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

Posted by Ahmed Abdelhay
Ahmed Abdelhay
6,025 Points

the same output but giving me there is an error

when running my script giving the same output for the problem but in the simulator give me there are a error

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}
# {'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}
# Lowercase the string to make it easier.

def word_count(sentence):
    words = sentence.lower().split(" ")
    words_dict = {}
    for word in words:
        try:
            if words_dict[word]:
                words_dict[word] += 1
        except KeyError:
            words_dict[word] = 1

    return words_dict

2 Answers

Louise St. Germain
Louise St. Germain
19,424 Points
Louise St. Germain
Louise St. Germain
19,424 Points

Hello Ahmed,

You're very close! The only thing is that it wants you to split over all whitespace (spaces, tabs, new lines, etc). Right now you're splitting only on spaces. If you change the second line of your code to this (i.e., use split() with nothing inside the parentheses), everything will work, since the default for the split method is to split on all whitespace.

words = sentence.lower().split()

I hope this helps!

Ahmed Abdelhay
Ahmed Abdelhay
6,025 Points
Ahmed Abdelhay
Ahmed Abdelhay
6,025 Points

Thank you Louise St. Germain for the help