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

Evgeniy Vdovenko
Evgeniy Vdovenko
1,346 Points
Posted by Evgeniy Vdovenko
Evgeniy Vdovenko
1,346 Points

word_count not working

Hi! Im not sure what im doing wrong that system doesn't lets me through. I tried different variants, but nothing is working. in particular this one i created in PyCharm, tested it and then just copied over here, but in PyCharm it worked, but not here.

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(someexp):
    new = {}
    string = someexp.lower().split()
    count = 0
    for word in string:
        if word in new:
            count += 1
            new.update({word: count})
        else:
            count = 1
            new.update({word: count})
    return new

2 Answers

Frank Campos
Frank Campos
4,175 Points
Frank Campos
Frank Campos
4,175 Points

I am not sure, but I believe that we can use the update() function only when you want to add another dictionary that already exists. the following code passed it.

def word_count(someexp):
    new = {}
    someexp=someexp.lower()
    string = someexp.split()

    for word in string:
        if word in new:
            #new[word] + 1 hold the value of the first dictionarie or second, or so on,  added it
            new[word]=new[word]+1
        else:

            new[word]=1
    return new
Frank Campos
Frank Campos
4,175 Points
Frank Campos
Frank Campos
4,175 Points

I am not sure, but I believe that we can use the update() function only when you want to add another dictionary that already exists. the following code passed it.

def word_count(someexp):
    new = {}
    someexp=someexp.lower()
    string = someexp.split()

    for word in string:
        if word in new:
            #new[word] + 1 hold the value of the first dictionarie or second, or so on,  added it
            new[word]=new[word]+1
        else:

            new[word]=1
    return new