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

Adam Teale
Adam Teale
10,989 Points

Test not passing - seems to work fine in workspaces

Hey Guys from what I can see in workspaces and in the python interpreter this script should pass. Any ideas? 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(astring):
    b = astring.lower().split(" ")
    wordsfound = {}
    for word in b:
        if word in wordsfound.keys():
            wordsfound[word] += 1
        else:
            wordsfound[word] = 1
    return wordsfound

He Adam,

did you realize if you call your programm it return every time a different order ?

{'do': 1, 'am': 1, 'it': 1, 'i': 2, 'sam': 1, 'like': 1, 'not': 1}
treehouse:~/workspace$ python wordcounting.py
{'i': 2, 'am': 1, 'it': 1, 'sam': 1, 'like': 1, 'do': 1, 'not': 1}
treehouse:~/workspace$ python wordcounting.py
{'it': 1, 'i': 2, 'not': 1, 'sam': 1, 'do': 1, 'like': 1, 'am': 1}
treehouse:~/workspace$ python wordcounting.py
{'i': 2, 'it': 1, 'not': 1, 'sam': 1, 'do': 1, 'am': 1, 'like': 1}
treehouse:~/workspace$ python wordcounting.py
{'sam': 1, 'do': 1, 'i': 2, 'not': 1, 'like': 1, 'it': 1, 'am': 1}

This is might a problem for the tests. Often there run in test is multiply times and expect always the the answer.

But this is just an fast idea.

Greetings Julian

3 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,426 Points

You are very close! Split on WHITESPACE instead of a literal SPACE. This is the default mode when using .split() without any arguments.

Adam Teale
Adam Teale
10,989 Points

G'day Chris & Julian! Thank you both for your help! In the end the solution was as you suggested Chris. Thank you!

This is another approach.

def word_count(arg):
    words = arg.lower().split()
    keys = set(words)
    ret = dict(zip(keys, [0 for  _ in range(len(keys))]))
    for key in words:
        if ret.get(key) is not None:
            ret[key] += 1
    return ret