Python Collections - wordcount.py challenge not accepting my answer (but it works!)

Question

When I tried this with the sample string provided in the comments I get the exact same results as the second line of comments. It's in a slightly different order, which is expected since it's a dictionary and so unordered, however the results the correct. I keep getting a "bummer" error reminding me to lower case everything and split on white spaces (which my code has always done). 
I'm not sure what result is expected here if my function isn't producing it as to me nothing looks wrong. I tested it in the REPL and it worked fine, help!
```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 wordcount(mystring):
    mystring = mystring.lower()
    mywords = mystring.split(" ")
    my_dict = {}
for words in my_words:
    my_dict[words] = my_words.count(words)

return my_dict
```

Jennifer Nordell · Accepted Answer

Hi there, @Adam Kingsley (https://teamtreehouse.com/adamkingsley) !  I received your request for assistance. Yes, it does work, but only sometimes. The key here is in the Bummer! where it asks you to split on all whitespace. Currently, you are only splitting on spaces, but there are also tabs and new line characters to consider as well. Your test data likely doesn't have any of the latter so you are getting back a false positive.
This splits on spaces:
```py
.split(" ")
```
But this splits on all whitespace:
```py
.split()
```
Simply removing the quotation marks as arguments from your split method causes this to pass with flying colors! :sparkles:

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Adam Kingsley

Adam Kingsley

Python Collections - wordcount.py challenge not accepting my answer (but it works!)

1 Answer

Jennifer Nordell

Jennifer Nordell

Adam Kingsley

Adam Kingsley