How to find a vector in a matrix?

Hi all, I just wonder whether there are Matlab functions which can deal with the problem. Say,

A=[1 1 1 1; ...
   1 1 2 3; ...
   1 2 3 1; ...
   1 1 1 2];
B=[1 2 3]

So B appears twice in A.

I tried ismember, but it finds all the elements of B in A, regardless of the sequence.

Thanks!

[Shashank Prasanna]

Just to clarify, you don't want to do this using a for loop? that would be fairly straightforward

Answer 1

[Matt J]

    A=[1 1 1 1; 1 1 2 3; 1 2 3 1; 1 1 1 2];
    B=[1 2 3]; 

    [j,i]=ind2sub(fliplr(size(A)),  strfind(reshape(A.',1,[]),B).' );

    >> [i,j]

    ans =

         2     2
         3     1

Answer 2

[Andrei Bobrov]

s = size(B) - 1;
n = size(A) - s;
out = [];
for ii = 1:n(1)
    i1 = ii : ii + s(1);
    for jj = 1:n(2)
        if isequal(A(i1,jj : jj + s(2)),B)
            out = [out;[ii,jj]];
        end
    end
end

Answer 3

[José-Luis]

A=[1 1 1 1; 1 1 2 3; 1 2 3 1; 1 1 1 2];
B=[1 2 3]; 

isSubStr = @(x,y) ~isempty(strfind(x,y));
getStr = @(x) sprintf('%d',x);
numVals = size(A,2);

logical_index = arrayfun(@(idx) isSubStr(getStr(A(idx,:)),getStr(B)),1:numVals)';

Answer 4

[Matt J]

 N=sqrt( conv2(A.^2,ones(size(B)), 'valid' ));

[i,j]=find(abs( conv2(A,fliplr(B)/norm(B),'valid')./N - 1)<=1e-6)

Answer 5

[Roger Stafford]

Another way:

 [m,n] = size(A);
 k = length(B(:));
 p = hankel(1:k,k:n);
 f = find(all(bsxfun(@eq,reshape(A(:,p).',k,n-k+1,m),B(:)),1));
 [c,r] = ind2sub([n-k+1,m],f);

where c gives the column indices and r the row indices in A where a match with B begins. In your example these would be

 r   c
 -   -
 2   2
 3   1

Answer 6

[Raymond]

Thanks all! All the solutions work, but Matt's looks simpler and more efficient.

I want to avoid for loops since I have to deal with big matrix.

[Sean de Wolski]

@Raymond, The for-loops will smoke ind2sub() any day of the week for large arrays. ind2sub() is convenient not fast.

[Matt J]

Don't know what ind2sub has to do with anything, but the for-loop approach is definitely not faster:

    A=[1 1 1 1; 1 1 2 3; 1 2 3 1; 1 1 1 2];
    B=[1 2 3];
    A(10000,1)=0;

    tic;
        [j,i]=ind2sub(fliplr(size(A)),  strfind(reshape(A.',1,[]),B).' );
    toc
    %Elapsed time is 0.001019 seconds.

####################

     tic
    s = size(B) - 1;
    n = size(A) - s;
    out = [];
    for ii = 1:n(1)
        i1 = ii : ii + s(1);
        for jj = 1:n(2)
            if isequal(A(i1,jj : jj + s(2)),B)
                out = [out;[ii,jj]];
            end
        end
    end
    toc;
    %Elapsed time is 0.039563 seconds.