Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

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

Huston Petty
Huston Petty
2,833 Points
Posted by Huston Petty
Huston Petty
2,833 Points

Is there a cleaner more concise way of doing this?

I got this problem correct, but I am wondering if there is a better, more concise way of doing this? I don't really see the correlation to packing/unpacking here. This seems to be what everyone else comes up with as well.

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(string):
    new_dict = {}
    for word in string.lower().split():
        if word in new_dict:
            new_dict[word] += 1
        else:
            new_dict[word] = 1

    return new_dict

1 Answer

Steven Parker
Steven Parker
230,995 Points
Steven Parker
Steven Parker
230,995 Points

This challenge is not about packing/unpacking. And your solution is pretty much the "classic" one. Just for comparison, here's a slightly more concise one. I haven't benchmarked them but I doubt this one is any more efficient (it may even be less so):

def word_count(string):
    new_dict = {}
    words = string.lower().split()
    for word in set(words):
        new_dict[word] = words.count(word)
    return new_dict