在2D矩阵中对块进行求和 - MATLAB

Question

我正在研究Matlab,并想知道如何在一个大矩阵中添加术语.具体来说,我有一个4914x4914矩阵,并希望创建一个189x189矩阵,其中每个项等于每个26x26子集中的项的总和.

为了说明,我说魔法4x4矩阵如下:

[16 2  3  13;

 5  11 10 8;

 9  7  6  12;

 4  14 15 1]

我想创建一个2x2等于2x2原始魔法中每个矩阵之和的矩阵4x4,即:

[(16+2+5+11)   (3+13+10+8);

(9+7+4+14)  (6+12+15+1)]

感谢任何建议!谢谢杰克

Answer 1

假设A是输入4914x4914矩阵,这可能是一种有效的(在运行时方面)方法 -

sublen = 26; %// subset length
squeeze(sum(reshape(sum(reshape(A,sublen,[])),size(A,1)/sublen,sublen,[]),2))

对于通用块大小,让我们有一个功能 -

function out = sum_blocks(A,block_nrows, block_ncols)
out = squeeze(sum(reshape(sum(reshape(A,block_nrows,[])),...
                    size(A,1)/block_nrows,block_ncols,[]),2));
return

样品运行 -

>> A = randi(9,4,6);
>> A
A =
     8     2     4     9     4     5
     3     3     8     3     6     8
     9     6     6     7     1     9
     4     5     5     7     1     2
>> sum_blocks(A,2,3)
ans =
    28    35
    35    27
>> sum(sum(A(1:2,1:3)))
ans =
    28
>> sum(sum(A(1:2,4:6)))
ans =
    35
>> sum(sum(A(3:4,1:3)))
ans =
    35
>> sum(sum(A(3:4,4:6)))
ans =
    27

如果你想避免squeeze-

sum(permute(reshape(sum(reshape(A,sublen,[])),size(A,1)/sublen,sublen,[]),[1 3 2]),3)

标杆

希望您关心性能,以下是此处发布的所有解决方案的基准测试结果.我使用的基准代码 -

num_runs = 100; %// Number of iterations to run benchmarks
A = rand(4914);
for k = 1:50000
    tic(); elapsed = toc(); %// Warm up tic/toc
end

disp('---------------------- With squeeze + reshape + sum')
tic
for iter = 1:num_runs
    sublen = 26; %// subset length
    out1 = squeeze(sum(reshape(sum(reshape(A,sublen,[])),...
                                   size(A,1)/sublen,sublen,[]),2));
end
time1 = toc;
disp(['Avg. elapsed time = ' num2str(time1/num_runs) ' sec(s)']), clear out1 sublen

disp('---------------------- With kron + matrix multiplication')
tic
for iter = 1:num_runs
    n = 189; k = 26;
    B = kron(speye(k), ones(1,n));
    result = B*A*B';
end
time2 = toc;
disp(['Avg. elapsed time = ' num2str(time2/num_runs) ' sec(s)']),clear result n k B

disp('---------------------- With accumarray')
tic
for iter = 1:num_runs
    s = 26; n = size(A,1)/s;
    subs = kron(reshape(1:(n^2), n, n),ones(s));
    out2 = reshape(accumarray(subs(:), A(:)), n, n);
end
time2 = toc;
disp(['Avg. elapsed time = ' num2str(time2/num_runs) ' sec(s)']),clear s n subs out2

基准测试结果我得到了我的系统 -

---------------------- With squeeze + reshape + sum
Avg. elapsed time = 0.050729 sec(s)
---------------------- With kron + matrix multiplication
Avg. elapsed time = 0.068293 sec(s)
---------------------- With accumarray
Avg. elapsed time = 0.64745 sec(s)

Answer 2

如果您没有图像处理工具箱,那么您可以使用accumarray以下方法执行此操作:

s = 26;
n = size(A,1)/s;
subs = kron(reshape(1:(n^2), n, n),ones(s)); 
reshape(accumarray(subs(:), A(:)), n, n) 

如果您决定聚合某种方式而不是简单的总和,例如中位数,则可以重复使用:

reshape(accumarray(subs(:), A(:), [], @median), n, n) 

Answer 3

你当然可以使用矩阵乘法:

n = 26;
k = 189;
B = kron(speye(k), ones(1,n));
result = B*A*B';

Answer 4

另一种方法是将整个矩阵重新整形为4D矩阵,并在第一维和第三维上对元素求和:

result = squeeze(sum(sum(reshape(A,26,189,26,189),1),3));