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

Posted by Gabor Szilagyi
Gabor Szilagyi
3,283 Points

my code is rejected

my code does the job but it gets rejected

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(word):
    word = word.lower()
    word_list = []
    word_list = word.split(" ")  
    word_collection = []
    for word in word_list:
        if word not in word_collection:
            word_collection.append(word)
    word_count = []    
    for word in word_collection:
        word_count.append(word_list.count(word))
    return dict(zip(word_collection, word_count))

2 Answers

Christopher Shaw
seal-mask
PLUS
.a{fill-rule:evenodd;}techdegree seal-36
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 Points
Christopher Shaw
.a{fill-rule:evenodd;}techdegree seal-36
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 Points

It is probably your use of split. split(' '), splits on a single white-space, while split() will handle single and multiple white-spaces.

Vibhutha Kumarage
Vibhutha Kumarage
3,095 Points

Thank you... !!!!

Gabor Szilagyi
Gabor Szilagyi
3,283 Points
Gabor Szilagyi
Gabor Szilagyi
3,283 Points

Thanks, that did the trick!