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

Ronit Mankad
Ronit Mankad
12,166 Points

Tell me my mistake!

Plz tell me my mistake!

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(st):
    word_dict = {}
    for word in st.split():
        word_dict[word.lower()] = st.count(word)
    return word_dict

2 Answers

Modou Sawo
Modou Sawo
13,141 Points

Hi Ronit, Using an if condition and an "in" statement is perhaps a cleaner way. Create two variables: one you split the string and lowercase it & the other which would contain the returned output(my case I called it new_dict)

def word_count(arg):
    output = arg.lower().split()
    new_dict = {}
    for item in output:
        if item in new_dict:
            new_dict[item] += 1
        else:
            new_dict[item] = 1
    return new_dict
Doc Collins
Doc Collins
7,087 Points

Hi Ronit:

Your code worked for me.

Microsoft Windows [Version 10.0.10586] (c) 2015 Microsoft Corporation. All rights reserved.

C:\WINDOWS\system32>python Python 3.5.2 (v3.5.2:4def2a2901a5, Jun 25 2016, 22:18:55) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information.

def word_count(st): ... word_dict = {} ... for word in st.split(): ... word_dict[word.lower()] = st.count(word) ... return word_dict ... word_count("hello world") {'world': 1, 'hello': 1} word_count("hello world. it's time to go") {'go': 1, 'hello': 1, 'to': 1, 'world.': 1, "it's": 1, 'time': 1}