How do I convert a 2X2 matrix to 4X4 matrix in MAT

问题:

I need some help in converting a 2X2 matrix to a 4X4 matrix in the following manner:

A = [2 6;
     8 4]

should become:

B = [2 2 6 6;
     2 2 6 6;
     8 8 4 4;
     8 8 4 4]

How would I do this?

回答1:

A = [2 6; 8 4];
% arbitrary 2x2 input matrix

B = repmat(A,2,2);
% replicates rows & columns but not in the way you want

B = B([1 3 2 4], :);
% swaps rows 2 and 3

B = B(:, [1 3 2 4]);
% swaps columns 2 and 3, and you're done!

回答2:

In newer versions of MATLAB (R2015a and later) the easiest way to do this is using the repelem function:

B = repelem(A, 2, 2);

For older versions, a short alternative to the other (largely) indexing-based solutions is to use the functions kron and ones:

>> A = [2 6; 8 4];
>> B = kron(A, ones(2))

B =

     2     2     6     6
     2     2     6     6
     8     8     4     4
     8     8     4     4

回答3:

Can be done even easier than Jason's solution:

B = A([1 1 2 2], :);  % replicate the rows
B = B(:, [1 1 2 2]);  % replicate the columns

回答4:

Here's one more solution:

A = [2 6; 8 4];
B = A( ceil( 0.5:0.5:end ), ceil( 0.5:0.5:end ) );

which uses indexing to do everything and doesn't rely on the size or shape of A.

回答5:

This works:

A = [2 6; 8 4];
[X,Y] = meshgrid(1:2);
[XI,YI] = meshgrid(0.5:0.5:2);
B = interp2(X,Y,A,XI,YI,'nearest');

This is just two-dimensional nearest-neighbor interpolation of A(x,y) from x,y ∈ {1,2} to x,y ∈ {0.5, 1, 1.5, 2}.

Edit: Springboarding off of Jason S and Martijn's solutions, I think this is probably the shortest and clearest solution:

A = [2 6; 8 4];
B = A([1 1 2 2], [1 1 2 2]);

回答6:

Here's a method based on simple indexing that works for an arbitrary matrix. We want each element to be expanded to an MxN submatrix:

A(repmat(1:end,[M 1]),repmat(1:end,[N 1]))

Example:

>> A=reshape(1:6,[2,3])

A =

     1     3     5
     2     4     6

>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))

ans =

     1     1     1     1     3     3     3     3     5     5     5     5
     1     1     1     1     3     3     3     3     5     5     5     5
     1     1     1     1     3     3     3     3     5     5     5     5
     2     2     2     2     4     4     4     4     6     6     6     6
     2     2     2     2     4     4     4     4     6     6     6     6
     2     2     2     2     4     4     4     4     6     6     6     6

To see how the method works, let's take a closer look at the indexing. We start with a simple row vector of consecutive numbers

>> m=3; 1:m

ans =

     1     2     3

Next, we extend it to a matrix, by repeating it M times in the first dimension

>> M=4; I=repmat(1:m,[M 1])

I =

     1     2     3
     1     2     3
     1     2     3
     1     2     3

If we use a matrix to index an array, then the matrix elements are used consecutively in the standard Matlab order:

>> I(:)

ans =

     1
     1
     1
     1
     2
     2
     2
     2
     3
     3
     3
     3

Finally, when indexing an array, the 'end' keyword evaluates to the size of the array in the corresponding dimension. As a result, in the example the following are equivalent:

>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))
>> A(repmat(1:2,[3 1]),repmat(1:3,[4 1]))
>> A(repmat([1 2],[3 1]),repmat([1 2 3],[4 1]))
>> A([1 2;1 2;1 2],[1 2 3;1 2 3;1 2 3;1 2 3])
>> A([1 1 1 2 2 2],[1 1 1 1 2 2 2 2 3 3 3 3])

回答7:

There is a Reshape() function that allows you to do this...

For example:

reshape(array, [64, 16])

And you can find a great video tutorial here

Cheers