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

Sam Wederell
Sam Wederell
20,276 Points

My solution works on a sample but gets marked as incorrect

My code works for the Dr. Seuss sample text and a few others that I've tried, however, it doesn't seem to work for the string being used to test to pass. What am I missing? It converts to lowercase and splits on whitespace.

# 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(string):
    word_list = string.split()
    word_dict = {}
    for word in word_list:
            word_dict[word] = word_dict[word] + 1
        except KeyError:
            word_dict[word] = 1
    return word_dict

1 Answer

Jennifer Nordell
Jennifer Nordell
Treehouse Teacher

Hi there! Yes, it does convert to lower case, but it doesn't save that conversion anywhere. It's just sort of floating out there in Python limbo. So when you do the string.split() it's back to uppercase again.

Where you wrote:


You should have:

string = string.lower()

This will lowercase the string and then save the results back into the string variable.

Hope this helps! :sparkles:

Sam Wederell
Sam Wederell
20,276 Points

Thanks. works perfectly now