如何知道列表中元素的个数是偶数还是奇数 [英] How to know if a list has an even or odd number of elements

问题描述

如何确定任意列表中元素的数量是偶数还是奇数.

How can I find out if there is even, or odd, number of elements in an arbitrary list.

我尝试过list.index()来获取所有索引...但是我仍然不知道如何告诉程序什么是偶数和什么是奇数.

I tried list.index() to get all of the indices... but I still don't know how I can tell the program what is an even and what is an odd number.

推荐答案

您可以为此使用内置函数len().

You can use the built in function len() for this.

Python Doc- -len()

Python Doc -- len()

获取任意列表的长度(元素数).

Gets the length (# of elements) of any arbitrary list.

myList = [0,1,2,3,4,5]

if len(myList) % 2 == 0:
    print ("even")
else
    print ("odd")

定义返回布尔值(真或假)的函数.

def is_even(myList):

    if len(myList) % 2 == 0:
        return true
    else:
        return false

main():

    myList = [0,1,2,3]
    theListIsEven = is_even(myList)  # will be true in this example
                                     # because 4 items in myList

    if theListIsEven(myList) == True:
        # do something
    else:
        # do something else

    return 0

模运算符 % 给出余数.

EX:7％2 = 1

EX: 7 % 2 = 1

• 2将平均除以的最接近7的数字是6
• 7点距离1点.
• 因此，剩余的1代表7％2.

EX:4％2 = 0

EX: 4 % 2 = 0

• 当n％2时，任何偶数n将给出0作为余数
• 因为n已被2除以平均数

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