Is it possible to find corresponding row from other matrix's row??

y = [ 90    90   -45     0     0    45    45     0   -45    15    15;
  90    90   -45     0     0    45    45     0   -45    15    15;
  90    90   -45     0     0    45    45     0   -45    15    0;
  90    90   -45     0     0    45    45     0   -45    15    0;
  90    90   -45     0     0    45    45     0   -45    15    15;
  90    90   -45     0     0    45    45     0   -45    15    15;];

    y0 = sort(y,2);
  y = y(sum([ones(size(y,1),1),diff(y0,[],2)~=0],2) >= numel(unique(y)),:);
    y = permute(y ,[3 2 1]);

 for  k = 1:size(y,3)

    [x0(:,:,k), x00(:,:,k)] = hist(y(:,:,k), unique(y(:,:,k)));

 end
 x0 =permute(x0,[3 2 1]);  x00 =permute(x00,[3 2 1]);
 x0(any(x0<2,2),:) = [];

I have to find corresponding row of "x00"

[Matthew Eicholtz]

Your question is a little confusing. Can you try rewording for clarity?

Where do you actually need help? In the last part?

[Stephen23]

Continuation of this question:

http://www.mathworks.com/matlabcentral/answers/272233-how-can-i-find-deleted-rows-from-matrix

Answer 1

[Stephen23]

It is easy, you just need to assign the logical conditions to an index variable. So instead of this:

x0(any(x0<2,2),:) = [];

you should allocate that logical condition to a variable:

idx = any(x0<2,2);

and then you can use idx for any of the matrices:

>> xdel = x0(idx,:)
xdel =
   2   4   1   2   2
   2   4   1   2   2
>> xnew = x0(~idx,:)
xnew =
   2   3   2   2   2
   2   3   2   2   2
   2   3   2   2   2
   2   3   2   2   2

and now you can try x00(idx,:) and x00(~idx,:) yourself!

[Triveni]

Ya right...Thank you. ...

[Triveni]

    y = [ 90    90   -45     0     0    45    45     0   -45    15    15;
    90    90   -45     0     0    45    45     0   -45    15    15;
    90    90   -45     0     0    45    45     0   -45    15    0;
    90    90   -45     0     0    45    45     0   -45    15    0;
    90    90   -45     0     0    45    45     0   -45    15    15;
    90    90   -45     0     0    45    45     0   -45    15    15;];

    y0 = sort(y,2);
  y = y(sum([ones(size(y,1),1),diff(y0,[],2)~=0],2) >= numel(unique(y)),:);
    y = permute(y ,[3 2 1]);

 for  k = 1:size(y,3)

    [x0(:,:,k), x00(:,:,k)] = hist(y(:,:,k), unique(y(:,:,k)));

 end
 x0 =permute(x0,[3 2 1]);  x00 =permute(x00,[3 2 1]);
  y = permute(y ,[3 2 1]);
 idx = any(x0<2,2);
 xdel = x0(idx,:)
 xnew = x0(~idx,:)
 y = y(~idx,:)

Answer 2

[Ced]

I am not sure I understood the full question, but in short, you want to delete rows in which there is an element < 2 ? You basically already answered the question. Since you like one liners, I think this should do the trick:

x(any(x<2,2),:) = [];

This finds all rows (DIM 2) in which there is ANY element smaller than 2, selects these rows, and deletes them.

If you want to match your centers, you can do:

rows_del = any(x<2,2);
x(rows_del,:) = [];
z(rows_del,:) = [];