在python中检查数字是奇数还是偶数
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Check if a number is odd or even in python
提问by user3320350
I'm trying to make a program which checks if a word is a palindrome and I've gotten so far and it works with words that have an even amount of numbers. I know how to make it do something if the amount of letters is odd but I just don't know how to find out if a number is odd. Is there any simple way to find if a number is odd or even?
我正在尝试制作一个程序来检查一个单词是否是回文，到目前为止我已经得到了它并且它可以处理具有偶数数量的单词。如果字母数量是奇数，我知道如何让它做一些事情，但我只是不知道如何找出一个数字是否是奇数。有什么简单的方法可以找到一个数字是奇数还是偶数？
Just for reference, this is my code:
仅供参考，这是我的代码：
a = 0
while a == 0:
print("\n \n" * 100)
print("Please enter a word to check if it is a palindrome: ")
word = input("?: ")
wordLength = int(len(word))
finalWordLength = int(wordLength / 2)
firstHalf = word[:finalWordLength]
secondHalf = word[finalWordLength + 1:]
secondHalf = secondHalf[::-1]
print(firstHalf)
print(secondHalf)
if firstHalf == secondHalf:
print("This is a palindrom")
else:
print("This is not a palindrom")
print("Press enter to restart")
input()
Thanks
谢谢
采纳答案by DeadChex
if num % 2 == 0:
pass # Even
else:
pass # Odd
The %sign is like division only it checks for the remainder, so if the number divided by 2has a remainder of 0it's even otherwise odd.
该%标志是象师只是它检查剩余，因此，如果除以数量2有剩余0，它甚至否则奇怪。
Or reverse them for a little speed improvement, since any number above 0 is also considered "True" you can skip needing to do any equality check:
或者反转它们以提高一点速度，因为任何大于 0 的数字也被认为是“真”，您可以跳过需要进行任何相等性检查：
if num % 2:
pass # Odd
else:
pass # Even
回答by Esteban Aliverti
One of the simplest ways is to use de modulus operator %. If n % 2 == 0, then your number is even.
最简单的方法之一是使用 de 模数运算符 %。如果 n % 2 == 0，那么您的数字是偶数。
Hope it helps,
希望能帮助到你，
回答by NPE
The middle letter of an odd-length word is irrelevant in determining whether the word is a palindrome. Just ignore it.
奇数长度单词的中间字母与确定该单词是否为回文无关。只是忽略它。
Hint: all you need is a slight tweak to the following line to make this work for all word lengths:
提示：您只需要对以下行稍作调整即可使其适用于所有字长：
secondHalf = word[finalWordLength + 1:]
P.S. If you insist on handling the two cases separately, if len(word) % 2: ...would tell you that the word has an odd number of characters.
PS 如果你非要分开处理这两种情况，if len(word) % 2: ...会告诉你这个词有奇数个字符。
回答by kylie.a
It shouldn't matter if the word has an even or odd amount fo letters:
单词的字母数量是偶数还是奇数都没有关系：
def is_palindrome(word):
if word == word[::-1]:
return True
else:
return False
回答by Maxime Lorant
Use the modulo operator:
使用模运算符：
if wordLength % 2 == 0:
print "wordLength is even"
else:
print "wordLength is odd"
For your problem, the simplest is to check if the word is equal to its reversed brother. You can do that with word[::-1], which create the list from wordby taking every character from the end to the start:
对于你的问题，最简单的就是检查单词是否等于它的反向兄弟。你可以用 来做到这一点word[::-1]，它word通过从结尾到开头获取每个字符来创建列表：
def is_palindrome(word):
return word == word[::-1]
回答by lejlot
Similarly to other languages, the fastest "modulo 2" (odd/even) operation is done using the bitwise andoperator:
与其他语言类似，最快的“模 2”（奇数/偶数）运算是使用bitwise and运算符完成的：
if x & 1:
return 'odd'
else:
return 'even'
