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

*[SOLVED]* Bummer! Even though my function works as intended.

Am I missing something here?

It works when I run it locally on my computer but the test says "Bummer! Didnt get the expected results...."

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(indata):
    custom_dic = {}
    word_feeded = indata.lower().split(" ")
    for word in word_feeded:
        if word in custom_dic.keys():
            count = custom_dic[word] + 1
            custom_dic[word] = count
        else:
            custom_dic[word] = 1

    return custom_dic

Moderator edit: Marked as solved, as per comment by Poster

Ok, I figured it out myself.

I used split on spaces split(" ") when instead you should not pass any arguments to the split method so that it splits on all white space like so: split().

Steven Parker
Steven Parker
216,012 Points

Congratulations on resolving your issue! :+1:
And many folks encounter the same issue, evidenced by the othe questions in the forum archives. :wink: