1D 索引的 4D 位置?

问题描述

我需要从一维数组中提取一个 4D 位置.我可以看到它在 2D 和 3D 中的表现，但我很难将头环绕在第 4 维..

I need to extract a 4D position from a 1D array. I can see how it goes for 2D and 3D but I'm having a hard time wrapping my head around the 4th dimension..

对于 2D:

int* array = new int[width * height];
int index = y * width + x;
int x = index / height
int y = index - x * height;

对于 3D:

int* array = new int[width * height * depth];
int index = z * width * height + y * width + z;
int x = index / (height * depth);
int y = index - (x * height * depth) / depth;
int z = index - (x * height * depth) - (y * depth);

对于 4D?

int* array = new int[width * height * depth * duration];
int index = w * width * height * depth + z * width * height + y * width + w;
int x = index / (height * depth * duration);
int y = ??

推荐答案

索引公式是任何给定的维度值乘以之前所有维度的乘积.

The indexing formula is given by the multiplication of any given dimension value with the product of all the previous dimensions.

Index = xn ( D1 * ... * D{n-1} ) + x{n-1} ( D1 * ... * D{n-2} ) + ... + x2 * D1 + x1

所以对于 4D

index = x + y * D1 + z * D1 * D2 + t * D1 * D2 * D3;
x = Index % D1;
y = ( ( Index - x ) / D1 ) %  D2;
z = ( ( Index - y * D1 - x ) / (D1 * D2) ) % D3;
t = ( ( Index - z * D2 * D1 - y * D1 - x ) / (D1 * D2 * D3) ) % D4;
/* Technically the last modulus is not required,
   since that division SHOULD be bounded by D4 anyways... */

通用公式的形式为

xn = ( ( Index - Index( x1, ..., x{n-1} ) ) / Product( D1, ..., D{N-1} ) ) % Dn

这篇关于1D 索引的 4D 位置?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！