使用Matlab查找矩阵中相等元素的位置

Question

Suppose I have:

m = [1,2,3;1,4,5;6,4,7]

I want to get a list containing the positions of the elements in the matrix m so that the positions of equal elements are grouped together. The output for matrix m must be:

{{1,1},{2,1}},{{2,2},{3,2}},{1,2},{1,3},{2,3},{3,1},{3,3}
%     1             2         3     4     5     6     7

We can see here that the positions for the elements that are all equal to each other are grouped together.

Answer 1

The simplest way would be to loop through every unique value and determine the row and column positions that match each value. Something like this could work:

val = unique(m);
pos = cell(1, numel(val));
for ii = 1 : numel(val)
    [r,c] = find(m == val(ii));
    pos{ii} = [r,c];
end

pos would be a cell array containing all of the positions for each unique value. We can show what these positions are by:

>> format compact; celldisp(pos)
pos{1} =
     1     1
     2     1
pos{2} =
     1     2
pos{3} =
     1     3
pos{4} =
     2     2
     3     2
pos{5} =
     2     3
pos{6} =
     3     1
pos{7} =
     3     3

This of course is not meaningful unless you specifically show each unique value per group of positions. Therefore, we can try something like this instead where we can loop through each element in the cell array as well as display the corresponding element that each set of positions belongs to:

 for ii = 1 : numel(val)
     fprintf('Value: %f\n', val(ii));
     fprintf('Positions:\n');
     disp(pos{ii});
 end

What I get is now:

Value: 1.000000
Positions:
     1     1
     2     1
Value: 2.000000
Positions:
     1     2
Value: 3.000000
Positions:
     1     3
Value: 4.000000
Positions:
     2     2
     3     2
Value: 5.000000
Positions:
     2     3
Value: 6.000000
Positions:
     3     1
Value: 7.000000
Positions:
     3     3

Answer 2

This gives you what you want, except for the fact that indices of unique elements are also wrapped in cell twice, just like the indices of repeating elements:

m = [1,2,3;1,4,5;6,4,7];

[~, idx] = ismember(m(:), unique(m(:)));
linInd = 1:numel(m);
[i,j] = ind2sub(size(m), linInd);
res = accumarray(idx, linInd, [], @(x) {num2cell([i(x);j(x)]',2)});

Result:

>> celldisp(res)

res{1}{1} =
     2     1
res{1}{2} =
     1     1
res{2}{1} =
     1     2
res{3}{1} =
     1     3
res{4}{1} =
     2     2
res{4}{2} =
     3     2
res{5}{1} =
     2     3
res{6}{1} =
     3     1
res{7}{1} =
     3     3