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

Posted by mamadou diene
.a{fill-rule:evenodd;}techdegree
mamadou diene
Python Web Development Techdegree Student 1,971 Points

Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it. I need

i am trying to solve this one my way, but i am not successful at it yet. I wold appreciate some pointers please.

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 = {}
    key_count = []
    string = string.lowercase()
    word = key
    for key in string.key():
        key_count = 1
        if key in key_count:
            key_count +=1
        else:
            continue


    return {new_dict}

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points

Your latest code is close, but needs the following corrections:

  • the for loop should loop over string_list not string_list() as it is not a function
  • the if should look for word in new_dict
  • when referring to elements of new_dict use square brackets: new_dict[word]
  • on style, since word_count is only used as the constant 1, the number can be used directly, or define the variable to be more representative of the constant, say, init_word_count and move the definition to the beginning of the function.

Post back if you need more help. Good luck!!!

behar
behar
10,799 Points
behar
behar
10,799 Points

Hey mamadou! Big tip here! Strings dont have a .key attribute! Also .lowercase does not exist in python (should be .lower())

mamadou diene
.a{fill-rule:evenodd;}techdegree
mamadou diene
Python Web Development Techdegree Student 1,971 Points

i made more correction here, and still not working

def word_count(string):
    new_dict = {}
    word_count = []
    string_list = string.lower().split()


    for word in string_list():
        word_count = 1
        if word in word_count:
            new_dict(word) +=1

        else:
            new_dict(word)= word_count

    return new_dict

[MOD: added ```python formatting -cf]