Python 中的 if/else 循环问题 [英] Issues with a if/else loop in Python

问题描述

我正在尝试用 Python 制作这个 Pig 拉丁语翻译器，它运行良好，直到我试图缩小它的大小.

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

有人可以看看这段代码并告诉我为什么，当我在开头输入一个没有元音的单词时，它仍然会在这个 if 语句中打印元音"代码?

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?

代码:

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'

推荐答案

您需要单独检查所有元音.

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

目前，您的 if 条件评估为:-

Currently, your if condition is evaluated as: -

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

or 返回其条件中的第一个真值，此处为 True 或 e，具体取决于您的第一个条件是 True 还是不是.现在，由于 'e' 被评估为 True，所以两个值都是 true，因此您的条件将始终为 true.

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.

你应该这样做:-

if low_original[0] in 'aeiou':

或:-

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

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