How to enlarge matrix by putting average of surrounding numbers in between of every number of original matrix

Hi everyone,

i would like to create a larger matrix from a small one by putting average of surrounding numbers. so for a matrix

   a = [1 2; 3 4]
a =

       1     2
       3     4

i'd like to create matrix of

b = [1 0 2;0 (1+2+3+4)/4 0;3 0 4]

b =

      1.0000         0    2.0000
           0    2.5000         0
      3.0000         0    4.0000

then to put average to the places where is zero to creat a 3 by 3 matrix from original 2 by 2:

 c = [1 (1+2+2.5)/3 2;(1+3+2.5)/3 2.5 (2+4+2.5)/3; 3 (3+4+2.5)/3 4]

c =

    1.0000    1.8333    2.0000
    2.1667    2.5000    2.8333
    3.0000    3.1667    4.0000

And now i want to do so over and over to create matrices of 2x2 to 3x3 to 5x5 to 8x8 to 13x13 to 24x24 to 46x46 to 92x92 to 184x184. The reason to do so is to create some kind of map, i.e. brainmap.

So any idea how to make it some elegant. I can only thing of using some 4 for cyclus to do so like in paper i need to solve out some PC way. Thanks for any idea in advance

Jan

[Andrew Newell]

I think you have the numbers wrong there: the dimensions should be 2,3,5,9,17,33,... (after each n comes n + (n-1) = 2*n-1).

[Sean de Wolski]

Andrew, they did have the numbers wrong; I agree. That's a minor error and Jan still gets a vote for a well written and described question. +1

[Andrew Newell]

Absolutely! I really appreciate it because it is so rare.

[Jan]

Thank you guys all for your help. I'll need some time to go trought the codes but i really appriciate your time. I'll also add some pictures to describe what is this good for, i.e. brainmapping.

Here are some pictures that describes a proces of creating a brainmap from eeg data:

https://picasaweb.google.com/109325916488505014236/Brainmap?feat=directlink

Answer 1

[Sean de Wolski]

Here's a fully vectorized method. Call it n number of times and it will repeat each time! (It will also work on non-square matrices)

function out = expandWmean(in)
%SCd 04/06/2011
%
  M1 = false((2*size(in))-1);  %Map of points we have
  M1(1:2:end,1:2:end) = true;
  sz = size(M1);
  M2 = false(sz);
  M2(2:2:end,2:2:end) = true; %Map of points we get on first iteration
  out = zeros(sz);
  out(M2) = conv2(in,ones(2)./4,'valid'); %Mean for first iteration (center of boxes)
  out(M1) = in;
  N = ones(sz)*4; %number of elements in the valid mean calculation
  N([1 end],:) = 3;
  N(:,[1 end]) = 3;
  C2 = conv2(out,ones(3),'same')./N; %Second iteration
  out(~M1&~M2) = C2(~M1&~M2); 
end

EDIT: There are 4 valid points in the middle parts on the second iteration, not 5.

[Andrew Newell]

Nice!

[Jan]

I am impressed. Work just as i needed to. Just curious, how would you change it to let the matrix grown to the bigger one. I mean the input would be let say 5x6 (or whatever) matrix and one would like to have much bugger one like a 1000x1000 matrix. I know just time the function again and again.

Answer 2

[Andrew Newell]

You may not be able to get what you want using just interp2. The following code

nb = 2*size(a,1)-1;
[ia,ja] = meshgrid(1:2:nb,1:2:nb);
[ib,jb] = meshgrid(1:nb,1:nb);
c = interp2(ia,ja,a,ib,jb,'*linear')

returns the matrix

c =

    1.0000    1.5000    2.0000
    2.0000    2.5000    3.0000
    3.0000    3.5000    4.0000

As you can see, the middle components are averages of just two neighbors ( e.g, 1.5 = (1+2)/2 ). Unless you have a compelling reason to do it the way you described, this would be much easier.

Answer 3

[Chris]

This made a nice brain teaser to work on for a few minutes every hour or so. So, here's the hard way, if so inclined. It could be cleaned up plenty, and you could add some fixes to allow an 'a' with a dimension smaller than 2, but this will work for all rectangles, not just squares, 2x2 and up.

% Supply your own N-by-M matrix 'a', at least 2x2
S = size(a,1);
T = size(a,2);
U = 2*S-1;
V = 2*T-1;

%Preallocate 'b'
b = zeros(U,V);

%Calc 'b'
for n = 1:size(a,1)
    for m = 1:size(a,2)
        W = 2*n;
        X = 2*m;
        Y = 2*n-1;
        Z = 2*m-1;
        if n < size(a,1) && m < size(a,2)
        b(W,X) = (a(n,m)+a(n+1,m)+a(n,m+1)+a(n+1,m+1))/4;
        end
        b(Y,Z) = a(n,m);
    end
end

%Preallocate 'c'
c = b;

%Calc 'c'
for n = 1:size(a,1)
    for m = 1:size(a,2)
        W = 2*n;
        X = 2*m;
        Q = 2*n-2;
        R = 2*m-2;
        Y = 2*n-1;
        Z = 2*m-1;
        O = 2*n+1;
        P = 2*m+1;
        if n == 1 && m == 1
            c(W,Z) = (b(Y,Z)+b(Y+2,Z)+b(W,X))/3;
            c(Y,X) = (b(Y,Z)+b(Y,Z+2)+b(W,X))/3;
        elseif m == 1 && n < size(a,1)
            c(W,Z) = (b(Y,Z)+b(Y+2,Z)+b(W,X))/3;
            c(Y,X) = (b(Y,Z)+b(Y,Z+2)+b(W,X)+b(Q,X))/4;
        elseif n == 1 && m < size(a,2)
            c(Y,X) = (b(Y,Z)+b(Y,Z+2)+b(W,X))/3;
            c(W,Z) = (b(Y,Z)+b(Y+2,Z)+b(W,X)+b(W,R))/4;
        elseif n > 1 && n < size(a,1) && m > 1 && m < size(a,2)
            c(W,Z) = (b(Y,Z)+b(Y+2,Z)+b(W,X)+b(W,R))/4;
            c(Y,X) = (b(Y,Z)+b(Y,Z+2)+b(W,X)+b(Q,X))/4;
        elseif m == size(a,2) && n < size(a,1)
            c(W,Z) = (b(Y,Z)+b(W,R)+b(O,Z))/3;
        elseif n == size(a,1) && m < size(a,2)
            c(Y,X) = (b(Y,Z)+b(Q,X)+b(Y,P))/3;
        end
    end
end