how to split a large matrix into smaller matrices and assign random values

I have a 180x180 matrix of 0's and I was to assign different range of random integers to each 45x45 set in the matrix. I have show the example output that I want for a much smaller matrix (10x10) that breaks it up into 5x5 matrices of different random values below. Any easy way to do this that doesn't involve a million for loops? I just want to expand this code to an N = zeros(180,180) type deal with 45x45 subparts.

N = zeros(10,10);
for i = 1:5
    for j = 1:5
        N(i,j) = randi([1,10]);
    end
end
for i = 1:5
    for j = 6:10 
        N(i,j) = randi([15,100])
    end
end
for i = 6:10
    for j = 6:10 
        N(i,j) = randi([1,17])
    end
end
for i = 6:10
    for j = 1:5 
        N(i,j) = randi([100,250])
    end
end
N

Answer 1

[Walter Roberson]

low_vals = [1 15 1 100;
            8 -4 17 3;
            1 2 9 15;
            0 -8 3 72];
high_vals = [10 100 17 250;
             14 11 84 9;
             33 11 99 222;
             4 -3 88 105];

N = cell2mat( arrayfun(@(L, H) randi([L, H], 45, 45), low_vals, high_vals, 'Uniform', 0) );

The above would create 4 x 4 of 45 x 45 subblocks.

[Andrew Poissant]

This is not exactly what I want. Sorry, the problem statement is a little unclear. I want to make a 180x180 matrix and specify different ranges of random numbers for every 45x45 block in the 180x180 matrix. So there will be 16 separate blocks in the 180x180 matrix that have different ranges of random integers. Is there a way to easily make that? I just showed the 10x10 matrix to illustrate a simplified version of what I want.

[Walter Roberson]

Did you run the above code? It creates a 180 x 180 matrix in which each 45 x 45 (non-overlapped) sub-block is based on randi() between the corresponding low value and high value. For example, the third sub-block over on the second row of sub-blocks would be based on randi([17, 84], 45, 45) in the above demonstration code.

[Andrew Poissant]

Nvm, I was able to use the cell2mat for my question. Thanks!