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

Milind Wankhede
Milind Wankhede
3,571 Points
Posted by Milind Wankhede
Milind Wankhede
3,571 Points

I'm trying to submit wordcount challenge. However, the console is erroring out. I tried my code locally and it worked.

First, it was complaining about the method get_key. I see that we can use that method with dict. Now, it just gives error 'Try again!' no details so, don't know how to make it run successfully. Any pointers would be appreciated! :)

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):
    dataList = str.lower().split(' ')
    wordCount = {}
    for item in dataList:
        if wordCount.has_key(item):
            wordCount[item] += 1
        else:
            wordCount[item] = 1
    return wordCount

1 Answer

JD Vila
JD Vila
8,095 Points
JD Vila
JD Vila
8,095 Points

The line if wordCount.has_key(item) won't work, since dictionaries in python don't have that function. You could refactor that into something like:

def word_count(str):
    dataList = str.lower().split(' ')
    wordCount = {}
    for item in dataList:
        if item in wordCount.keys():
            wordCount[item] += 1
        else:
            wordCount[item] = 1
    return wordCount

I faced the same problem earlier, and a moderator pointed out that I was passing a space, or " " as a parameter in the split function, like split(" "), when it should have been split() without any arguments, that way I could catch all forms of whitespace!

Also, your parameter str is a class type, like int or list, so you might reconsider changing that to something like "string", or something else that's less confusing. Hope that helps!