CODEWITHSUNDEEP
pythonpython-3.xlist
Robo Bot
Robo Bot

Reputation: 51

Unable to findout vowel count using python

I am trying to find out ID numbers where the number of vowels in the body is an even number for the URL: https://jsonplaceholder.typicode.com/posts

My code is given below:

import json

import requests

   
url ='https://jsonplaceholder.typicode.com/posts'
content = requests.get(url).content

j = json.loads(content)
for each in j:

        print(each['id'],each['body'])

I am now able to print the body for each user ID but unable to find out the number of vowels in the body which is an even number. Need help

Upvotes: 2

Views: 56

Answers (2)

I'mahdi
I'mahdi

Reputation: 24059

You can count vowels with this code:

print(*map(each['body'].lower().count, "aeiou"))

whole code:

import json

import requests

   
url ='https://jsonplaceholder.typicode.com/posts'
content = requests.get(url).content

id_even = []
j = json.loads(content)
for each in j:
    cnt_vwl = 0
    for c in "aeiou":
        cnt_vwl += each['body'].lower().count(c)
    if cnt_vwl%2==0:
        id_even.append(each['id'])
id_even

Output: (id that each['body'] have even vowels)

[1, 3, 4, 5, 6, 7, 10,...]

Upvotes: 3

mozway
mozway

Reputation: 261860

Here is a solution using a list comprehension and re.sub to keep only the vowels:

import re

ids_even_vowels = [i['id']
                   for i in j if i['body']
                   if not len(re.sub('[^aeiouy]', '', i['body'], flags=re.I))%2==1
                  ]

output:

[1, 3, 4, 5, 6, 7, 10, 12, 15, 16, 18, 20, 22, 24, 27, 31, 32, 33, 34, 35, 40, 45, 46, 48, 49, 50, 53, 55, 56, 57, 58, 61, 63, 65, 66, 67, 68, 73, 76, 78, 82, 84, 90, 92, 95, 96, 97, 99]

how counting the vowels works:

>>> my_string = 'AbCDefGHI'

>>> re.sub('[^aeiouy]', '', my_string, flags=re.I)
'AeI'   # vowels only

>>> len(re.sub('[^aeiouy]', '', my_string, flags=re.I))
3       # number of vowels

>>> len(re.sub('[^aeiouy]', '', my_string, flags=re.I))%2
1       # remainder of division by 2: 1 is odd, 0 is even

>>> not len(re.sub('[^aeiouy]', '', my_string, flags=re.I))%2
False   # is it even?

Upvotes: 1

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