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

Abdulkadir Kollere
Abdulkadir Kollere
1,723 Points
Posted by Abdulkadir Kollere
Abdulkadir Kollere
1,723 Points

Be sure you are not splitting only on spaces

My code works fine in workspaces. When I try it on the challenge, it gives me the error "make sure you are not splitting only on spaces" Anyone for the rescue?

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):
    splitted_word = word.split( )
    dic = {}
    for word in splitted_word:
        dic[word] = splitted_word.count(word)
    return dic

1 Answer

Andrew Mlamba
Andrew Mlamba
15,642 Points
Andrew Mlamba
Andrew Mlamba
15,642 Points

Hi Abdulkadir,

The grader seems to be very picky in this particular case. If it works on your local machine or workspaces, it does not necessarily mean it will pass the grader tests.

Remember that the function named word_count takes a string (of words). Lowercase and split the string and you will be okay.

I hope this helps. If it does, please feel free to mark it as best answer.

def word_count(word): # <-- correction needed here
    splitted_word = word.split( ) # <-- correction needed here
    dic = {}
    for word in splitted_word:
        dic[word] = splitted_word.count(word)
    return dic
Colby Work
Colby Work
20,422 Points

I added a string.replace(", " , " ") before splitting with split(), and that worked. For some reason it would not pass by using string.split(" "), even though that worked fine on my own computer, getting the same results. This particular course has been really hard for some reason with the challenges.

Andrew Mlamba
Andrew Mlamba
15,642 Points

Hi there Colby,

There is a hint in the example that was given, that we are supposed to Lowercase the string to make it easier. I think that's what is tripping people up.