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

Adam Stefański
Adam Stefański
14,255 Points

Possibly a bug in Python Collections wordcount challenge.

It says it doesn't return what it's suppose to, but I checked in the Workspaces and it does return exactly that.

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(x):
    y=x.lower()
    z=y.split(" ")
    dic={}
    for word in z:
        if word not in dic:
            dic[word]=1
        else:
            dic[word]+=1
    return dic

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,457 Points

You are very close! Your code is splitting on a literal space, whereas the challenge checker may be testing using a string with Tabs and spaces mixed. It is best to split on Whitespace. That is, use split() with no arguments.

Post back if you have more questions. Good Luck!!

Adam Stefański
Adam Stefański
14,255 Points

Thanks Chris! It was driving me crazy. I'm sure going to remember that "Tabs and spaces" thing for future challenges.