Simple Dictionary

Question

The question asks to write a function creating a dictionary with the count of each word in the string and only remove the punctuation if it is the last character in the word. I've been trying to solve the punctuation part of the problem. For the assignment I need to identify the punctuation using isalpha, but

I'm not sure if using the word[-1] is helping to identify if the last character is punctuation and
I don't know how to write the else statement to get the full dictionary to produce.

def word_distribution(s):

    s = s.lower()
    s = s.split()
    wordfreq = {}
    for word in s:
        if word[-1].isalpha():
            if word in wordfreq:
                wordfreq[word] += 1 
            else:
                wordfreq[word] = 1

    return wordfreq

Example of what my code is producing...

word_distribution("Why's it always the nice guys?")

Out : {'always': 1, 'it': 1, 'nice': 1, 'the': 1}

Example of what it should be producing.....

Out : {'always': 1, 'it': 1, 'nice': 1, 'the': 1, 'why's': 1, 'guys': 1}

Format your question, add input and desired output example(s). — Dinac23, Feb 28 '20 at 22:54
The problem here is that it is treating guys? as a single word and filtering out the response. — Anurag Reddy, Feb 28 '20 at 22:56
In this situation, punctuation can be left in the dictionary if it is not the last character of the word. @AnuragReddy — Anfieldisonfire, Feb 28 '20 at 22:58
@Dinac23 My apologies, I'm brand new to stack overflow. I think I have that in there? what else would you like? — Anfieldisonfire, Feb 28 '20 at 22:59
Also, I have a question @Anfieldisonfire, how are you comparing with respect to an empty dictionary? — Anurag Reddy, Feb 28 '20 at 23:00
Sorry, didn't quite catch the example at the end of the post. Make sure to include the _def_ part of the code with the rest of the code. — Dinac23, Feb 28 '20 at 23:00

score 1 · Answer 1 · answered Feb 28 '20 at 23:19

Generally for counting stuff, you should use the collections.Counter class, and for checking if an element is a , or ? you should use string.punctuation that contains them, for example:

import string
from collections import Counter
txt = "Why's it always the nice guys?"

counted = Counter(
    word if not word[-1] in string.punctuation else word[:-1] for word in txt.split()
)
print(counted)

>>> Counter({"Why's": 1, 'it': 1, 'always': 1, 'the': 1, 'nice': 1, 'guys': 1})

Now if you really need a dictionary for output just do:

print(dict(counted))

>>> {"Why's": 1, 'it': 1, 'always': 1, 'the': 1, 'nice': 1, 'guys': 1}

Elidor00 · Answer 2 · 2020-02-28T23:21:43.303

Hello and welcome on SO.

If I understand your problem, I think you simply wrote the construct if then else in the wrong way.

import string

def word_distribution(s):
    s = s.lower()
    s = s.split()
    wordfreq = {}
    for word in s:
        if word[-1] in string.punctuation:
            if word[-1] in wordfreq:
                wordfreq[word[:-1]] += 1 
            else:
                wordfreq[word[:-1]] = 1
        else:
            if word in wordfreq:
                wordfreq[word] += 1 
            else:
                wordfreq[word] = 1

    return wordfreq

x = word_distribution("Why's it always the nice guys?")
print(x)

I used string.punctuation to check if the last character is a punctuation symbol.

The output is: {"why's": 1, 'it': 1, 'always': 1, 'the': 1, 'nice': 1, 'guys': 1}

Edit: Another solution using isalpha:

def word_distribution(s):
    s = s.lower()
    s = s.split()
    wordfreq = {}
    for word in s:
        if word[-1].isalpha():
            if word[-1] in wordfreq:
                wordfreq[word] += 1 
            else:
                wordfreq[word] = 1
        else:
            if word in wordfreq:
                wordfreq[word[:-1]] += 1 
            else:
                wordfreq[word[:-1]] = 1

    return wordfreq

x = word_distribution("Why's it always the nice guys?")
print(x)

The output is: {"why's": 1, 'it': 1, 'always': 1, 'the': 1, 'nice': 1, 'guys': 1}

Simple Dictionary

2 Answers2