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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Posted by Janusz Pachołek
Janusz Pachołek
3,334 Points

Is there better solution for this challenge?

I solved challenge with code attached below. I didn't used dictionary packing mentioned in movie before. It also checks every key even if it has been checked already. Is there better, more pythonic way to solve it then?

wordcount.py 
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.

def word_count(stringy):
    words = (stringy.lower()).split()
    diction = {}
    for x in words:
        counter = 0
        for i in words:
            if i == x:
                counter += 1
        diction.update({x: counter})
    return diction

1 Answer

Jeremiah Bushau
Jeremiah Bushau
24,061 Points
Jeremiah Bushau
Jeremiah Bushau
24,061 Points

I would not say that this is a 'better' way but it is another way. Using dictionary comprehension and the count method. Hope this helps give ya some ideas! :) Feel free to let me know if you'd like further explanation.

def word_count(string_arg):
    words = string_arg.lower().split()
    return { word: words.count(word) for word in words }
Janusz Pachołek
Janusz Pachołek
3,334 Points

Thanks, it's much better than my solution. I just wonder why words that are multiple times in a words list are not counted multiple times? Are they excluded from for loop if they were counted before? I hope I'm clear...