CODEWITHSUNDEEP
pythonstringfor-loop
Hypothetical Ninja
Hypothetical Ninja

Reputation: 4077

Check presence of vowels in a string

I need to check whether a vowel is present in a word. If it is, an operation should be carried out on the word say op(word). I want to avoid a for loop because I thought of this:

for char in word:
    if char in 'aeiou':
#confused here...

Please, recommend a method that is low in cost when it comes to execution time. Also, help me correct the above approach too.

Upvotes: 0

Views: 54179

Answers (6)

Simon
Simon

Reputation: 1

Or you could just do this. It's very simple.

line = input('Enter text: ')
frequency = {'a': 0, 'e': 0, 'i': 0, 'o': 0, 'u': 0}
for i in line:
    if i in 'aeiou':
        frequency[character] = frequency[character] + 1
for vowel in 'aeiou':
    print(vowel + ': ' + str(frequency[vowel]))

Upvotes: 0

Ehayf2016
Ehayf2016

Reputation: 11

def vowelz(a):
        vowels = ["a", "e", "i", "o", "u"]
        vowel = False
        for vowell in vowels:
                if vowell in a:
                        vowel = True
        print vowel
vowelz(raw_input("Enter a word:"))

Upvotes: 1

thefourtheye
thefourtheye

Reputation: 239473

vowels = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"}
if any(char in vowels for char in word):
   ...

Note: This is better because it short circuits, as soon as it finds the vowel in the word. So, it doesn't have to check all the characters unless there are no vowels in the string.

Edit: Ran a timeit test and found that, @falsetru's answer is extremely fast, but with few optimizations, the re version beats everything else.

import re

vowels = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"}
pattern = re.compile("[AEIOUaeiou]")

def intersection():
    return bool(vowels.intersection("TWYNDYLLYNGS"))

def any_version():
    return any(char in vowels for char in "TWYNDYLLYNGS")

def re_version():
    return bool(pattern.search("TWYNDYLLYNGS"))

def disjoint():
    return vowels.isdisjoint("TWYNDYLLYNGS")

from timeit import timeit

print timeit("intersection()", "from __main__ import intersection, vowels")
print timeit("any_version()", "from __main__ import any_version, vowels")
print timeit("re_version()", "from __main__ import re_version, vowels")
print timeit("disjoint()", "from __main__ import disjoint, vowels")

Upvotes: 9

falsetru
falsetru

Reputation: 369074

Using set.isdisjoint (This method returns as soon as it found match):

>>> vowels = set('aeiou') # set('aeiouAEIOU') if you want case-insensitivty
>>> not vowels.isdisjoint('bcd')
False
>>> not vowels.isdisjoint('hello')
True

Upvotes: 4

Jayanth Koushik
Jayanth Koushik

Reputation: 9894

You could use regex.

import re

if re.search('[AEIOU]', word, flags=re.I):
    # contains vowels
else:
    # does not

Upvotes: 0

mhlester
mhlester

Reputation: 23221

You can avoid the for loop using a set intersection

if set('aeiou').intersection(word.lower()):

Upvotes: 6

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