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

Qiong Li
Qiong Li
4,192 Points
Posted by Qiong Li
Qiong Li
4,192 Points

Is something wrong with it?

I have tested it, it could get right result. but it always show try again

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):

    string = (string.lower()).split(' ')

    # go through each word and add them to the dictionary with the value of 1
    response = {}
    for item in string:
        if item in response.keys():
            # if you come across a word that is already in the dictionary, increment the value by one
            response[item] += 1
        else:
            response[item] = 1
    # return the string
    print(response)

1 Answer

Philip Schultz
Philip Schultz
11,437 Points
Philip Schultz
Philip Schultz
11,437 Points

Hello, Here is how accomplished the task

def word_count(string1):
    string1 = string1.lower().split()
    dict1 = {}
    for item in string1:
        dict1[item]  = string1.count(item)
    return dict1
Philip Schultz
Philip Schultz
11,437 Points
Philip Schultz
Philip Schultz
11,437 Points

let me know if you have questions

Here is some info on the count method https://www.tutorialspoint.com/python/list_count.htm