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

Hung Chuan Yu
Hung Chuan Yu
7,491 Points
Posted by Hung Chuan Yu
Hung Chuan Yu
7,491 Points

The output looks fine but I still can't pass the challenge...

Here is my output (test in IDE):

word_count("I do not like it Sam I Am") {'it': 1, 'like': 1, 'sam': 1, 'not': 1, 'i': 2, 'do': 1, 'am': 1}

looks like as challenge ask for but still showing error. anyone could help?? thanks....

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(get_string):
    word_count_dict = {}
    get_string = get_string.lower()
    words = get_string.split(' ')
    for i in words:
         word_count_dict[i] = 0
    for i in words:
        if i in word_count_dict.keys():
        word_count_dict[i]+=1
    return word_count_dict

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

2 Answers

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

:warning: Be careful about testing a challenge in an external REPL or IDE.

If you have misunderstood the challenge, it's also very likely that you will misinterpret the results.

I see two (of three) issues:

  • don't split on just spaces — leave the argument empty to split on all white space
  • statements controlled by an if must be indented more than the if itself
  • but you don't really need the if since you preset all words in the first loop

Also, you only need to define the function, not call it.

Jeff Vig
Jeff Vig
16,345 Points
Jeff Vig
Jeff Vig
16,345 Points

Yup. I was having the same issue as you until I replaced the split argument to look like this

words = get_string.split()