[英]Convert a 1d array index to a 3d array index?
I've got a int that I want to convert to 3 ints for the index of a 3d array, here's an example of what I'm on about. 我有一个int想要转换为3个int用于3d数组的索引,这里是我所关注的一个例子。
byte[,,] array = new byte[XSize, YSize, ZSize];
int i = 0;
//other code
array[#,#,#] = cur;
//other code
I don't know how to get the correct numbers for the #,#,# from just i. 我不知道如何从#,#,#中获取正确的数字。
Supposing you want to iterate through Z, then Y, and then X. . 假设您想要遍历Z,然后是Y,然后是X. . 。
int zDirection = i % zLength;
int yDirection = (i / zLength) % yLength;
int xDirection = i / (yLength * zLength);
You have to make an assumption about the orientation of the array-space. 您必须对阵列空间的方向做出假设。 Assuming you are mapping the numbers with 0 corresponding to 0, 0, 0, and that you iterate z, then y, then x, the math is fairly simple: 假设您将数字映射为0,对应于0,0,0,并且您迭代z,则y,然后x,数学非常简单:
int x;
int y;
int z;
if (i < XSize * YSize * ZSize)
{
int zQuotient = Math.DivRem(i, ZSize, out z);
int yQuotient = Math.DivRem(zQuotient, YSize, out y);
x = yQuotient % XSize;
}
Note that this saves some redundant operations over BlackVegetable's solution. 请注意,这比BlackVegetable的解决方案节省了一些冗余操作。 For a (3, 3, 3) matrix, this yields the set: 对于(3,3,3)矩阵,这会得到集合:
i (x, y, z)
0 (0, 0, 0)
1 (0, 0, 1)
2 (0, 0, 2)
3 (0, 1, 0)
4 (0, 1, 1)
5 (0, 1, 2)
6 (0, 2, 0)
7 (0, 2, 1)
8 (0, 2, 2)
9 (1, 0, 0)
10 (1, 0, 1)
11 (1, 0, 2)
12 (1, 1, 0)
13 (1, 1, 1)
14 (1, 1, 2)
15 (1, 2, 0)
16 (1, 2, 1)
17 (1, 2, 2)
18 (2, 0, 0)
19 (2, 0, 1)
20 (2, 0, 2)
21 (2, 1, 0)
22 (2, 1, 1)
23 (2, 1, 2)
24 (2, 2, 0)
25 (2, 2, 1)
26 (2, 2, 2)
Also, this is reversible with i = z + y * ZSize + x * ZSize * YSize
. 此外,这是可逆的, i = z + y * ZSize + x * ZSize * YSize
。
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