First uncategorized solution for Striped Words by qria - python coding challenges

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First solution in Uncategorized category for Striped Words by qria

import re
ALPHABETS = ['AEIOUYaeiouy', 'BCDFGHJKLMNPQRSTVWXZbcdfghjklmnpqrstvwxz']
def checkio(text):
    words = re.findall(r"[\w]+", text)
    return sum( len(w) > 1 and ( all(w[i] in ALPHABETS[i%2] for i in range(len(w))) 
                              or all(w[i] in ALPHABETS[(i+1)%2] for i in range(len(w)))) for w in words)

Oct. 8, 2013

Comments:

Amachua on Oct. 8, 2013, 2:26 p.m.
Nice ! :)

qria on Oct. 8, 2013, 2:33 p.m.
Haha, thanks.

veky on Oct. 8, 2013, 7:59 p.m.
Nice, but can be compressed much further. First, you don't need twice as long items in ALPHABETS, just say text.upper() at end of line 4. Second, once you know that only \w are in your words (and BTW you don't need []s), you only need one item in ALPHABETS, another one can be deduced with [A-Z] and not the first one. Third, "len(w) > 1 and" part can be elegantly coded in re: just say r"\w{2,}" (or even r"\w\w+") there. BTW, I solved it without re, and it's still shorter than yours. Beat that. ;-)

qria on Oct. 9, 2013, 4:31 a.m.
Ah, thanks :) I managed to shrink it down a bit (see my 'A bit hacky' solution), but without regex? I can't beat that :(

Strings Theory

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