我高度建议你使用MMX工具箱 matlab。 它可以尽可能快地增加单维矩阵。

</strong>的优点是:

1. It is easy to use.
2. Multiply n-dimensional matrices (actually it can multiply arrays of 2-D matrices)
3. It performs other matrix operations (transpose, Quadratic Multiply, Chol decomposition and more)
4. It uses C compiler and multi-thread computation for speed up.

For this problem, you just need to write this command:

C=mmx( mul ,A,B);

我在@Amro的答复中增加了以下职能:

%# mmx toolbox
function C=func6(A,B,n,m,p)
    C=mmx( mul ,A,B);
end

www.un.org/Depts/DGACM/index_spanish.htm

    1.6571 # FOR-loop
    4.3110 # ARRAYFUN
    3.3731 # NUM2CELL/FOR-loop/CELL2MAT
    2.9820 # NUM2CELL/CELLFUN/CELL2MAT
    0.0244 # Loop Unrolling
    0.0221 # MMX toolbox  <===================

我使用了@Amro的代码来管理基准。

我现在进行了一些时间测试,这是2x2x2x2的最快途径。 注

C = A;
C(1,1,:) = A(1,1,:).*B(1,1,:) + A(1,2,:).*B(2,1,:);
C(1,2,:) = A(1,1,:).*B(1,2,:) + A(1,2,:).*B(2,2,:);
C(2,1,:) = A(2,1,:).*B(1,1,:) + A(2,2,:).*B(2,1,:);
C(2,2,:) = A(2,1,:).*B(1,2,:) + A(2,2,:).*B(2,2,:);

在一般情况下,选择住宿实际上是最快的(登机忘记预先分配C,但!)。

Should one already have the result as cell-array of matrices though, using cellfun is the fastest choice, it is also faster than looping over the cell elements:

C = cellfun(@mtimes, A, B,  UniformOutput , false);

但是,必须打上rel=“nofollow”>num2cell 首先(Ac = num2cell(A, [1 2])和cell2mat,用于3d-array案废物,时间过长。

这里,我是随机挑选的2×2×1e4:

 array-for: 0.057112
 arrayfun : 0.14206
 num2cell : 0.079468
 cell-for : 0.033173
 cellfun  : 0.025223
 cell2mat : 0.010213
 explicit : 0.0021338

Explicit refers to using direct calculation of the 2 x 2 matrix elements, see bellow. The result is similar for new random arrays, cellfun is the fastest if no num2cell is required before and there is no restriction to 2x2xN. For general 3d-arrays looping over the third dimension is indeed the fastest choice already. Here s the timing code:

n = 2;
m = 2;
l = 1e4;

A = rand(n,m,l);
B = rand(m,n,l);

% naive for-loop:
tic
%Cf = nan(n,n,l);
Cf = A;
for jl = 1:l
    Cf(:,:,jl) = A(:,:,jl) * B(:,:,jl);
end;
disp([  array-for:   num2str(toc)]);

% using arrayfun:
tic
Ca = arrayfun(@(k) A(:,:,k)*B(:,:,k), 1:size(A,3),  UniformOutput ,false);
Ca = cat(3,Ca{:});
disp([  arrayfun :   num2str(toc)]);

tic
Ac = num2cell(A, [1 2]);
Bc = num2cell(B, [1 2]);
disp([  num2cell :   num2str(toc)]);

% cell for-loop:
tic
Cfc = Ac;
for jl = 1:l
    Cfc{jl} = Ac{jl} * Bc{jl};
end;
disp([  cell-for :   num2str(toc)]);

% using cellfun:
tic
Cc = cellfun(@mtimes, Ac, Bc,  UniformOutput , false);
disp([  cellfun  :   num2str(toc)]);

tic
Cc = cell2mat(Cc);
disp([  cell2mat :   num2str(toc)]);

tic
Cm = A;
Cm(1,1,:) = A(1,1,:).*B(1,1,:) + A(1,2,:).*B(2,1,:);
Cm(1,2,:) = A(1,1,:).*B(1,2,:) + A(1,2,:).*B(2,2,:);
Cm(2,1,:) = A(2,1,:).*B(1,1,:) + A(2,2,:).*B(2,1,:);
Cm(2,2,:) = A(2,1,:).*B(1,2,:) + A(2,2,:).*B(2,2,:);
disp([  explicit :   num2str(toc)]);

disp(   );

这里是我的基准测试,比较了https://stackoverflow.com/questions/6580656/matlab-how-to-vector-multiply-second-arrays-of-matrices/6580802#65802>@TobiasKienzler。 答案。 我正在使用 TPASIT功能,以获得更准确的时间。

function [t,v] = matrixMultTest()
    n = 2; m = 2; p = 1e5;
    A = rand(n,m,p);
    B = rand(m,n,p);

    %# time functions
    t = zeros(5,1);
    t(1) = timeit( @() func1(A,B,n,m,p) );
    t(2) = timeit( @() func2(A,B,n,m,p) );
    t(3) = timeit( @() func3(A,B,n,m,p) );
    t(4) = timeit( @() func4(A,B,n,m,p) );
    t(5) = timeit( @() func5(A,B,n,m,p) );

    %# check the results
    v = cell(5,1);
    v{1} = func1(A,B,n,m,p);
    v{2} = func2(A,B,n,m,p);
    v{3} = func3(A,B,n,m,p);
    v{4} = func4(A,B,n,m,p);
    v{5} = func5(A,B,n,m,p);
    assert( isequal(v{:}) )
end

%# simple FOR-loop
function C = func1(A,B,n,m,p)
    C = zeros(n,n,p);
    for k=1:p
        C(:,:,k) = A(:,:,k) * B(:,:,k);
    end
end

%# ARRAYFUN
function C = func2(A,B,n,m,p)
    C = arrayfun(@(k) A(:,:,k)*B(:,:,k), 1:p,  UniformOutput ,false);
    C = cat(3, C{:});
end

%# NUM2CELL/FOR-loop/CELL2MAT
function C = func3(A,B,n,m,p)
    Ac = num2cell(A, [1 2]);
    Bc = num2cell(B, [1 2]);
    C = cell(1,1,p);
    for k=1:p
        C{k} = Ac{k} * Bc{k};
    end;
    C = cell2mat(C);
end

%# NUM2CELL/CELLFUN/CELL2MAT
function C = func4(A,B,n,m,p)
    Ac = num2cell(A, [1 2]);
    Bc = num2cell(B, [1 2]);
    C = cellfun(@mtimes, Ac, Bc,  UniformOutput , false);
    C = cell2mat(C);
end

%# Loop Unrolling
function C = func5(A,B,n,m,p)
    C = zeros(n,n,p);
    C(1,1,:) = A(1,1,:).*B(1,1,:) + A(1,2,:).*B(2,1,:);
    C(1,2,:) = A(1,1,:).*B(1,2,:) + A(1,2,:).*B(2,2,:);
    C(2,1,:) = A(2,1,:).*B(1,1,:) + A(2,2,:).*B(2,1,:);
    C(2,2,:) = A(2,1,:).*B(1,2,:) + A(2,2,:).*B(2,2,:);
end

The results:

>> [t,v] = matrixMultTest();
>> t
t =
      0.63633      # FOR-loop
      1.5902       # ARRAYFUN
      1.1257       # NUM2CELL/FOR-loop/CELL2MAT
      1.0759       # NUM2CELL/CELLFUN/CELL2MAT
      0.05712      # Loop Unrolling

正如我在评论中解释的那样,简单易行是最佳解决办法(,就最后一种情况而言,只有2-by-2 matrices才能做到这一点)。