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

Harris Handoko
Harris Handoko
3,932 Points
Posted by Harris Handoko
Harris Handoko
3,932 Points

word_count challenge task not working correctly: it works on workspace, but it didn't pass the checker

Adding the following to call the function:

string = "I do not like it Sam I Am"
print(word_count(string))

I got:

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

Please tell me where I went wrong?

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):
    word_list = (string.lower()).split()
    print(word_list)
    word_count = {}
    for word in word_list:
        count = 1
        try:
            if word_count[word] >= 1:
                word_count[word] = count + 1 
        except KeyError:
            word_count[word] = count
    return word_count

1 Answer

Renato Guzman
Renato Guzman
51,436 Points
Renato Guzman
Renato Guzman
51,436 Points

Try testing with: "I like people, I like pets and I like ice cream". Notice that count is always 1 in every iteration. Also I recommend to change the name of your variable word_count since it is the same as your method. Hope it helps!

Harris Handoko
Harris Handoko
3,932 Points
Harris Handoko
Harris Handoko
3,932 Points

Thanks! count = 1 was not correct, how could I forget the age-old += 1 ?!!! ;)

def word_count(string):
    word_list = (string.lower()).split()
    print(word_list)
    word_count = {}
    for word in word_list:
        try:
            if word_count[word] >= 1:
                word_count[word] += 1 
        except KeyError:
            word_count[word] = 1
    return word_count

string = "I like people, I like pets and I like ice cream"
print(word_count(string))