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!
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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
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
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)';
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
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.
@Raymond, The for-loops will smoke ind2sub() any day of the week for large arrays. ind2sub() is convenient not fast.
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.
1 Comment
Direct link to this comment:
http://mathworks.com/matlabcentral/answers/59025#comment_123034
Just to clarify, you don't want to do this using a for loop? that would be fairly straightforward