我如何计算坐标，反之亦然一个3D的线性指标？ [英] How do I compute the linear index of a 3D coordinate and vice versa?

问题描述

如果我有一个点（x，y z）的，我怎么找到线性指标，我该点？我的编号方案将是（0,0,0）为0，（1，0,0）是1，。 。 ，（0，1，0）是最大-x维度，....
另外，如果我有一个线性的坐标，我，我怎么觉得（X，Y，Z）？
我似乎无法找到这对谷歌，所有的结果都充满了其他无关的东西。谢谢！

If I have a point (x, y z), how do I find the linear index, i for that point? My numbering scheme would be (0,0,0) is 0, (1, 0, 0) is 1, . . ., (0, 1, 0) is the max-x-dimension, .... Also, if I have a linear coordinate, i, how do I find (x, y, z)? I can't seem to find this on google, all the results are filled with other irrelevant stuff. Thank you!

推荐答案

有几个方法来映射一个三维坐标到一个单一的数字。这里有一种方法。

There are a few ways to map a 3d coordinate to a single number. Here's one way.

一些函数f（x，Y，Z）给出坐标的线性索引（X，Y，Z）。它有我们希望派生，所以我们可以写一个有用的转换函数一些常量A，B，C，D。

some function f(x,y,z) gives the linear index of coordinate(x,y,z). It has some constants a,b,c,d which we want to derive so we can write a useful conversion function.

f(x,y,z) = a*x + b*y + c*z + d

您已经指定了（0,0,0）映射到0.所以：

You've specified that (0,0,0) maps to 0. So:

f(0,0,0) = a*0 + b*0 + c*0 + d = 0
d = 0
f(x,y,z) = a*x + b*y + c*z

这是公司的D解决。
您已经指定了（1,0,0）映射到1，所以：

That's d solved. You've specified that (1,0,0) maps to 1. So:

f(1,0,0) = a*1 + b*0 + c*0 = 1
a = 1
f(x,y,z) = x + b*y + c*z

这是一个解决。
让我们任意决定后下一个最高数目（MAX_X，0，0）是（0,1,0）。

That's a solved. Let's arbitrarily decide that the next highest number after (MAX_X, 0, 0) is (0,1,0).

f(MAX_X, 0, 0) = MAX_X
f(0, 1, 0) = 0 + b*1 + c*0 = MAX_X + 1
b = MAX_X + 1
f(x,y,z) = x + (MAX_X + 1)*y + c*z

这'S B解决。
让我们任意决定后下一个最高数目（MAX_X，MAX_Y，0）是（0,0,1）。

That's b solved. Let's arbitrarily decide that the next highest number after (MAX_X, MAX_Y, 0) is (0,0,1).

f(MAX_X, MAX_Y, 0) = MAX_X + MAX_Y * (MAX_X + 1)
f(0,0,1) = 0 + (MAX_X + 1) * 0  + c*1 = MAX_X + MAX_Y * (MAX_X + 1) + 1
c = MAX_X + MAX_Y * (MAX_X + 1) + 1
c = (MAX_X + 1) + MAX_Y * (MAX_X + 1)
c = (MAX_X + 1) * (MAX_Y + 1)

现在我们知道A，B，C和D，我们可以按如下方式编写功能：

now that we know a, b, c, and d, we can write your function as follows:

function linearIndexFromCoordinate(x,y,z, max_x, max_y){
    a = 1
    b = max_x + 1
    c = (max_x + 1) * (max_y + 1)
    d = 0
    return a*x + b*y + c*z + d
}

您可以通过类似的逻辑，线性指标的坐标。我有这样一个真正了不起的演示，此页面是太小，无法容纳。所以我会跳过数学讲座，只是给你的最终方法​​。

You can get the coordinate from the linear index by similar logic. I have a truly marvelous demonstration of this, which this page is too small to contain. So I'll skip the math lecture and just give you the final method.

function coordinateFromLinearIndex(idx, max_x, max_y){
    x =  idx % (max_x+1)
    idx /= (max_x+1)
    y = idx % (max_y+1)
    idx /= (max_y+1)
    z = idx
    return (x,y,z)
}

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