CODEWITHSUNDEEP
pythonstringlist
Ramya
Ramya

Reputation: 321

Function to return a list filtered by vowel count in Python

I need to write a function for returning a list by filtering it by vowel count.

I tried this but output is incorrect:

def filter_by_vowel_count(input, count):

    for words in input:
        for p in words:
            if p in 'aeiou':
                value +=1
                if value == count:
                    list1.append(words)

Upvotes: 1

Views: 1841

Answers (5)

jonrsharpe
jonrsharpe

Reputation: 122106

Your problems are with indentation, resetting and returning:

for words in input:
    value = 0 # reset vowel count
    for p in words:
        if p in 'aeiou':
            value +=1
    if value == count: # should be outside for loop
        list1.append(words)
return list1 # return the new list

As others have pointed out, there are also neater ways to achieve what you want.

Upvotes: 0

James Sapam
James Sapam

Reputation: 16940

Using regex:

import re
def filter( ls, c ):
    find = [ (s,len(re.findall("[aeiou]", s))) for s in ls ]
    return [ x for x in find if x[1] == c ]

fruits = ['banana', 'apple', 'lemon', 'pineapple', 'coconut']
print filter(fruits, 2)
print filter(fruits, 3)

Output:

[('apple', 2), ('lemon', 2)]
[('banana', 3), ('coconut', 3)]

Upvotes: 0

Saullo G. P. Castro
Saullo G. P. Castro

Reputation: 58975

I can greatly simplify your life using the built-in collections.Counter class:

from collections import Counter

def filter(fruits, num):
    num_of_vowels = [sum([c[k] for k in 'aeiou']) for c in [Counter(f) for f in fruits]] 
    return [f for i,f in enumerate(fruits) if num_of_vowels[i]==num]

Upvotes: 0

idanshmu
idanshmu

Reputation: 5271

words = [] # list of strings
vowels = "aeiou"
filter(lambda x: sum(x.count(c) for c in vowels) == count, words )

Upvotes: 0

falsetru
falsetru

Reputation: 369334

Using sum and generator expression:

>>> def filter_by_vowel_count(words, count):
...     result = []
...     for word in words:
...         if sum(p in 'aeiou' for p in word) == count:
...             result.append(word)
...     return result
...
>>> fruits = ['banana', 'apple', 'lemon', 'pineapple', 'coconut']
>>> filter_by_vowel_count(fruits, 2)
['apple', 'lemon']
>>> filter_by_vowel_count(fruits, 3)
['banana', 'coconut']

Upvotes: 2

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