2 ways: set & dict/zip clear solution for Popular Words by Striga - python coding challenges

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2 ways: set & dict/zip solution in Clear category for Popular Words by Striga

def popular_words(text: str, words: list) -> dict:
    t=text.lower().split()
    return {w : t.count(w) for w in words}
    
#Solution 2:  
def popular_words(text: str, words: list) -> dict:
    l=[]
    t=text.lower().split()
    for w in words:
        l.append(t.count(w))
    return dict(zip(words, l))


if __name__ == '__main__':
    assert popular_words('''
When I was One
I had just begun
When I was Two
I was nearly new
''', ['i', 'was', 'three', 'near']) == {
        'i': 4,
        'was': 3,
        'three': 0,
        'near': 0
    }

Oct. 11, 2020

Comments:

PetrStar on Dec. 18, 2020, 6:11 p.m.
<p>You can use Counter.</p>

bqlz2o3 on Jan. 30, 2021, 6:22 p.m.
<p>This was useful to me. I didn't know I could use split without arguments, so I first did `split(' ')` and then I had to remove all the linebreaks `\n`. By just using `split()` you do that all in once. That's great!</p>

Look up in the Dict()

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