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

Sebastiaan van Vugt
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Sebastiaan van Vugt
Python Development Techdegree Graduate 13,554 Points
Posted by Sebastiaan van Vugt
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Sebastiaan van Vugt
Python Development Techdegree Graduate 13,554 Points

Why is my word count function not accepted? I think it works.

I believe that the following function can put all words of a phrase in a dictionary in lower case and count the number of duplicates. Still, my answer is not accepted. Does anyone have an idea why?

def word_count(stringy):
    wcd = {}
    stringy = stringy.lower()
    wcdl = stringy.split(" ")
    for word in wcdl:
        wcd[word] = 0
    for word in wcdl:
        wcd[word] += 1
    return(wcd)

Thank you.

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):
    wcd = {}
    stringy = stringy.lower()
    wcdl = stringy.split(" ")
    for word in wcdl:
        wcd[word] = 0
    for word in wcdl:
        wcd[word] += 1
    return(wcd)

1 Answer

Steven Parker
Steven Parker
229,708 Points
Steven Parker
Steven Parker
229,708 Points

The error message contains a hint: "Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To split on "all whitespace", you should leave the argument to "split" empty. Providing a space as the argument causes it to split only on explicit space characters.

Sebastiaan van Vugt
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Sebastiaan van Vugt
Python Development Techdegree Graduate 13,554 Points
Sebastiaan van Vugt
.a{fill-rule:evenodd;}techdegree seal-36
Sebastiaan van Vugt
Python Development Techdegree Graduate 13,554 Points

Thanks a lot. I indeed noticed that with my function it goes wrong when multiple (connected) spaces are used. Leaving the "split" empty solves this issue.