# How to permute elements of jth column of a matrix iteratively

조회 수: 3(최근 30일)
Darren 2022년 9월 25일
답변: David Hill 2022년 9월 25일
I am trying to write a function which performs some calculations on a data set A. I want the function to return d (number of dimensions of A) matrices like A but with the jth column elements permuted:
A=[1,2,3 ; 7,8,9 ; 13,14,15]
perms_of_(A)
function perms = perms_of_(A)
[n,d]=size(A);
for i = 1:d
A(:,i) = A(randperm(n),i)
end
end
I want matrices like:
A=[7,2,3 ; 1,8,9 ; 13,14,15]
A=[1,14,3 ; 7,2,9 ; 13,8,15]
A=[1,2,9 ; 7,8,3 ; 13,14,15]
A=[7,2,3 ; 1,8,9 ; 13,14,15]
A=[7,14,3 ; 1,2,9 ; 13,8,15]
A=[7,14,15 ; 1,2,9 ; 13,8,3]
In other words, I want matrices exactly like the ORIGINAL matrix A but with JUST the jth column permuted. Somehow at the beginning of each iteration I need the matrix A to be reset to the original matrix defined outside the function.

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### 채택된 답변

David Hill 2022년 9월 25일
A=[1,2,3 ; 7,8,9 ; 13,14,15];
perms_of_(A,2)
ans =
ans(:,:,1) = 1 14 3 7 8 9 13 2 15 ans(:,:,2) = 1 14 3 7 2 9 13 8 15 ans(:,:,3) = 1 8 3 7 14 9 13 2 15 ans(:,:,4) = 1 8 3 7 2 9 13 14 15 ans(:,:,5) = 1 2 3 7 14 9 13 8 15 ans(:,:,6) = 1 2 3 7 8 9 13 14 15
function A_perms = perms_of_(A,j)
p=perms(A(:,j));
for i = 1:length(p)
A_perms(:,:,i)=A;
A_perms(:,j,i) = p(i,:)';
end
end

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### 추가 답변(1개)

Walter Roberson 2022년 9월 25일
A=[1,2,3 ; 7,8,9 ; 13,14,15]
A = 3×3
1 2 3 7 8 9 13 14 15
perms_of_(A)
ans =
ans(:,:,1) = 1 2 3 7 8 9 13 14 15 ans(:,:,2) = 1 14 3 7 2 9 13 8 15 ans(:,:,3) = 1 2 15 7 8 3 13 14 9
function perms = perms_of_(A)
[rows, cols] = size(A);
perms = repmat(A, 1, 1, cols);
for i = 1 : cols
perms(:,i,i) = A(randperm(rows),i);
end
end
The result is a 3D matrix with as many planes as there are columns. In the K'th plane, the K'th column is the one permuted.

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