checking if the first letter of a word is a vowel

问题

I am trying to use python to write a function that checks whether the first letter of a given word, for instance "ball" is a vowel in either uppercase or lowercase. So for instance:

#here is a variable containing a word:
my_word = "Acrobat"

#letters in vowel as a list
the_vowel = ["a","e","i","o","u"]

How do a check that the first letter in "Acrobat" is one of the vowels in the list? I also need to take into consideration whether it is upper or lowercase?

回答1:

try my_word[0].lower() in the_vowel

回答2:

I don't know if it is better than the answers already posted here, but you could also do:

vowels = ('a','e','i','o','u','A','E','I','O','U')
myWord.startswith(vowels)

回答3:

Here are some hints to help you figure it out.

To get a single letter from a string subscript the string.

>>> 'abcd'[2]
'c'

Note that the first character is character zero, the second character is character one, and so forth.

The next thing to note is that an upper case letter does not compare equal to a lower case letter:

>>> 'a' == 'A'
False

Luckily, python strings have the methods upper and lower to change the case of a string:

>>> 'abc'.upper()
'ABC'
>>> 'a' == 'A'.lower()
True

To test for membership in a list us in:

>>> 3 in [1, 2, 3]
True
>>> 8 in [1, 2, 3]
False

So in order to solve your problem, tie together subscripting to get a single letter, upper/lower to adjust case, and testing for membership using in.

回答4:

my_word = "Acrobat"
the_vowel = "aeiou"

if myword[0].lower() in the_vowel:
    print('1st letter is a vowel')
else:
    print('Not vowel')

回答5:

My code looks like this.

original = raw_input("Enter a word:")
word = original.lower()
first = word[0]
vowel = "aeiou"

if len(original) > 0 and original.isalpha():
    if first in vowel:
        print word
        print first
        print "vowel!"
    else:
        print word
        print first
        print "consonant

回答6:

x = (input ("Enter any word: "))
vowel = "aeiouAEIOU"
if x[0] in vowel:
    print ("1st letter is vowel: ",x)       
else:
    print ("1st letter is consonant: ",x)

回答7:

thanks to alexvassel who provided me the missing part ;)

so here is the solution to the exercice on codecadmy.com

original = raw_input('Enter a word:')

word = original.lower()

first = word[0]

vowel = "aeiou"



if len(original) > 0 and original.isalpha():

   if first in vowel:

       print 'vowel'

       else:   

       print 'consonant'

else:

  print 'empty'

回答8:

Here's how I did it since the inputted word needs to be checked first before storing it as a variable:

original = raw_input('Enter a word:')

if len(original) > 0 and original.isalpha():
    word = original.lower()
    first = word[0]
    if first in ['a','e','i','o','u']:
        print "vowel"
    else:
        print "consonant"
else:
    print 'empty'

回答9:

Will it not be (slightly) faster to define the_vowel as a dictionary than a list?

the_vowel = {"a":1,"e":1,"i":1,"o":1,"u":1}
my_word[0].lower() in the_vowel

回答10:

anti vowel Function

def anti_vowel(text):
    vowel = ["a","e","i","o","u"]
    new_text = ''
    for char in text:
        if char.lower() in vowel:
            continue
        else:
            new_text += char
    print new_text
    return new_text

回答11:

x = raw_input("Enter a word: ")  
vowels=['a','e','i','o','u']
for vowel in vowels:
    if vowel in x:
        print "Vowels"
    else:
        print "No vowels"

This would print out 5 lines, if any of those includes a line that says Vowels then that means there is a vowel. I know this isn't the best way but it works.

回答12:

changes:

if my_word[0] in ('a','e','i','o','u'):
   print(' Yes, the first letter is vowel ')
else:
   print(' No, the first letter is not vowel ')

So, Here is the simple code for finding out that the first letter is either vowel or not!! If you have any further query in python or js, then comment it down.

来源：https://stackoverflow.com/questions/13379243/checking-if-the-first-letter-of-a-word-is-a-vowel

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