# How to compare rows of an array with other rows?

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lucksBi on 19 Jan 2018
Commented: lucksBi on 24 Jan 2018
Hey.. How to compare rows of an array with rows defined in x?
array={[1,2,3,4,6,13,18,9];[13];[12,1,2,5];[3,4];[1,5,6]}
x=[1;3]
comparison will be based on x. e.g. first element in x is 1 so it will compare 1st row of array with all other rows. 2nd element is 3 so 3rd row will be compared will all other rows. Comparison will get intersection of elements like:
result{1,1}= {[13;[1,2];[3,4];[1,6]]}
result{2,1}= {[[1,2];0;0;[1,6];[1,5]]}
0 means no common element was found.
Jan on 19 Jan 2018
Edited: Jan on 19 Jan 2018
The question is not clear. The elements of the cell array called "array" have one row only. Do you mean, that their first columns are compared? Or do you mean the 1st and 3rd element of the cell array?
Note that there is a missing ] in
result{1,1}= {[13;[1,2];[3,4];[1,6]]}
and this is not valid also:
result{2,1}= {[[1,2];0;0;[1,6];[1,5]]}
because you cannot concatenate vectors of different width vertically.
Jos (10584) on 19 Jan 2018
and you do realise that you only have to process the unique values in x?

Jan on 19 Jan 2018
Edited: Jan on 19 Jan 2018
The question is not clear, but I guess a code which produces the output:
array = {[1,2,3,4,6,13,18,9];[13];[12,1,2,5];[3,4];[1,5,6]}
x = [1;3];
Result = cell(1, numel(x));
n = numel(array);
for ix = 1:numel(x)
index = x(ix);
A = array{index};
B = cell(1, n - 1);
iB = 0;
for ia = 1:n
if ia ~= index
iB = iB + 1;
Inter = intersect(A, array{ia});
if isempty(Inter)
Inter = 0;
end
B{iB} = Inter;
end
end
Result{ix} = B;
end
It might help to save time, if you sort the arrays at first:
for k = 1:numel(array)
array{k} = sort(array{k});
end
lucksBi on 24 Jan 2018
Edited: lucksBi on 24 Jan 2018
ThankYou so much. Working good on smaller matrix but It is showing following error when i run on actual dataset. Can you please help on his i have tried almost everything i know to resolve this but still not successful:
Index exceeds matrix dimensions.
Error in test2 (line 54)
ai = unique(array{x(i)})
lucksBi on 24 Jan 2018
Problem solved. thanks alot for your time.

Birdman on 19 Jan 2018
Edited: Birdman on 19 Jan 2018
Another approach for your problem, which is shorter:
for i=1:numel(x)
temp=setdiff(1:size(array,1),x(i));
for j=1:numel(temp)
result{j,i}=intersect(array{x(i),1},array{temp(j),1});
end
end
Jan on 19 Jan 2018
Edited: Jan on 19 Jan 2018
Yes, it is shorter. But it omits the pre-allocation of result and using 0 as output, if no intersections are found. It is slower, because array{x(i),1} is addressed in each iteration.
Another idea, which avoids setdiff, which is an overkill for a scalar input:
n = size(array, 1);
v = 1:n;
result = cell(n-1, numel(x));
result(:) = {0};
for i = 1:numel(x)
temp = v(v ~= i); % Short form of: setdiff(v, i)
ai = array{x(i)}; % Or maybe: ai = unique(array{x(i)})
for j = 1:n-1
r = intersect(ai, array{temp(j)});
if ~isempty(r)
result{j,i} = r;
end
end
end
This might be an efficient mix of our solutions.
lucksBi on 20 Jan 2018