dict speedy solution for Popular Words by David_Jones - python coding challenges

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dict solution in Speedy category for Popular Words by David_Jones

def popular_words(text, words):
    counter = {word: 0 for word in words}
    for word in text.lower().split():
        if word in counter:
            counter[word] += 1
    return counter

June 17, 2019

Comments:

zaplily on June 18, 2019, 8:17 a.m.
Good easy-to-read code. Thanks to you, now I know how to make dictionary without special method!

ProgrammerBruce on April 4, 2020, 6:42 p.m.
+1 for clear solution, with n+k performance. Prefer casefold() instead of lower(). Also, casefold the keys (in words) for best matching.

vodka-bears on Dec. 20, 2021, 3:18 p.m.
{word: 0 for word in words} maybe better be replaced with dict.fromkeys(words, 0)

Look up in the Dict()

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