# How to look up a smaller array in a larger array while preserving shape

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Can Ozger on 26 Nov 2022
Commented: Image Analyst on 27 Nov 2022
I have a logical array. I want to look for a 2x2 "square" of 1's in this array, and return whether this square is present in this array or not.
LargeArray= [0,0,0,0;1,0,0,0;1,0,0,0;1,1,0,0;1,1,0,0;1,1,0,0;0,0,0,0;0,0,0,0]
LargeArray = 8×4
0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0
FindArray= [1,1;1,1]
FindArray = 2×2
1 1 1 1
When I use ismember, I get the Large array as the answer, but I am looking to get the answer of whether the smaller FindArray is present in the LargeArray or not in True or False so I can use it to tag my data. Thanks!

Image Analyst on 26 Nov 2022
Edited: Image Analyst on 26 Nov 2022
A simple brute force for loop will do it:
LargeArray= [0,0,0,0;1,0,0,0;1,0,0,0;1,1,0,0;1,1,0,0;1,1,0,0;0,0,0,0;0,0,0,0]
LargeArray = 8×4
0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0
FindArray= [1,1;1,1]
FindArray = 2×2
1 1 1 1
[rL, cL] = size(LargeArray);
[rt, ct] = size(FindArray);
foundIt = false;
for col = 1 : cL-ct
for row = 1 : rL-rt
subArray = LargeArray(row:row+rt-1, col:col+ct-1);
if isequal(subArray, FindArray)
foundIt = true;
fprintf('Found it at row %d, column %d.\n', row, col);
end
end
end
Found it at row 4, column 1. Found it at row 5, column 1.
Image Analyst on 27 Nov 2022
OK, I just thought that when you said, in the comment to your original post, "On case the pattern array contains only 1s," and gave simplified code, that the original post would handle any numbers. But anyway, thanks for giving generalized solution that works for any numbers. 🙂

Bruno Luong on 26 Nov 2022
Edited: Bruno Luong on 26 Nov 2022
Use convolution to detect matching
% I modified it to make example more interesting
LargeArray= [0,0,0,0;1,0,0,0;1,0,0,0;1,1,0,0;1,1,1,0;0,1,1,0;0,0,0,0;0,0,0,0]
LargeArray = 8×4
0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0
FindArray= [1,1;1,1]
FindArray = 2×2
1 1 1 1
shiftfun = @(B) 2*B-1; % transform 0/1 respectively to -1/1
c = conv2(shiftfun(LargeArray),rot90(shiftfun(FindArray),2),'valid');
% Match (upper-left) indexes in LargeArray
[row,col] = find(c==numel(FindArray));
MatchIndex = table(row,col)
MatchIndex = 2×2 table
row col ___ ___ 4 1 5 2
Bruno Luong on 26 Nov 2022
On case the pattern array contains only 1s, the code can be simplified in single-line
% I modified it to make example more interesting
LargeArray= [0,0,0,0;1,0,0,0;1,0,0,0;1,1,0,0;1,1,1,0;0,1,1,0;0,0,0,0;0,0,0,0]
LargeArray = 8×4
0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0
FindArray= ones(2,2)
FindArray = 2×2
1 1 1 1
% Match (upper-left) indexes in LargeArray
[row,col] = find(conv2(LargeArray,FindArray,'valid')==numel(FindArray));
MatchIndex = table(row,col)
MatchIndex = 2×2 table
row col ___ ___ 4 1 5 2

DGM on 26 Nov 2022
Edited: DGM on 26 Nov 2022
I'm going to demonstrate a couple ways you can do this using neighborhood operations. The case of a solid 2x2 nhood is a bit of a simplified case, but these can be extended to more general binary pattern matching. For these examples, I'm going to shamelessly steal Bruno's improved test array.
If you have IPT, you can use bwlookup().
LargeArray = [0,0,0,0; 1,0,0,0; 1,0,0,0; 1,1,0,0; 1,1,1,0; 0,1,1,0; 0,0,0,0; 0,0,0,0];
nhood = [1 1; 1 1]; % any 2x2 neighborhood
f = @(x) isequal(x,nhood); % function that describes matching behavior
lut = makelut(f,2); % create LUT for a 2x2 nhood
mk = bwlookup(LargeArray,lut) % logical map of matches
mk = 8×4
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[r c] = find(mk) % convert logical mask to row,col subscripts
r = 2×1
4 5
c = 2×1
1 2
Alternatively, you can use basic linear filters.
LargeArray = [0,0,0,0; 1,0,0,0; 1,0,0,0; 1,1,0,0; 1,1,1,0; 0,1,1,0; 0,0,0,0; 0,0,0,0];
nhood = [1 1; 1 1]; % any 2x2 neighborhood pattern
seb = 2.^([1 3; 2 4]-1); % index weighting array
mk = imfilter(double(LargeArray),seb) == sum(sum(seb.*nhood)) % logical map of matches
mk = 8×4 logical array
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[r c] = find(mk) % convert logical mask to row,col subscripts
r = 2×1
4 5
c = 2×1
1 2

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