How to compare rows of an array with other rows?

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.

please help.

[lucksBi]

Also comparison is quite large. no of rows are 1508 in 'array' and nearly 900 in 'x'

[Jos (10584)]

Just to be sure. X is a numerical column vector and array is a cell array (row vector) where each cell holds a numerical column vector?

[Jan]

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)]

and you do realise that you only have to process the unique values in x?

Answer 1

[Jan]

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]

Thanks alot @Jan Simon. I am using this solution. It is accurate on these arrays. I applied it on the larger data set and it is taking too much time. Almost 20 min now and still its running. Does it is something related to processor speed or because of loops?

[Jan]

How large is the cell array and how large are the vectors in average? Did you sort the vectors initially? Do the vectors contain repeated elements?

intersect has a lot of overhead. If the vectors are small (I did not run tests to check how large this is), this may be faster:

for k = 1:numel(array)
  array{k} = sort(array{k});
end

n = size(array, 1);
v = 1:n;
result = cell(n-1, numel(x));
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
      result{j,i} = ShortIntersect(ai, array{temp(j)});
  end
end

function AB = ShortIntersect(A, B)
% INPUT: A, B: Double vectors without NaN values
%              B is sorted already!

% AI = find(ismembc(A(:), B(:)));  % On older Matlab versions
AI = find(builtin('_ismemberhelper', A(:), B(:));
if ~isempty(AI)
   AB = A(AI);
else
   AB = 0;
end
end

Or perhaps for small vectors this is faster:

function AB = ShortIntersect(A, B)
AI = any(bsxfun(@eq, A(:), B(:).', 1);
if ~isempty(AI)
   AB = A(AI);
else
   AB = 0;
end
end

[lucksBi]

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]

Problem solved. thanks alot for your time.

Answer 2

[Birdman]

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]

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]

Thank You so much for your time. this is really helpful.