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

Chang Hyeon Lee
Chang Hyeon Lee
2,008 Points
Posted by Chang Hyeon Lee
Chang Hyeon Lee
2,008 Points

I need help

I got type error for this ...

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(a_string):
    string_dict = 0
    for word in a_string.split():
        if word in string_dict:
            string_dict[word] += 1
        else:
            string_dict[word] = 1
    return string_dict

1 Answer

Stuart Wright
Stuart Wright
41,120 Points
Stuart Wright
Stuart Wright
41,120 Points

The type error is because you initialised the variable 'string_dict' to 0, which is an integer. You should create a blank dictionary instead.

There is one other edit to make before your code will pass the challenge. You should convert the string to lowercase as the challenge instructs you to.

You can do the split() and lower() on separate lines if you prefer, but it works this way too:

def word_count(a_string):
    string_dict = {}
    for word in a_string.lower().split():
        if word in string_dict:
            string_dict[word] += 1
        else:
            string_dict[word] = 1
    return string_dict