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

Brecht Philips
Brecht Philips
8,863 Points
Posted by Brecht Philips
Brecht Philips
8,863 Points

Why doesn't it work?

If i do this code in pycharm it works correctly however in the exercise it doesn't work. Does anybody see the mistake i'm making or something else??

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(a_string):
    string_dict = {}
    sentence_list = a_string.lower().split(" ")
    for word in sentence_list:
        if word in string_dict:
            string_dict[word] += 1
        else:
            string_dict[word] = 1
    return string_dict

1 Answer

james south
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.a{fill-rule:evenodd;}techdegree seal-36
james south
Front End Web Development Techdegree Graduate 33,271 Points
james south
.a{fill-rule:evenodd;}techdegree seal-36
james south
Front End Web Development Techdegree Graduate 33,271 Points

change your split(" ") to split(). the error message says split on all whitespace, but (" ") is just the space character, there are other whitespace characters such as newline, carriage return, line feed etc.

Chris Grazioli
Chris Grazioli
31,225 Points
Chris Grazioli
Chris Grazioli
31,225 Points

@James South Why would that white space make a difference in a string like, "I do not like it Sam I Am". I can't imagine the grader would throw in carriage returns or line feeds on a test case. I don't know maybe it does??