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How to find a vector in a matrix?

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Raymond
Raymond 2013-1-16
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!
  1 个评论
Shashank Prasanna
Shashank Prasanna 2013-1-16
Just to clarify, you don't want to do this using a for loop? that would be fairly straightforward

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采纳的回答

Matt J
Matt J 2013-1-16
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

更多回答（5 个）

Andrei Bobrov
Andrei Bobrov 2013-1-17
编辑：Andrei Bobrov 2013-1-17
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
José-Luis
José-Luis 2013-1-16
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)';
Matt J
Matt J 2013-1-16
编辑：Matt J 2013-1-16
N=sqrt( conv2(A.^2,ones(size(B)), 'valid' ));
[i,j]=find(abs( conv2(A,fliplr(B)/norm(B),'valid')./N - 1)<=1e-6)
Roger Stafford
Roger Stafford 2013-1-17
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
Raymond
Raymond 2013-1-17
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.
  2 个评论
Sean de Wolski
Sean de Wolski 2013-1-17
@Raymond, The for-loops will smoke ind2sub() any day of the week for large arrays. ind2sub() is convenient not fast.
Matt J
Matt J 2013-1-17
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.

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