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

Posted by Debajyoti Kar
.a{fill-rule:evenodd;}techdegree
Debajyoti Kar
Front End Web Development Techdegree Student 2,244 Points

code gave an error in the test but works in the workspace

the code below is what I wrote for the challenge, it gives an error saying that it didn't give the expected output which is supposed to be

{'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}

if I use the string "I do not like it Sam I Am"

however, using the same code in the treehouse workspace prints exactly what is desired. I literally copy pasted the code with the exception that I use print function in the workspace to print the dictionary to check if the output I get is same as what I wanted.

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):
    lower_case_string = string.lower()
    string_array = lower_case_string.split(" ")
    string_dict = {}
    for word in string_array:
        if word in string_dict.keys():
            string_dict[word] += 1
        else:
            string_dict[word] = 1
    return string_dict

1 Answer

Kyle Knapp
Kyle Knapp
21,526 Points

This one is a bit of a trick question. Your function breaks if there is additional whitespace between any of the words in the string.

Example:

#function
word_count("I\t  \n do not like it Sam I Am")
#result
{'i\t': 1, '': 1, '\n': 1, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'i': 1, 'am': 1}

So you'll need to remove ALL whitespace between each word in the string.

Debajyoti Kar
.a{fill-rule:evenodd;}techdegree
Debajyoti Kar
Front End Web Development Techdegree Student 2,244 Points

Oh!! completely slipped my mind. It makes sense. Thanks, Kyle.