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

Dan Brubaker
Dan Brubaker
15,084 Points

Dict out of order? Solving WordCount challenge

My dict is not in the same order as the example. Can someone point me in the right direction for this? I've been stuck on it for a while...

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(string):
    keys = string.lower().split(" ")

    counts = []

    for word in keys:
        count = [keys.count(word)]
        counts = counts + count

    dictionary = dict(zip(keys, counts))

    return dictionary

4 Answers

Steven Parker
Steven Parker
230,995 Points

Dictionaries are orderless collections, by definition. When ordering is important, Python has another object type called "OrderedDict", but you don't need those for this challenge. The mechanism that checks your result will not be concerned about order.

But you're probably closer than you think, and the error message is giving you a hint: "Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To split on "all whitespace" the argument to "split" should be left empty or set to "None". Check the documentation on str.split() for further details.

Dan Brubaker
Dan Brubaker
15,084 Points

My code returns: {'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}

I get a check error of Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!

Not sure what else to do... 🤷🏻‍♂️

Dan Brubaker
Dan Brubaker
15,084 Points

Ah, just saw your post edit. Checking it out... ^

Dan Brubaker
Dan Brubaker
15,084 Points

Yup! The .split() got it. Thank you!