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

1 visualización (últimos 30 días)
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"
  2 comentarios
Matthew Eicholtz
Matthew Eicholtz el 8 de Mzo. de 2016
Your question is a little confusing. Can you try rewording for clarity?
Where do you actually need help? In the last part?

Iniciar sesión para comentar.

Respuesta aceptada

Stephen23
Stephen23 el 9 de Mzo. de 2016
Editada: Stephen23 el 9 de Mzo. de 2016
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!
  2 comentarios
Triveni
Triveni el 9 de Mzo. de 2016
Ya right...Thank you. ...
Triveni
Triveni el 9 de Mzo. de 2016
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,:)

Iniciar sesión para comentar.

Más respuestas (1)

Ced
Ced el 8 de Mzo. de 2016
Editada: Ced el 8 de Mzo. de 2016
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,:) = [];

Categorías

Más información sobre Logical en Help Center y File Exchange.

Etiquetas

Productos

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by