Question

我试图以“矢量化”的方式将大（2x2m）矩阵的（2x2）子矩阵相乘，以消除环路并提高速度。目前，我重塑为（2x2xm）然后使用for循环来执行此操作：

for n = 1:1e5
    m = 1e4;
    A = rand([2,2*m]);     % A is a function of n
    A = reshape(A,2,2,[]);
    B = eye(2);
    for i = 1:m
        B = A(:,:,i)*B;    % multiply the long chain of 2x2's
    end
end

函数目标类似于@prod，但是使用矩阵乘法而不是元素标量乘法。 @multiprod似乎很接近，但是将两个不同的nD矩阵作为参数。我想象一个解决方案使用一个非常大的2D数组的多个子矩阵，或者一个2x2m {xn}数组来消除一个或两个for循环。

提前谢谢，乔

Answer 1

我认为你必须以不同的方式重塑你的矩阵来进行矢量化乘法，就像在下面的代码中一样。这段代码也使用循环，但我认为应该更快

MM      = magic(2);
M0      = MM;
M1      = rot90(MM,1);
M2      = rot90(MM,2);
M3      = rot90(MM,3);


MBig1           = cat(2,M0,M1,M2,M3);
fprintf('Original matrix\n')
disp(MBig1)
MBig2           = zeros(size(MBig1,2));
MBig2(1:2,:)    = MBig1;
for k=0:3
    c1 =  k   *2+1;
    c2 = (k+1)*2+0;
    MBig2(:,c1:c2) = circshift(MBig2(:,c1:c2),[2*k 0]);
end
fprintf('Reshaped original matrix\n')
disp(MBig2)

fprintf('Checking [ M0*M0 M0*M1 M0*M2 M0*M3 ] in direct way\n')
disp([ M0*M0 M0*M1 M0*M2 M0*M3 ])
fprintf('Checking [ M0*M0 M0*M1 M0*M2 M0*M3 ] in vectorized way\n')
disp( kron(eye(4),M0)*MBig2 )


fprintf('Checking [ M0*M1*M2*M3 ] in direct way\n')
disp([ M0*M1*M2*M3 ])
fprintf('Checking [ M0*M1*M2*M3 ] in vectorized way\n')
R2 = MBig2;
for k=1:3
    R2 = R2 * circshift(MBig2,-[2 2]*k);
end
disp(R2)

输出

Original matrix
     1     3     3     2     2     4     4     1
     4     2     1     4     3     1     2     3

Reshaped original matrix
     1     3     0     0     0     0     0     0
     4     2     0     0     0     0     0     0
     0     0     3     2     0     0     0     0
     0     0     1     4     0     0     0     0
     0     0     0     0     2     4     0     0
     0     0     0     0     3     1     0     0
     0     0     0     0     0     0     4     1
     0     0     0     0     0     0     2     3

Checking [ M0*M0 M0*M1 M0*M2 M0*M3 ] in direct way
    13     9     6    14    11     7    10    10
    12    16    14    16    14    18    20    10

Checking [ M0*M0 M0*M1 M0*M2 M0*M3 ] in vectorized way
    13     9     0     0     0     0     0     0
    12    16     0     0     0     0     0     0
     0     0     6    14     0     0     0     0
     0     0    14    16     0     0     0     0
     0     0     0     0    11     7     0     0
     0     0     0     0    14    18     0     0
     0     0     0     0     0     0    10    10
     0     0     0     0     0     0    20    10

Checking [ M0*M1*M2*M3 ] in direct way
   292   168
   448   292

Checking [ M0*M1*M2*M3 ] in vectorized way
   292   168     0     0     0     0     0     0
   448   292     0     0     0     0     0     0
     0     0   292   336     0     0     0     0
     0     0   224   292     0     0     0     0
     0     0     0     0   292   448     0     0
     0     0     0     0   168   292     0     0
     0     0     0     0     0     0   292   224
     0     0     0     0     0     0   336   292

Answer 2

以下功能可以解决部分问题。它被称为“mprod”与prod，类似于时代与mtimes。通过一些重塑，它递归地使用multiprod。通常，递归函数调用比循环慢。 Multiprod声称速度提高了100倍，所以应该补偿不止。

function sqMat = mprod(M)
    % Multiply *many* square matrices together, stored
    % as 3D array M. Speed gain through recursive use 
    % of function 'multiprod' (Leva, 2010).

    % check if M consists of multiple matrices
    if size(M,3) > 1
        % check for odd number of matrices
        if mod(size(M,3),2)
            siz = size(M,1);
            M = cat(3,M,eye(siz));
        end
        % create two smaller 3D arrays
        X = M(:,:,1:2:end); % odd pages
        Y = M(:,:,2:2:end); % even pages
        % recursive call
        sqMat = mprod(multiprod(X,Y));
    else
        % create final 2D matrix and break recursion
        sqMat = M(:,:,1);
    end
end

我没有测试此功能的速度或准确性。我相信这比循环要快得多。它没有“矢量化”操作，因为它不能用于更高的尺寸;任何重复使用此功能必须在循环内完成。

编辑以下是似乎运行得足够快的新代码。对函数的递归调用很慢并占用堆栈内存。仍然包含一个循环，但通过log（n）/ log（2）减少循环次数。此外，还增加了对更多维度的支持。

function sqMats = mprod(M)
    % Multiply *many* square matrices together, stored along 3rd axis.
    % Extra dimensions are conserved; use 'permute' to change axes of "M".
    % Speed gained by recursive use of 'multiprod' (Leva, 2010).

    % save extra dimensions, then reshape
    dims = size(M);
    M = reshape(M,dims(1),dims(2),dims(3),[]);
    extraDim = size(M,4);

    % Check if M consists of multiple matrices...
    % split into two sets and multiply using multiprod, recursively
    siz = size(M,3);
    while siz > 1
        % check for odd number of matrices
        if mod(siz,2)
            addOn = repmat(eye(size(M,1)),[1,1,1,extraDim]);
            M = cat(3,M,addOn);
        end
        % create two smaller 3D arrays
        X = M(:,:,1:2:end,:); % odd pages
        Y = M(:,:,2:2:end,:); % even pages
        % recursive call and actual matrix multiplication
        M = multiprod(X,Y);
        siz = size(M,3);
    end

    % reshape to original dimensions, minus the third axis.
    dims(3) = [];
    sqMats = reshape(M,dims);
end

MATLAB - 矩阵在单个矩阵中乘法子矩阵

2 个答案: