将矩阵行块的矢量化shuffle

Question

我有一个简单的 5x15矩阵，称为 mat_orig。我想将此矩阵分为非重叠的块，每个块的长度为3。等于将三个非重叠的行放入一个块中(即有5个大小3x5的块(。

> 。

然后，我想洗牌mat_orig并生成一个新的矩阵。我正在使用randi随机绘制五个块。

我成功地使用以下代码完成了任务。但是，我想知道我是否可以摆脱循环？我可以应用矢量化吗？

mat_orig = reshape(1:75, 5, 15)';
block_length = 3;
num_blocks = size(mat_orig, 1) / block_length;
rand_blocks = randi(num_blocks, num_blocks, 1);
mat_shuffled = nan(size(mat_orig));
for r = 0 : num_blocks - 1
    start_row_orig = r * block_length + 1;
        end_row_orig = r * block_length + block_length;
    start_row_random_blocks = ...
        rand_blocks(r + 1) * block_length - block_length + 1;
    end_row_random_blocks = ...
        rand_blocks(r + 1) * block_length;
    mat_shuffled(start_row_orig:end_row_orig, :) = ...
        mat_orig(start_row_random_blocks:end_row_random_blocks, :);
end

Answer 1

您可以使用隐式扩展来创建块的索引，请参阅代码注释以获取详细信息。

mat_orig = reshape(1:75, 5, 15)';
block_length = 3;
num_blocks = size(mat_orig,1) / block_length;
% Get the shuffle order (can repeat blocks)
shuffIdx = randi( num_blocks, num_blocks, 1 ).';
% Expand these indices to fill the blocks
% This uses implicit expansion to create a matrix, and 
% will only work in R2016b or newer.
% For older versions, use 'bsxfun'
shuffIdx = block_length * shuffIdx - (block_length-1:-1:0).';
% Create the shuffled output
mat_shuffled = mat_orig( shuffIdx(:), : );