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

Posted by Mahyar Najizad
Mahyar Najizad
5,863 Points

Problem

The computer keeps telling me to try again!!

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):
    dictionary={}
    words=string.split()
    for eachWord in words:
        dictionary.update(eachWord:words.count(eachWord))
    return dictionary
Jason Davis
Jason Davis
4,498 Points
Jason Davis
Jason Davis
4,498 Points

a couple of things: you need to change all the words to lowercase so the I and i will show as the same word. also the syntax for update is dict.update({value:key}) you need to add the curly brackets.

see if that fixes it for you.

1 Answer

Dan Garrison
Dan Garrison
22,457 Points
Dan Garrison
Dan Garrison
22,457 Points

There are a few more things you need to do.

First, you need to lowercase the letters in the incoming string. (Hint: You can lowercase the letters and split them in a single line of code by using the lower() and then the split() methods).

Second, you need to check if the key already exists in your dictionary. If it does than you need to increment the count by 1. If it doesn't exist, then you need to create the key in the dictionary and set the value to 1. So you need an if/else conditional statement in there

Third, there does appear to be an issue with your update method. However, there are multiple ways to increment a value. I would recommend simplifying it a bit by just doing the following: 

dictionary[eachWord] += 1

You can also set values this way with: 

dictionary[eachWord] = 1