当变量变为偶数时，如何执行循环执行，并在奇数变量时执行其他操作？

Question

我目前正在开发一个井字游戏，我需要一些帮助。

如果变量agged = df.groupby('title').agg(dict(rank='idxmin', stores='sum')) df.loc[agged['rank'], ['title', 't_no', 't_descr', 'rank']].join(agged.stores, on='title') title t_no t_descr rank stores 0 A 1 a 1 1000 1 B 1 a 1 1800 3 C 2 b 2 800 4 D 1 a 1 1800 7 E 3 c 3 700 6 F 4 d 4 500 是偶数（2,4,6,8,10等），我试图让我的循环运行，就像制作一个X并做一些不同的事情，如果它是奇数（ 1,3,5,7,9等等，就像制作O一样。现在我只想弄清楚如何换掉轮流。我已经查看了其他问题，并且搜索了这个主题没有运气。

xoro

我是否必须使用int[] even = {2,4,6,8,10,12,14,16,18,20}; int xoro = 1; if (xoro == even[]) { // troublesome line System.out.println("an even number is " + xoro); XORO++; } else { System.out.println("an odd number is " + xoro); XORO++; } } 等？我知道必须有一种更简单的方法来做到这一点。

Answer 1

在Dave Newton的帮助下，我修复了程序，答案是这样的：

int XORO = 1;

    for(int i=0;i<20;i++) {
        if(XORO%2==0){
            System.out.println("an even number is " + XORO);
            XORO++;
            }
        else {
            System.out.println("an odd number is " + XORO);
            XORO++;
            }
        }

输出是我想要的：

an odd number is 1
an even number is 2
an odd number is 3
an even number is 4
an odd number is 5
an even number is 6
an odd number is 7
an even number is 8
an odd number is 9
an even number is 10
an odd number is 11
an even number is 12
an odd number is 13
an even number is 14
an odd number is 15
an even number is 16
an odd number is 17
an even number is 18
an odd number is 19
an even number is 20

Answer 2

     int[] even = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 21};
            for (int XORO = 0; XORO < even.length; XORO++) {
                if (even[XORO] % 2 == 0) {
                    System.out.println("an even number is " + even[XORO]);
                } else {
                    System.out.println("an odd number is " + even[XORO]);
                }
            }

output:-

an even number is 2
an even number is 4
an even number is 6
an even number is 8
an even number is 10
an even number is 12
an even number is 14
an even number is 16
an even number is 18
an even number is 20
an odd number is 21

Answer 3

您可以为此

使用简单的for循环

int[] even = {2,4,6,8,10,12,14,16,18,20};
for(int i=0;i<even.length;i++)
{
if(even[i]%2==0)
{ System.out.println("an even number is " + XORO);}
else
{System.out.println("an odd number is " + XORO);}
}