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

Posted by Ryan Hallett
Ryan Hallett
5,796 Points

word_count code challenge

I run this in workspaces and it gives me exactly what I'm looking for. Why is this not passing?

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(st):
    my_dict = {}
    words = st.lower().split(" ")
    for word in words:
        my_dict.update({word: words.count(word)})
    return my_dict

2 Answers

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

When I tried your code I got this hint message: "Bummer! Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To split on all whitespace, the argument to split should be empty. Supplying an argument of a space causes it to split only on spaces.

Ryan Hallett
Ryan Hallett
5,796 Points

Ah awesome, thanks. I thought you needed to specify what to split on, ie " " being a single space, but I guess that doesn't cover two spaces etc etc. Nice!

Dawit Tsegaye
Dawit Tsegaye
950 Points
Dawit Tsegaye
Dawit Tsegaye
950 Points

def word_count(string): word_count = {} string = string.lower().split()

 for word in string:
     if word in word_count:
         word_count[word] += 1
     else:
         word_count[word] = 1

retrun word_count