identify a matrix within a matrix

I'm pretty new to this. What I am wondering is how to test a few different matrices within a larger one.

this would be an example of what I want to do:

    A = 9 7 6 5 4    A = [9 7 6 5 4;6 7 5 4 2;7 7 5 5 5;7 6 4 4 3]
        6 7 5 4 2
        7 7 5 5 5 
        7 6 4 4 3

    B = 7 7          B = [7 7;7]
        7 

    C = 7            C = [7;7;7]
        7 
        7

    D = 5 5 5        D = [5 5 5]

and then show me where in the original one they are, replacing the non-matches with 0's.

Cheers.

Any direction as to what functions I need to look into would be greatly appreciated

Answer 1

[Andrei Bobrov]

Bad variant

A = [9 7 6 5 4;6 7 5 4 2;7 7 5 5 5;7 6 4 4 3]

B = [7 7;7 0];
p = abs(filter2(B,A) - norm(B(:))^2) < eps(1e4);
out1 = imdilate(p,B>0).*A;

C = [7;7;7];
p2 = abs(filter2(C,A) - norm(C(:))^2) < eps(1e4);
out2 = imdilate(p2,C>0).*A;

D = [5 5 5];
p3 = abs(filter2(D,A) - norm(D(:))^2) < eps(1e4);
out2 = imdilate(p3,D>0).*A;

other variant:

use m-file findarray.m:

function [idx,arrfnd] = findarray(A,B)
    [m,n] = size(A);
    Ai = reshape(1:n*m,[m,n]);
    [mb,nb] = size(B);
    B = B(:);
    t = ~isnan(B);
    pb = bsxfun(@plus,(0:mb-1)',(0:nb-1)*m);
    pb = pb(t);
    Av = reshape(Ai(1:m-mb+1,1:n-nb+1),1,[]);
    i0 = bsxfun(@plus,Av,pb(:));
    ii = all(bsxfun(@eq,A(i0),B(t)));
    idx = i0(:,ii);
    arrfnd = zeros(m,n);
    arrfnd(idx) = A(idx); 
end

example of use

>> A
A =

             9           7           6           5           4
             6           7           5           4           2
             7           7           5           5           5
             7           6           4           4           3

>> B1
B1 =

             7           7
             7   NaN

>> [idx,arrfnd] = findarray(A,B1)
idx =

             3
             4
             7

arrfnd =

             0           0           0           0           0
             0           0           0           0           0
             7           7           0           0           0
             7           0           0           0           0

>> C
C =

             7
             7
             7

>> [idx,arrfnd] = findarray(A,C)
idx =

             5
             6
             7

arrfnd =

             0           7           0           0           0
             0           7           0           0           0
             0           7           0           0           0
             0           0           0           0           0

>> D
D =

             5           5           5

>> [idx,arrfnd] = findarray(A,D)
idx =

            11
            15
            19

arrfnd =

             0           0           0           0           0
             0           0           0           0           0
             0           0           5           5           5
             0           0           0           0           0

>>|

[byron goodship]

This worked great thanks, I changed the matrix A to incorporate some more of the patterns and it showed them as well. I am new to this software and am wondering if you could explain how you got to this code, if you have time? I am curious as to how you put it together. I have been pulling it apart to find out how it works, but still a bit puzzled. Cheers, Byron.

[Andrei Bobrov]

Hi! Byron! What part of the code is not clear?

[byron goodship]

p = abs(filter2(B,A) - norm(B(:))^2) < eps(1e4);
out1 = imdilate(p,B>0).*A;

As previously stated, I am fairly new, I've followed through some an introduction course on Coursera.org but I'm still left without a great knowledge. Just wondering if you could break it down, again if you've got the time. Cheers.

[Stephen23]

The single quote is the complex conjugate transpose operator:

https://www.mathworks.com/help/matlab/ref/ctranspose.html

The function bsxfun is used here to add the row (0:nb-1)*m to the column (0:mb-1)', creating a matrix. You can try it on some simple example to see what it is doing:

 >> bsxfun(@plus,(0:3)',0:5)
 ans =
   0   1   2   3   4   5
   1   2   3   4   5   6
   2   3   4   5   6   7
   3   4   5   6   7   8

[Satuk Bugrahan]

Thank you for this clear answer, though I just deleted my message because understand the deal myself howewer I am still a little bit confused about how did we add two matrix with different sizes together with 'bsxfun' . When I look at the input matrices still I cant estimate the output . At any case , thank you for the answer ...

[Andrei Bobrov]

Hi Byron! Coursera.org is very good but I would started with MATLAB Documentation.

Please first read about functions used in these expressions:

filter2,

norm,

imdilate from the Image Processing Toolbox.

[Arne T]

Hi Andrei!

Your code works perfectly, but I want to use it in a 3D Matrix. Unfortunately your Code only works in 2D cause bsxfun respectivly the plus operator only works in 2D with this result. Do you know a way to expand this function that it would work in 3D.

Cause Im writing a recursiv function with up to 300Mio iteratons the runtime is very important.

Thanks!

[Arne T]

I solved the problem with the following code. This works in 3D Matrix.

[m,n,o] = size(A);
Ai = reshape(1:n*m*o,[m,n,o]);
[mb,nb,ob] = size(B);
B = B(:);
t = ~isnan(B);
pb = (0:mb-1)'+(0:nb-1)*m+reshape([0:ob-1],1,1,ob)*n*m;
pb = pb(t);
Av = reshape(Ai(1:m-mb+1,1:n-nb+1,1:o-ob+1),1,[]);
i0 = Av+pb(:);
ii = all(bsxfun(@eq,A(i0),B(t)));
idx = i0(:,ii);

Answer 2

[Matt J]

This looks applicable,

http://www.mathworks.com/matlabcentral/fileexchange/33340-wncc-weighted-normalized-cross-correlation

Answer 3

[KSSV]

Note that B = [7 7;7] is not a matrix.

You can find your required numbers from matrix A using find or ==.

Eg:

idx = A==7 ;
A(idx)

idx = find(A==5)
A(idx)