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

Posted by tmsmth
tmsmth
10,543 Points

work count challenge

Hello this code works in my terminal but not in the challenge, do you know why ? Thanks !

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(string):
    string_list = string.split()
    dico_string = {}
    for words in string_list:
        dico_string[words.lower()] = string.count(words) 
    return dico_string

1 Answer

Kurt Maurer
Kurt Maurer
16,725 Points

Hey there,

You need to lowercase your string at the beginning you can do this by chaining methods together:

    string_list = string.lower().split()

Then, your for loop needs to reference string_list, not string. The way you have it written, your for loop is counting words that are not capitalized and not counting words that are capitalized. I tested those corrections on your code block and passed the challenge. Hope this helps!

Kurt Maurer
Kurt Maurer
16,725 Points

For clarity:

def word_count(string):
    string_list = string.lower().split() # lowercase everything in the string, then split it by all whitespace.
    dico_string = {}
    for words in string_list:
        dico_string[words] = string_list.count(words)  # Removed .lower() and referenced string_list instead of string
    return dico_string
tmsmth
tmsmth
10,543 Points

Thanks ! So if i got it right, in "string.count(words)", words i lowered because i did words.lower() just before ?