问题
What silly mistake am I making here that is preventing me from determining that the first letter of user input is a consonant? No matter what I enter, it allows evaluates that the first letter is a vowel.
original = raw_input(\'Enter a word:\')
word = original.lower()
first = word[0]
if len(original) > 0 and original.isalpha():
if first == \"a\" or \"e\" or \"i\" or \"o\" or \"u\":
print \"vowel\"
else:
print \"consonant\"
else:
print \"empty\"
回答1:
Change:
if first == "a" or "e" or "i" or "o" or "u":
to:
if first in ('a', 'e', 'i', 'o', 'u'): #or `if first in 'aeiou'`
first == "a" or "e" or "i" or "o" or "u" is always True because it is evaluated as
(first == "a") or ("e") or ("i") or ("o") or ("u"), as an non-empty string is always True so this gets evaluated to True.
>>> bool('e')
True
回答2:
What you are doing in your if statement is checking if first == "a" is true and then if "e" is true, which it always is, so the if statement always evaluates to true.
What you should do instead is:
if first == "a" or first == "e" ...
or better yet:
if first in "aeiou":
回答3:
Your issue is that first == "a" or "e" is being evaluated as (first == "a") or "e", so you're always going to get 'e', which is a True statement, causing "vowel" to be printed. An alternative is to do:
original = raw_input('Enter a word:')
word = original.lower()
first = word[0]
if len(original) > 0 and original.isalpha():
if first in 'aeiou':
print "vowel"
else:
print "consonant"
else:
print "empty"
来源：https://stackoverflow.com/questions/20226110/detecting-vowels-vs-consonants-in-python