re.sub one-liner clear solution for Bird Language by a354913 - python coding challenges

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re.sub one-liner solution in Clear category for Bird Language by a354913

import re

VOWELS = "aeiouy"

def translation(phrase):
    return re.sub(r'([^{v} ])[{v}]|([{v}])\2\2|( )'.format(v=VOWELS), r'\1\2\3', phrase)

if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert translation("hieeelalaooo") == "hello", "Hi!"
    assert translation("hoooowe yyyooouuu duoooiiine") == "how you doin", "Joey?"
    assert translation("aaa bo cy da eee fe") == "a b c d e f", "Alphabet"
    assert translation("sooooso aaaaaaaaa") == "sos aaa", "Mayday, mayday"

March 30, 2019

Comments:

ceho.main on April 15, 2019, 7:41 a.m.
Wow, can you please explain me how does this regexp work?

a354913 on April 15, 2019, 9:47 a.m.
Sure: <ol> <li>"Evaluate" format to see the final regexp more clear: `([^aeiouy ])[aeiouy]|([aeiouy])\2\2|( )`</li> <li>Spot "alternation" operators. You can divide expression there into easier to understand independent sub-parts: `([^aeiouy ])[aeiouy]` or `[aeiouy])\2\2` or `( )`</li> <li>You see it looks for matches that can be either of 3 things: <ol> <li>`([^aeiouy ])[aeiouy]` Consonant followed by vowel. Note space included into `[^aeiouy ]` part. Consonant captured into the first capturing group.</li> <li>`[aeiouy])\2\2` Vowel followed by two same vowels. Vowel captured in the second group. I used `\2` numeric reference to reference the second group.</li> <li>`( )` just space. Captured into the third group.</li> </ol> Only one group will contain something captured, and two others will be empty due to "alternation" operator. For example, if `\2` is not empty then `\1` and `\3` will be empty. So we can concatenate them as `\1\2\3` without much trouble.</li> <li>Then it keeps matching and replacing until the end of string.</li> </ol> Hope it helps. For faster regex prototyping, I use regex editor <a href="https://regexr.com/">https://regexr.com/</a>

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