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

Abhishek Dokania
Abhishek Dokania
893 Points
Posted by Abhishek Dokania
Abhishek Dokania
893 Points

Not sure why this code is now passing on Treehouse platform. It ran fine on local machine

Please advise

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(str1):
    list1 = str1.split()
    for i in range(len(list1)):
        list1[i] = list1[i].lower()


    count_dict = {}
    for i in range(len(list1)):
        if list1[i] not in count_dict:
            count_dict[list1[i]] = 1

        j= i+1
        while (j < len(list1)):


                if list1[j] == list1[i]:
                    count_dict[list1[i]]= count_dict[list1[i]]+1
                j=j+1   

    return(count_dict)

1 Answer

Kenneth Love
STAFF
Kenneth Love
Treehouse Guest Teacher
Kenneth Love
Kenneth Love
Treehouse Guest Teacher

Hmm, you're doing a lot of work here. More than I would say is required for this particular challenge.

if list1[i] not in count_dict:
            count_dict[list1[i]] = 1

        j= i+1
        while (j < len(list1)):


                if list1[j] == list1[i]:
                    count_dict[list1[i]]= count_dict[list1[i]]+1
                j=j+1

I'm not 100% certain, but won't this cause each word to be counted twice? The first time it's encountered, it's set as occurring once. Then you're going through the list of words again and, if you find it, incrementing the count. Do I have that right?