Rearrange elements of matrix based on an index matrix

조회 수: 2 (최근 30일)
Hossein Kazemi
Hossein Kazemi 2024년 8월 27일
댓글: Hossein Kazemi 2024년 8월 27일
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>

채택된 답변

Stephen23
Stephen23 2024년 8월 27일
편집: Stephen23 2024년 8월 27일
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>
  댓글 수: 1
Hossein Kazemi
Hossein Kazemi 2024년 8월 27일
Thanks. The first solution appears to be the fastest (my matrix is 3400x67).

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Operating on Diagonal Matrices에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by