Python problem 9 of 30

Count the words

Python
Easy

Write word_counts(text) that returns a dictionary mapping each word to how many times it appears. Treat upper and lower case as the same word.

word_counts("To be or not to be")
# {"to": 2, "be": 2, "or": 1, "not": 1}

Show hint

d.get(key, 0) returns 0 when the key is not in the dictionary yet.

Show solution

def word_counts(text):
    counts = {}
    for word in text.lower().split():
        counts[word] = counts.get(word, 0) + 1
    return counts

Lower-casing first makes "To" and "to" the same key. get with a default avoids a special case for new words. collections.Counter(text.lower().split()) does the same in one line.

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