将任何 N*M 矩阵拆分为相等的子矩阵部分

Question

我想将任何矩阵分成相等的子矩阵部分。我使用 mat2cell 来做到这一点。但它使大多数子矩阵成为所需的大小，但某些部分不相等。 例如：我有一个大小为 973*973 的矩阵。我想平均分成 216*216 个子矩阵。当我应用 mat2cell 时，它会生成一个 6*6 的子矩阵。其中前 5*5 是 216*216 个子矩阵。但最后一行和最后一列的块不是 216*216。它们是随机尺寸。这是不希望的。我希望每个子矩阵都具有相同数量的行和列。每个块将是 216*216 大小或用户指定的大小。

I=imread(file);
[rows columns numberOfColorBands] = size(I);
blockSizeR = 256; % Rows in desired block which can be anynumber.
blockSizeC = 256; % Columns in desired block which can be anynumber.
wholeBlockRows = floor(rows / blockSizeR);
blockVectorR = [blockSizeR * ones(1, wholeBlockRows), rem(rows, blockSizeR)];
% Figure out the size of each block in columns.
wholeBlockCols = floor(columns / blockSizeC);
blockVectorC = [blockSizeC * ones(1, wholeBlockCols), rem(columns, blockSizeC)];

ca = mat2cell(I, blockVectorR, blockVectorC);

Answer 1

如果您只是想用零填充空格，最简单的方法是在拆分之前填充初始矩阵：

I=imread(file);
[rows columns numberOfColorBands] = size(I);
blockSizeR = 256; % Rows in desired block which can be anynumber.
blockSizeC = 256; % Columns in desired block which can be anynumber.

% Figure out the size of each block in rows.
blockRows = ceil(rows / blockSizeR);
blockVectorR = blockSizeR * ones(1, blockRows);

% Figure out the size of each block in columns.
blockCols = ceil(columns / blockSizeC);
blockVectorC = blockSizeC * ones(1, blockCols);

% Pad with zeros
I(blockRows*blockSizeR,blockCols*blockSizeC) = 0;

% Split it up
ca = mat2cell(I, blockVectorR, blockVectorC);

我认为用零填充是最好的方法，因为否则你可能会遇到两个无法拆分的素数维度的 I

如果您想用 NaN 填充，只需将其替换为上面我用零填充的代码：

% Pad with NaN
I = [I, NaN(size(I,1), blockCols*blockSizeC-size(I,2)) ;...
NaN(blockRows*blockSizeR-size(I,1), size(I,2)) NaN(blockRows*blockSizeR-size(I,1),blockCols*blockSizeC-size(I,2))];