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

Posted by Hakim Rachidi
Hakim Rachidi
38,490 Points

Used varstring.lower().split(" ") and got "Be sure you're lowercasing the string and splitting on all whitespace!"

Runs as expected on my local machine:

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

returned: -->

{'like': 1, 'sam': 1, 'it': 1, 'am': 1, 'not': 1, 'i': 2, 'do': 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(st):
    word_dict = {}
    new_list = st.lower().split(" ")
    for element in set(new_list):
        word_dict[element] = new_list.count(element)
    return word_dict
Paul Reust
Paul Reust
6,708 Points

I could be wrong (since I don't actually know python), but typically whitespace means more than just a single space which is what you appear to be splitting at. If there is a tab instead of a space you won't get the split like the instructions say.

Doing a quick search it appears .split() with no arguments will split on all whitespace. So new_list = st.lower().split() should work. Apologies if this isn't correct!

1 Answer

Hakim Rachidi
Hakim Rachidi
38,490 Points
Hakim Rachidi
Hakim Rachidi
38,490 Points

Thanks, now it just works fine.