How to compare rows of an array with other rows?

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lucksBi 2018-1-19

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评论： lucksBi 2018-1-24

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.

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Jan 2018-1-19

编辑：Jan 2018-1-19

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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) 2018-1-19

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

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Answer 1

Jan 2018-1-19

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https://ww2.mathworks.cn/matlabcentral/answers/377900-how-to-compare-rows-of-an-array-with-other-rows#answer_300830

编辑：Jan 2018-1-19

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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

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Jan 2018-1-22

编辑：Jan 2018-1-22

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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 2018-1-24

编辑：lucksBi 2018-1-24

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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 2018-1-24

Problem solved. thanks alot for your time.

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Answer 2

Birdman 2018-1-19

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编辑：Birdman 2018-1-19

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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

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Jan 2018-1-19

编辑：Jan 2018-1-19

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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