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Start your free trialDavis Mercier
7,805 PointsWord Count Objective
Hello,
I've been working on the word count challenge. I have a solution that I'm pretty sure is correct (I've run it through the console with several strings and it always has produced the correct output), but every time I submit the answer is rejected. This function should take a string and return a dictionary where every word is a key and the assigned value is the number of times the words appears.
Example:
word_count("Hello hello is there anybody in there")
{'hello': 2, 'is': 1, 'there': 2, 'anybody': 1, 'in': 1}
Code is below:
def word_count(string): list_of_words = string.lower().split(" ")
dictionary = {}
for word in list_of_words:
count = 0
for word_again in list_of_words:
if word == word_again:
count += 1
dictionary[word] = count
return dictionary
Any help you can offer would be appreciated.
# 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):
list_of_words = string.lower().split(" ")
dictionary = {}
for word in list_of_words:
count = 0
for word_again in list_of_words:
if word == word_again:
count += 1
dictionary[word] = count
return dictionary
1 Answer
Jason Anders
Treehouse Moderator 145,860 PointsHey Davis,
You've pretty much got it exactly, except for just one tiny error.
Now, it's really not very clear in the instructions, nor is the error any indication, but the problem is the split()
method.
Think about everything that could be passed into the function. Someone may have used the enter key or the tab to space the words... So... you should probably split on all whitespace and not just spaces as it is now.
Just fix up the split()
call to include all whitespace (remember this is done by passing in no parameters to the method call).
Other than that... Awesome job!! :)
Davis Mercier
7,805 PointsDavis Mercier
7,805 PointsMakes perfect sense, thanks for the reply!