未知维矩阵的索引 [英] Indexing of unknown dimensional matrix

问题描述

我有一个非固定维矩阵M，我想从中访问单个元素. 元素的索引包含在向量J中.

I have a non-fixed dimensional matrix M, from which I want to access a single element. The element's indices are contained in a vector J.

例如:

M = rand(6,4,8,2);
J = [5 2 7 1];

output = M(5,2,7,1)

这次，M有4个维，但这是未知的.这取决于我正在编写的算法的设置.可能就是那样

This time M has 4 dimensions, but this is not known in advance. This is dependent on the setup of the algorithm I'm writing. It could likewise be that

M = rand(6,4);
J = [3 1];

output = M(3,1)

所以我不能简单地使用

output=M(J(1),J(2))

我当时正在考虑使用 sub2ind ，但这也需要使用变量以逗号分隔..

I was thinking of using sub2ind, but this also needs its variables comma separated..

@gnovice

这可行，但是我打算从矩阵M中大量使用这种元素提取.因此，如果每次访问M时都必须创建一个临时变量cellJ，这会不会极大地减慢计算速度?

this works, but I intend to use this kind of element extraction from the matrix M quite a lot. So if I have to create a temporary variable cellJ every time I access M, wouldn't this tremendously slow down the computation??

我也可以编写一个单独的函数

I could also write a separate function

function x= getM(M,J)
    x=M(J(1),J(2));
    % M doesn't change in this function, so no mem copy needed = passed by reference
end

，并将其调整为适用于算法的不同配置.当然，这是我在问题中没有提到的速度与灵活性的关系.

and adapt this for different configurations of the algorithm. This is of course a speed vs flexibility consideration which I hadn't included in my question..

但是:这仅可用于获取元素，用于设置除了实际使用索引(最好是线性索引)外没有其他方法.我仍然认为sub2ind是一个选择.我原本打算的最终结果是这样的:

BUT: this is only available for getting the element, for setting there is no other way than actually using the indices (and preferably the linear index). I still think sub2ind is an option. The final result I had intended was something like:

function idx = getLinearIdx(J, size_M)
    idx = ...
end

---

结果:

function lin_idx = Lidx_ml( J, M )%#eml
%LIDX_ML converts an array of indices J for a multidimensional array M to
%linear indices, directly useable on M
%
% INPUT
%   J       NxP matrix containing P sets of N indices
%   M       A example matrix, with same size as on which the indices in J
%           will be applicable.
%
% OUTPUT
%   lin_idx Px1 array of linear indices
%

% method 1
%lin_idx = zeros(size(J,2),1);
%for ii = 1:size(J,2)
%    cellJ = num2cell(J(:,ii)); 
%    lin_idx(ii) = sub2ind(size(M),cellJ{:}); 
%end

% method 2
sizeM = size(M);
J(2:end,:) = J(2:end,:)-1;
lin_idx = cumprod([1 sizeM(1:end-1)])*J;

end

方法2的速度是方法1的20倍(要转换的少量索引集(=P))到80倍(大量的索引集(=P))快.易于选择

method 2 is 20 (small number of index sets (=P) to convert) to 80 (large number of index sets (=P)) times faster than method 1. easy choice

推荐答案

对于J可以是任意长度的一般情况(我认为始终与M中的尺寸数匹配)，有几个选择你有:

For the general case where J can be any length (which I assume always matches the number of dimensions in M), there are a couple options you have:

1. 您可以使用J的每个条目放置在单元格阵列的单元格中="nofollow noreferrer"> num2cell 函数，然后创建冒号运算符:

1. You can place each entry of J in a cell of a cell array using the num2cell function, then create a comma-separated list from this cell array using the colon operator:

cellJ = num2cell(J);
output = M(cellJ{:});

您可以回避 sub2ind 函数并使用以下公式自己计算线性索引一点数学:

You can sidestep the sub2ind function and compute the linear index yourself with a little bit of math:

sizeM = size(M);
index = cumprod([1 sizeM(1:end-1)]) * (J(:) - [0; ones(numel(J)-1, 1)]);
output = M(index);

这篇关于未知维矩阵的索引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IT屋！