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

Joaquim Maurício
Joaquim Maurício
1,148 Points

Word Count Challenge - Why doesn't my solution work?

Hello, my code is attached.

When I use the function with the output that's suggested it works correctly. Can someone point me to what I'm doing wrong? Thanks in advance!

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(arg):
    mydict = {}
    arg = arg.lower().split(" ")
    for a in arg:
        mydict[a] = arg.count(a)
    return mydict

1 Answer

Steven Parker
Steven Parker
231,275 Points

The error message you get when you test this code contains a clue. It says: "Be sure you're lowercasing the string and splitting on all whitespace!"

To split on "all whitespace" you must leave the argument to "split" empty, or set it to None. Giving it a space makes it split only on explicit spaces. See the Python documentation for split for more details.

Joaquim Maurício
Joaquim Maurício
1,148 Points

Thanks a bunch, but I'm not understanding the difference! Could you provide an example of what would happen in the different situations?

Steven Parker
Steven Parker
231,275 Points

Let's say you started with: "one space   two   spaces   and   some   tabs".
If you split() (all whitespace) you get :point_right: ["one", "space", "two", "spaces", "and", "some", "tabs"]
But if you split(" ") you get :point_right: ["one", "space", "", "two", "", "spaces", "", "and some tabs"]

Joaquim Maurício
Joaquim Maurício
1,148 Points

Thanks again for your patience, helped me a lot.