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

Everett Yates
Everett Yates
9,280 Points
Posted by Everett Yates
Everett Yates
9,280 Points

I'm so lost, can someone help me?

I can't seem to figure out what I'm missing.

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(str):
    count_dict = {}
    for words in str.lower().split():
        count_dict[words] = str.count(words)
    return count_dict

2 Answers

Saugat Mukherjee
Saugat Mukherjee
3,272 Points
Saugat Mukherjee
Saugat Mukherjee
3,272 Points

This should work: 

def word_count(str):
    listitem=str.lower().strip().split()
    dicti={}
    for item in listitem:
        dicti[item]=listitem.count(item)
    return dicti
Saugat Mukherjee
Saugat Mukherjee
3,272 Points

sorry for the formatting (basically didn't read the Markdown cheatsheet), but if you format it right, it should work ;)

Andreas cormack
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points
Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

Hi Everett

Basically, lowercase the string and split on whitespace, create a dictionary and check if each word appears in that dictionary, if it does add it to the count. See below example

def word_count(mystring):
    words = mystring.lower().split()
    mydict = {}
    count = 1
    for word in words:
        if word in mydict:
            mydict[word]=count+1
        else:
            mydict[word]=count
    return mydict
Everett Yates
Everett Yates
9,280 Points
Everett Yates
Everett Yates
9,280 Points

hmm, your solution didn't seem to work.

Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

there is nothing wrong with my code, i ran it locally and it gives me exactly what's being asked. Will check, could be something to do with the code challenge interpreter itself. Will get back to you.

Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

sorry just realised my logic was off slightly. It can also be done like so.

def word_count(mystring):
    words = mystring.lower().strip().split()
    mydict = {}
    for word in words:
        if word in mydict:
            mydict[word]=mydict[word]+1
        else:
            mydict[word]= 1
    return mydict