First uncategorized solution for Striped Words by lazzaro.mike - python coding challenges

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First solution in Uncategorized category for Striped Words by lazzaro.mike

import re

vow = "AEIOUY"
con = "BCDFGHJKLMNPQRSTVWXZ"
pat = r"([%s][%s])+[%s]?$|([%s][%s])+[%s]?$" % (con, vow, con, vow, con, vow)

def checkio(text):
    filter = re.compile(pat, re.I).match
    striped = [ ( word, m.group() ) for word in re.findall(r"[\w]+",text) for m in (filter(word),) if m ]
    return len(striped)

#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"

Oct. 9, 2013

Comments:

Cjkjvfnby on Dec. 11, 2013, 10:20 p.m.
You don't need raw string in regexp, and can skip square braces when use \w You can remove stripped variable and return len(...) You did not care about list values, no need to make any calculation in list creation. <pre class='brush: python'>return len([1 for word in re.findall("\w+", text) for m in (filter(word),) if m]) </pre>

Strings Theory

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