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

I have a solution for wordcount.py that works in workspaces but when I check my work, it tells me it's wrong.

It tells me to split on whitespaces and to lowercase which I do! Is this a bug?

# 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(astring):
    bstring = astring.lower()
    alist = bstring.split(" ")
    adict = dict()
    for i in alist:
        if i not in adict:
            adict.update({i: 1})
    return adict

2 Answers

Steven Parker
Steven Parker
229,982 Points

There's a difference between "all whitespace" and explicit spaces. To split on all whitespace, just leave the argument to "split" empty.

What exactly is the difference?

Steven Parker
Steven Parker
229,982 Points

Well, if you provide a space, it will split only on space characters. But if you leave the argument empty, it will split on spaces, tabs, newlines, form feeds, perhaps a few other non-printing characters, and any combinations of those.

The challenge tests your code with a more complex sample than the one shown in the example.

Dylan Bourque
Dylan Bourque
2,700 Points

very impressive that you did this, i was very stumped!