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

Posted by Leandro Pezán
Leandro Pezán
1,596 Points

What am I doing wrong? This runs flawlessly on PyCharm...

The output (tried with several strings) is as desired! 

{'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}
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(str_input):
    lowercase_list = str_input.lower().split(" ")
    dict_output = {}

    for item in lowercase_list:
        item_count = lowercase_list.count(item)
        dict_output.update({item: item_count})

    return dict_output

1 Answer

Jennifer Nordell
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STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there! You're doing great! Your syntax and logic are spot on, but you happened to miss a tiny detail in both the instructions and in your test data sets. I'm guessing you didn't try any data sets that included a tab or a new line character. The instructions say to split on all whitespace, but currently, you are only splitting on spaces.

This, however, is easily fixed. Using the split method without any arguments causes the string to be split on all white space including tabs and new lines.

Take a look:

lowercase_list = str_input.lower().split()

If I simply remove your " " from the arguments, your code passes with flying colors! Good job! :sparkles:

Leandro Pezán
Leandro Pezán
1,596 Points

You're absolutely right! I didn't know about the split method without arguments. Thanks!