an easy line clear solution for Words Order by kalauroma7997 - python coding challenges

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an easy line solution in Clear category for Words Order by kalauroma7997

words_order = lambda text, words: list({v: None for v in text.split() if v in words}) == words


if __name__ == '__main__':
    print("Example:")
    print(words_order('hi world im here', ['world', 'here']))

    # These "asserts" are used for self-checking and not for an auto-testing
    assert words_order('hi world world im here', ['world', 'here']) == True
    assert words_order('hi world im here', ['here', 'world']) == False
    assert words_order('hi world im here', ['world']) == True
    assert words_order('hi world im here',
 ['world', 'here', 'hi']) == False
    assert words_order('hi world im here',
 ['world', 'im', 'here']) == True
    assert words_order('hi world im here',
 ['world', 'hi', 'here']) == False
    assert words_order('hi world im here', ['world', 'world']) == False
    assert words_order('hi world im here',
 ['country', 'world']) == False
    assert words_order('hi world im here', ['wo', 'rld']) == False
    assert words_order('', ['world', 'here']) == False
    print("Coding complete? Click 'Check' to earn cool rewards!")

Nov. 8, 2020

Comments:

Wakko_Warner on Jan. 8, 2021, 8 p.m.
I like how you put it in one line. Can you explain one part? <pre class='brush: python'>list({v: None for v in text.split() if v in words}) </pre> what is the v: None doing?

kalauroma7997 on Jan. 8, 2021, 10:14 p.m.
OK, cool)) That's just a dictionary (we need only the keys) that keep the defined order and doesn't have duplicates. Personally, I would use sets instead, but order, man))

jakubbortlik on Nov. 15, 2022, 2:33 p.m.
I like the way you use the default behaviour of `dict`: it keeps the insertion order (since v3.16 or v3.17 I believe) and when turned into a `list`, only the keys are considered. This solution should really get more attention.

List but not the least

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