Issues with a if/else loop in Python

Question

I am trying to make this Pig Latin translator in Python and it was working well until I tried to downsize it a bit.

Can someone please take a look at this code and tell me why, when I type in a word without a vowel at the beginning it will still print the "vowel" code in this if statement?

CODE:

pyg = 'ay'

original = raw_input('Enter a word: ')
low_original = original.lower()

if len(low_original) > 0 and low_original.isalpha():
        print low_original
        if low_original[0] == 'a' or 'e' or 'i' or 'o' or 'u':
                print "vowel"
                pyg_vowel = low_original + pyg
                print pyg_vowel
        else:
                print "consonant"
                pyg_cons = low_original[1: ] + low_original[0] + pyg
                print pyg_cons
else:
        print 'empty'

You might want to use `if low_original[0] in ['a','e','i','o','u']:`. Otherwise even if the first comparison is `False`, `'e'` is evaluated to `True` and because you have `or`, the `if` clause is always `True`. — Hristo Iliev, Nov 09 '12 at 09:34

Rohit Jain · Answer 1 · 2012-11-09T09:35:10.317

You need to do a check for all the vowels separately.

Currently, your if condition is evaluated as: -

if (low_original[0] == 'a') or 'e' or 'i' or 'o' or 'u':

or returns the first true value in its condition, which will either True or e here, depending upon your first condition is True or not. Now, since 'e' is evaluated to True, so both the values are true, hence your condition will always be true.

You should do it like this: -

if low_original[0] in 'aeiou':

or: -

if low_original[0] in ('a', 'e', 'i', 'o', 'u'):

alexvassel · Answer 2 · 2012-11-09T09:24:28.990

3

You should replace the string:

if low_original[0] == 'a' or 'e' or 'i' or 'o' or 'u':

with:

if low_original[0] in ('a', 'e', 'i', 'o', 'u'):

edited Nov 09 '12 at 09:24

answered Nov 09 '12 at 09:17

alexvassel

10,600
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score 0 · Answer 3 · answered Nov 09 '12 at 09:18

The if is alwasys returning True!

pyg = 'ay'

original = raw_input('Enter a word: ')
low_original = original.lower()

if len(low_original) > 0 and low_original.isalpha():
    print low_original
    if low_original[0] in ['a' , 'e' , 'i' , 'o' , 'u']:
            print "vowel"
            pyg_vowel = low_original + pyg
            print pyg_vowel
    else:
            print "consonant"
            pyg_cons = low_original[1: ] + low_original[0] + pyg
            print pyg_cons
else:
    print 'empty'

Artsiom Rudzenka · Answer 4 · 2012-11-09T09:23:26.683

The problem is in 'if low_original[0] == 'a' or 'e' or 'i' or 'o' or 'u':' - first is not pythonic second is not giving you not what you expect.

Try to update your code to:

pyg = 'ay'

original = raw_input('Enter a word: ')
low_original = original.lower()

if len(low_original) > 0 and low_original.isalpha():
    print low_original
    if low_original[0] in ('a', 'e', 'i', 'o', 'u'):
            print "vowel"
            pyg_vowel = low_original + pyg
            print pyg_vowel
    else:
            print "consonant"
            pyg_cons = low_original[1: ] + low_original[0] + pyg
            print pyg_cons
else:
    print 'empty'

score 0 · Answer 5 · answered Nov 09 '12 at 09:19

0

replace the if statement with

if low_original[0] in ['a', 'e', 'i', 'o', 'u']

answered Nov 09 '12 at 09:19

Marwan Alsabbagh

25,364
9
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score 0 · Answer 6 · answered Nov 09 '12 at 09:21

The condition in this if will always evaluate to True.

 if low_original[0] == 'a' or 'e' or 'i' or 'o' or 'u':
            print "vowel"
            pyg_vowel = low_original + pyg
            print pyg_vowel

It's the same as if (low_original[0] == 'a') or 'e' or 'i' or 'o' or 'u'

You should use something like if low_original[0] in 'aeiou'

Issues with a if/else loop in Python

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