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

Posted by Mark Chesney
Mark Chesney
11,747 Points

refactoring word_count (an intermediate Python solution)

Hi everyone,

I'm revisiting this challenge as I'm finishing up the intermediate Python track, and looking for a more Pythonic solution to this, just to demonstrate what I've learned. I came up with a dictionary comprehension. Would anyone have any good feedback on whether I could slim this down sensibly? I'm really curious if the for-loop can be discarded. Much thanks in advance!

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):
    list_of_keys = string.lower().split()
    dict = {key: 0 for key in list_of_keys}
    for word in list_of_keys:
        dict[word] += 1
    return dict
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

It great to revisit old challenges with new skills! Be careful not to add complexity for the sake of elegance. Can you say more about what was refactored?

This new solution may be less efficient that using an if statement to check if word is already in the dict. The dict comprehension is doing more work than the if in that it is reinitializing keys each time a repeated key is reached. It's the same as:

    for key in list_of_keys:
        dict[key] = 0
    for word in list_of_keys:
        dict[word] += 1
    return dict

1 Answer

kyle kitlinski
kyle kitlinski
5,619 Points
kyle kitlinski
kyle kitlinski
5,619 Points 
sentence = "I do not like it Sam I Am"
def word_count(string):
    list_of_keys = string.lower().split()
    dict = {key: 0 for key in list_of_keys}
    for word in list_of_keys:
        dict[word] += 1
    return dict
%timeit word_count(sentence)

>> 1.41 µs ± 5.79 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

def comprehension_count(string):
    list_of_keys = string.split()
    return {word: list_of_keys.count(word) for word in set(list_of_keys)}
%timeit comprehension_count(sentence)

>> 1.87 µs ± 2.11 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

# however if we use a larger string the comprehension count is a tad faster

%timeit word_count(sentence*999999)

>> 1.36 s ± 5.17 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


%timeit comprehension_count(sentence*999999)

>> 1.33 s ± 4.68 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

The comprehension solution posted is exploiting that the set(list_of_keys) is very small since using sentence*999999. If word_count added the set optimization, performance would equivalent to the comprehension:

def word_count(string):
    list_of_keys = string.lower().split()
    words = {key: 0 for key in set(list_of_keys)}
    for word in list_of_keys:
        words[word] += 1
    return words