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

Huston Petty
Huston Petty
2,833 Points
Posted by Huston Petty
Huston Petty
2,833 Points

Is there a cleaner more concise way of doing this?

I got this problem correct, but I am wondering if there is a better, more concise way of doing this? I don't really see the correlation to packing/unpacking here. This seems to be what everyone else comes up with as well.

Thanks

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(string):
    new_dict = {}
    for word in string.lower().split():
        if word in new_dict:
            new_dict[word] += 1
        else:
            new_dict[word] = 1

    return new_dict

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

This challenge is not about packing/unpacking. And your solution is pretty much the "classic" one. Just for comparison, here's a slightly more concise one. I haven't benchmarked them but I doubt this one is any more efficient (it may even be less so):

def word_count(string):
    new_dict = {}
    words = string.lower().split()
    for word in set(words):
        new_dict[word] = words.count(word)
    return new_dict