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

Posted by Teige Dougherty
Teige Dougherty
3,476 Points

I don't understand why this isn't right. It works in pycharm with the example given.

I feel like I have the right answer. What am I missing?

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):
    myDict = {}
    myString=string.lower().split()
    for word in myString:
        if word in myDict:
            myDict[word] += 1
        else:
            myDict[word] = 1
    print(myDict)

word_count("I do not like it Sam I Am")

2 Answers

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points

You are using print instead of return. For the task the function needs to return a dictionary

Fergus Clare
Fergus Clare
12,120 Points
Fergus Clare
Fergus Clare
12,120 Points

Also, you are lowercasing the dictionary. Are you sure the interpreter is expecting lowercase responses? Consider the following code:

def word_count(sentence):
   dictionary = {}
   for word in sentence.split():
      dictionary.update({word: sentence.count(word)})
   return dictionary

The code above will return the following when you enter word_count(sentence) into the interpreter:

{'I': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'Sam': 1, 'Am': 1}