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

Posted by Oluwatominiyin Adebanjo
Oluwatominiyin Adebanjo
3,223 Points

I'm facing an error while completing the dictionary challenge. Please Help!!

The challenge is to create a function named word_count. It should accept a single argument which will be a string. The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string, lowercased. The values will be how many times that particular word appears in the string.

I get this error when I try to submit ---> Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!.

My current solution is attached below

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(str):
    lis = str.lower().split(' ')
    dicts = {}
    for s in lis:
        if s in dicts:
            dicts[s] += 1
        else:
            dicts[s] = 1
    return dicts

1 Answer

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,972 Points

The error presented states to make sure you are splitting on all whitespace. This is accomplished by omitting the separator in the split function. Like this:

 lis = str.lower().split()
Oluwatominiyin Adebanjo
Oluwatominiyin Adebanjo
3,223 Points

Thanks a lot ! That fixed it.