O(n) (1 regex) uncategorized solution for Striped Words by DiZ - python coding challenges

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O(n) (1 regex) solution in Uncategorized category for Striped Words by DiZ

from re import I, findall

VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"

def checkio(text):
    pattern = r'(?:\b(?:[{}][{}])+[{}]?\b)'
    regex = '({}|{})'.format(*[pattern]*2).format(*[VOWELS, CONSONANTS]*3)
    return len(findall(regex, text, I))

#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert checkio("My name is ...") == 3, "All words are striped"
    assert checkio("Hello world") == 0, "No one"
    assert checkio("A quantity of striped words.") == 1, "Only of"
    assert checkio("Dog,cat,mouse,bird.Human.") == 3, "Dog, cat and human"

Dec. 23, 2013

Comments:

veky on Dec. 23, 2013, 7:32 a.m.
Wow, nested formats. This is painfully original. :-D And writing f(* [blah] * 2) instead of f(blah,blah) is the pinnacle of laziness. Not to mention the regrouping of 2 * 3 format specs into 3 * 2. At first I thought, "hey, this man could really use dict formatting", but then I realized you need precisely positional arguments, and anonymous {}s, for this mixing to work. :-D Seriously, nice solution. I like it. Minor nitpick: I don't really get why you have () around regex on line 8, except if you want it to look like a face. :-~

DiZ on Dec. 23, 2013, 10:29 a.m.
On the first thought, I wanted to use this following to give a better insight: <pre class='brush: python'>def checkio(text): pattern = r'(?:\b(?:[{}][{}])+[{}]?\b)' parse = lambda *_map: pattern.format(*['{'+str(i)+'}' for i in _map]) regex = '({}|{})'.format(parse(0,1,0), parse(1,0,1)) return len(findall(regex.format(VOWELS, CONSONANTS), text, I)) </pre> But as you realized, it is also based on sequence [0,1,0,1,0,1], or simply [0,1]*3, and here comes the current grouping. For f(* [foo] * 2), I did it to keep it "consistent" with f([VOWELS, CONSONANTS]3) (and also due to my incredible laziness :p). Finally the niptick: well, exercice asks to find a number of stripped words, but in fact, I wanted to find all the stripped word. Remove len(...) around the last line and it will do it. If you remove () on line 8, result will be a mess of strings and empty matches. For example: checkio("My name is...") will not return ["My", "name", "is"], but [("", "My"), ("", "name"), ("is", "")]. If you follow precisely the description, then code can be simplified by: <pre class='brush: python'>def checkio(text): pattern = r'\b([{}][{}])+[{}]?\b' regex = '|'.join([pattern]*2).format(*[VOWELS, CONSONANTS]*3) return len(findall(regex, text, I)) </pre>

veky on Dec. 23, 2013, 4:17 p.m.
About that last code: yeah, but that way you don't have nested formats. That's automatically one thumb less from me. :-D

Strings Theory

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