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

nicole lumpkin
PLUS
nicole lumpkin
Courses Plus Student 5,328 Points
Posted by nicole lumpkin
nicole lumpkin
Courses Plus Student 5,328 Points

Could someone point out what's wrong, this works in my IDE?

Any hints would be appreciated! :)

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(str1):
    my_dict = {}
    for word in str1.lower().split(' '):
        if word not in my_dict:
            my_dict[word] = 1
        else:
            my_dict[word] += 1
    return my_dict

1 Answer

Stuart Wright
Stuart Wright
41,119 Points
Stuart Wright
Stuart Wright
41,119 Points

The challenge is being a bit picky, but to pass you will need to just change your line that uses the split() method, from:

for word in str1.lower().split(' '):

to

for word in str1.lower().split():

The reason is that, by default, the split() method splits on all whitespace, and there are more types of whitespace characters than simply ' ' (e.g. tabs). The challenge wants you to split on all whitespace.

nicole lumpkin
nicole lumpkin
Courses Plus Student 5,328 Points
nicole lumpkin
nicole lumpkin
Courses Plus Student 5,328 Points

Thank you so much! That was quite informative re: white space. I did notice the wording but didn't know its significance! Have a great day (or night depending on where you are :)