Python 检查变量的多个值
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Checking multiple values for a variable
提问by Alexander Nilsson
original = raw_input('Enter a word:')
if len(original) > 0 and original.isalpha():
word = original.lower()
first = str(word)[0]
print first
if str(first) == "a" or "e" or "i" or "u" or "o":
print "vowel"
else:
print "consonant"
I want to check if a word starts with a vowel or consonant. However, this part does not work: if str(first) == "a" or "e" or "i" or "u" or "o"
我想检查一个单词是否以元音或辅音开头。但是，这部分不起作用：如果str(first) == "a" or "e" or "i" or "u" or "o"
So how would you check if the first letter is either "a" or "e" or "i" or "u" or "o"?
那么如何检查第一个字母是“a”还是“e”或“i”或“u”或“o”？
回答by Fernando Freitas Alves
You better use in
你最好用 in
if len(original) and original.isalpha():
word = original.lower()
first = word[0]
print first
if first in ('a','e','i','o','u'):
print "vowel"
else:
print "consonant"
also you are doing it wrong, if you are trying to use OR clause you must use like this BUT it's not the better pythonic way:
你也做错了，如果你试图使用 OR 子句，你必须像这样使用，但这不是更好的 pythonic 方式：
if first =='a' or first =='e' or first =='i' or first =='o' or first =='u':
回答by Yarkee
if str(first) == "a" or "e" or "i" or "u" or "o":
should moditied to
应该修改为
if str(first) in ("a", "e", "i", "o", "u"):
Python has a explicit demand on indent. Make sure you have a right indent.
Python 对缩进有明确的要求。确保你有一个正确的缩进。
original = raw_input('Enter a word:')
if len(original) > 0 and original.isalpha():
word = original.lower()
first = str(word)[0]
print first
if str(first) in ("a", "e", "i", "o", "u"):
print "vowel"
else:
print "consonant"
