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

parth bhardwaj
parth bhardwaj
4,353 Points
Posted by parth bhardwaj
parth bhardwaj
4,353 Points

Bummer! Hmm, didn't get the expected output. Be sure you're not splitting only on spaces!

My code runs on other editors, here it is showing an error

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.
import re
def word_count(a):
    dict1={}
    x=re.findall(r"[\w']+", a)
    for word in x:
        count=0
        for repword in x:
            if word==repword:

                count+=1
        dict1.update({word:count})
    return dict1
Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

Hi parth,

Please be patient and give others a chance to answer your question instead of posting the same question again. If you wanted to add another attempt, you can edit your original question and add your code to that one.

Also, you can search the community and see if others have had your same question.

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

I assume Jason is referring to this question, although the code in it is quite different.

1 Answer

Alexander Davison
Alexander Davison
65,469 Points
Alexander Davison
Alexander Davison
65,469 Points

You don't need a Regex expression. The split function splits on all whitespace.

Example:

>>> 'Alex likes programming'.split()
['Alex', 'likes', 'programming']

So, to solve this, you can do this:

def word_count(a):
    words = a.lower().split()
    result = {}
    for word in words:
        try:
            result[word] += 1
        except KeyError:
            result[word] = 1
    return result

I hope this helps. ~Alex