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

Woongcheol Yang
Woongcheol Yang
5,605 Points
Posted by Woongcheol Yang
Woongcheol Yang
5,605 Points

wordcount.py

What's problem?

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):
    words = string.lower().split(' ')
    res = {}
    for word in words:
        if word in res.keys():
            res[word] += 1
        else:
            res[word] = 1
    return res

2 Answers

Krishna Pratap Chouhan
Krishna Pratap Chouhan
15,203 Points
Krishna Pratap Chouhan
Krishna Pratap Chouhan
15,203 Points

Nothing in particular, but try split() rather than split(' ').

It should be working fine but i guess, its some kind of bug in the site.

# 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):
    words = string.lower().split()
    res = {}
    for word in words:
        if word in res.keys():
            res[word] += 1
        else:
            res[word] = 1
    return res
Stuart Wright
Stuart Wright
41,119 Points
Stuart Wright
Stuart Wright
41,119 Points

I've seen a few people have the same problem with this challenge. Your code will pass the challenge if you pass no arguments to the .split() method. This is because by default this method splits on all whitespace, and there is more than one type of whitespace character (the ' ' that you passed is just one of them, so your code won't split on other kinds of whitespace character - tabs for example).

def word_count(string):
    words = string.lower().split()  # <-- No arguments passed
    res = {}
    for word in words:
        if word in res.keys():
            res[word] += 1
        else:
            res[word] = 1
    return res