Rearrange elements of matrix based on an index matrix

2 visualizaciones (últimos 30 días)
I have a 5x3 matrix and I want to rearrange each row according to the correponding row of a 5x3 index matrix
x=randn(5,3)
x = 5x3
-0.2616 0.3522 -0.4451 -1.1699 0.8921 1.7676 1.4354 1.1977 -0.3197 -1.2002 0.9392 0.7452 -0.8374 -1.3827 0.2679
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
z=randn(5,3)
z = 5x3
-0.1687 0.4662 1.3644 0.1290 -3.2728 -0.8944 -0.7820 0.2506 -1.0295 0.0976 -0.0797 -0.2671 0.2607 -0.0966 -0.1679
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~,I]=sort(x,2)
I = 5x3
3 1 2 1 2 3 3 2 1 1 3 2 2 1 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Now I want to sort rows of z using the index matrix I. But using the following does not work. For example, I want the first row of zz to be sorted according to the first row of x, which should result in zz(1,:)= [1.3644, -0.1687, 0.4662].
zz=z(I)
zz = 5x3
-0.7820 -0.1687 0.1290 -0.1687 0.1290 -0.7820 -0.7820 0.1290 -0.1687 -0.1687 -0.7820 0.1290 0.1290 -0.1687 -0.7820
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Respuesta aceptada

Stephen23
Stephen23 el 27 de Ag. de 2024
Editada: Stephen23 el 27 de Ag. de 2024
Yes, it is awkward.
x=randn(5,3)
x = 5x3
0.9687 -0.7929 2.0516 -0.1236 0.2917 -0.9487 -1.4717 0.0275 1.1722 0.2800 1.5357 0.2596 -0.9693 -0.1854 0.3145
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
z=randn(5,3)
z = 5x3
-1.6696 0.6182 -1.4586 0.1528 1.6729 1.4683 0.3576 -1.0089 0.8616 2.1993 0.6213 -1.2327 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~,I] = sort(x,2)
I = 5x3
2 1 3 3 1 2 1 2 3 3 1 2 1 2 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Perhaps
S = size(I);
[R,~] = ndgrid(1:S(1),1:S(2));
J = sub2ind(S,R,I);
zz = z(J)
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Or
zz = z;
for k = 1:size(I,1)
zz(k,:) = zz(k,I(k,:));
end
zz
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Or
zz = cell2mat(cellfun(@(v,x)v(x),num2cell(z,2),num2cell(I,2),'uni',0))
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
  1 comentario
Hossein Kazemi
Hossein Kazemi el 27 de Ag. de 2024
Thanks. The first solution appears to be the fastest (my matrix is 3400x67).

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Operating on Diagonal Matrices en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by