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

Posted by Craig Steward
Craig Steward
9,100 Points

I need help with the word_count challenge in python collections

Hi,

I'm receiving the following error for my code - 'Bummer! Hmm, didn't get the expected output. Be sure you're not splitting only on spaces!

My understanding is that as I'm not specifying a separator, the default is to split by spaces. Where I'm lost is what I should actually be entering as the separator in split() ?

If someone could hint or point me in the right direction that'd be great!

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.

string_1 = "I do not like it Sam I Am"


def word_count(string_1):

    list_1 = string_1.split()
    dict_1 = {}
    count_1 = 0

    for key in list_1:
        if key in dict_1:
            count_1 += 1
            dict_1.update({key: count_1})
        else:
            count_1 = 1
            dict_1.update({key: count_1})

    return dict_1
john larson
john larson
16,594 Points

I'm getting that error also, local seems to be returning the proper output. Anyone resolve this? I am doing .lower() on the string.

stephenmelendy
stephenmelendy
3,578 Points
stephenmelendy
stephenmelendy
3,578 Points

John, make sure you are doing string = string.lower()

If you simply put string.lower(), it doesn't carry to future lines. I was getting the "splitting on spaces" error until I properly applied the .lower() method.

john larson
john larson
16,594 Points

Hi Steven, yes. Here is how I did .lower()

string = string.lower().split()
john larson
john larson
16,594 Points

Steven, since your code passed...do you feel like sharing it? My code works, I think there's just some quirk I'm missing

def word_count(string):
    dict_count = {}
    count = 0
    string = string.lower().split()



    for word in string:
        if word in dict_count:
            count +=1
        else:
            count =1
        dict_count.update({word: count})

    return dict_count
stephenmelendy
stephenmelendy
3,578 Points
stephenmelendy
stephenmelendy
3,578 Points

See my new answer below. This question seems a little funky, hopefully they don't mind me posting my code.

3 Answers

stephenmelendy
stephenmelendy
3,578 Points
stephenmelendy
stephenmelendy
3,578 Points

Since this question has weird behavior, here is my code (which worked). Although I don't remember Mr.Love covering the .count() method. 

def word_count(string):
    string = string.lower()
    list_1 = string.split()
    dict_1 = dict()

    for item in list_1:
        dict_1.update({item:list_1.count(item)})

    return dict_1
john larson
john larson
16,594 Points

Thanks Steven, string.count(word) did the trick, and made it a whole lot simpler. I'd still like to know what is wrong with mine the way I did it first. I passed in a different string and the counts were messed up.

stephenmelendy
stephenmelendy
3,578 Points
stephenmelendy
stephenmelendy
3,578 Points

Try making the string lowercase, my answer didn't take until I did:

string_1 = string_1.lower()

edit: simply putting string.lower() will only apply the method for that single line. You need the equals sign for it to apply to the string variable and carry forward.

Craig Steward
Craig Steward
9,100 Points

Tried that, still returning the same error. I knew I still had to lowercase everything but figured the error returning is specific to spaces so wouldn't be the problem.

Chris Walker
Chris Walker
5,004 Points
Chris Walker
Chris Walker
5,004 Points

I really hate these challenges. They aren't challenges to get the same output. They're challenges to do it exactly how they want. I write variation after variation that returns the same output (verified in repl.it) and still get errors.