[英]Issues with a if/else loop in Python
I am trying to make this Pig Latin translator in Python and it was working well until I tried to downsize it a bit. 我正在尝试使用Python制作此Pig Latin转换器,并且在我尝试缩小尺寸之前一直工作良好。
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? 有人可以看一下这段代码,然后告诉我为什么,当我在开头输入没有元音的单词时,仍然会在此if语句中打印“元音”代码吗?
CODE: 码:
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 need to do a check for all the vowels separately. 您需要分别检查所有元音。
Currently, your if condition is evaluated as: - 目前,您的if条件评估为:-
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. or
返回其条件中的第一个true值,在这里将为True
或e
,具体取决于您的第一个条件为True还是不为True。 Now, since 'e'
is evaluated to True
, so both the values are true
, hence your condition will always be true
. 现在,由于
'e'
被评估为True
,因此两个值都为true
,因此您的条件将始终为true
。
You should do it like this: - 您应该这样做:-
if low_original[0] in 'aeiou':
or: - 要么: -
if low_original[0] in ('a', 'e', 'i', 'o', 'u'):
You should replace the string: 您应该替换字符串:
if low_original[0] == 'a' or 'e' or 'i' or 'o' or 'u':
with: 与:
if low_original[0] in ('a', 'e', 'i', 'o', 'u'):
The if is alwasys returning True! 如果是alwasys,则返回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'
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. 问题出在'如果low_original [0] =='a'或'e'或'i'或'o'或'u':'-首先不是pythonic其次没有给您没有您期望的结果。
Try to update your code to: 尝试将您的代码更新为:
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'
将if语句替换为
if low_original[0] in ['a', 'e', 'i', 'o', 'u']
The condition in this if
will always evaluate to True
. 在这种状况
if
将始终评估为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'
与
if (low_original[0] == 'a') or 'e' or 'i' or 'o' or 'u'
You should use something like if low_original[0] in 'aeiou'
您应该
if low_original[0] in 'aeiou'
使用if low_original[0] in 'aeiou'
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