but I think this should work:

import re

words = ['ct', 'cat', 'cbbt', 'caaabbct', 'cbbccaat',  'cbcbbaat', 'caaccbabbt']
pat = re.compile(r'^c(?:([abc])\1*(?!.*\1))*t$')
for w in words:
    print w, "matches" if pat.match(w) else "doesn't match"

#ct matches
#cat matches
#cbbt matches
#caaabbct matches
#cbbccaat matches
#cbcbbaat doesn't match
#caaccbabbt doesn't match

This matches runs of a, b or c (that's the ([abc])\1* part), while the negative lookahead (?!.*\1) makes sure no other instance of that character is present after the run.

(edit: fixed a typo in the explanation)

Answer 2

Not sure how attached you are to regex, but here is a solution using a different method:

from itertools import groupby

words = ['ct', 'cat', 'cbbt', 'caaabbct', 'cbbccaat',  'cbcbbaat', 'caaccbabbt']
for w in words:
    match = False
    if w.startswith('c') and w.endswith('t'):
        temp = w[1:-1]
        s = set(temp)
        match = s <= set('abc') and len(s) == len(list(groupby(temp)))
    print w, "matches" if match else "doesn't match"

The string matches if a set of the middle characters is a subset of set('abc') and the number of groups returned by groupby() is the same as the number of elements in the set.