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

7,158 Points

wordcount.py, what is wrong?

What exactly is wrong? The elements of code seem to be right.

Many thanks, Tomas

# 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(my_str):
    my_dict = {}
    lower_str = my_str.lower()
    sliced_str = lower_str.split(" ")
    for item in sliced_str:  
        my_dict[item] = sliced_str.count(item)
    return my_dict

word_count("how do you do")

4 Answers

This question seems to have a syntax checker. If you use .split(" ") instead of .split() you fail, even if your output is correct.

I think the reason is to remind you that by default, split will break on all white space, (" ") is just if you want to break on spaces.

7,158 Points

Thank you Jon, I didnยดt know the thing about split().

7,158 Points

Very strange, a difference standing between Treehouse accepting the code as successfull is in indentation. I is OK when I changed it this way:

my_dict = {}; lower_str = my_str.lower() sliced_str = lower_str.split()

instead of

my_dict = {} lower_str = my_str.lower() sliced_str = lower_str.split()

And this have happened a several times in workspace.

I had the same problem as you, and I've been stuck trying to figure out what was wrong... Thanks Jon!