如何为 MATLAB 矩阵的对角线赋值？

发布于 2024-09-28 04:35:40 字数 492 浏览 7 评论 0原文

假设我有一个 NxN 矩阵 A、一个由数字 1:N 的子集组成的索引向量 V 和一个值 K，我想这样做：

 for i = V
     A(i,i) = K
 end

有没有一种方法可以在一个带有向量化的语句中做到这一点？

例如 A(something) = K

语句 A(V,V) = K 不起作用，它分配非对角线元素，这不是我想要的。例如：

>> A = zeros(5);
>> V = [1 3 4];
>> A(V,V) = 1

A =

 1     0     1     1     0
 0     0     0     0     0
 1     0     1     1     0
 1     0     1     1     0
 0     0     0     0     0

原文

Suppose I have an NxN matrix A, an index vector V consisting of a subset of the numbers 1:N, and a value K, and I want to do this:

 for i = V
     A(i,i) = K
 end

Is there a way to do this in one statement w/ vectorization?

e.g. A(something) = K

The statement A(V,V) = K will not work, it assigns off-diagonal elements, and this is not what I want. e.g.:

>> A = zeros(5);
>> V = [1 3 4];
>> A(V,V) = 1

A =

 1     0     1     1     0
 0     0     0     0     0
 1     0     1     1     0
 1     0     1     1     0
 0     0     0     0     0

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咿呀咿呀哟 2024-10-05 04:35:40

我通常使用 EYE ：

A = magic(4)
A(logical(eye(size(A)))) = 99

A =
    99     2     3    13
     5    99    10     8
     9     7    99    12
     4    14    15    99

或者，您可以只创建列表线性索引，因为从一个对角元素到下一个，需要 nRows+1 步：

[nRows,nCols] = size(A);
A(1:(nRows+1):nRows*nCols) = 101
A =
   101     2     3    13
     5   101    10     8
     9     7   101    12
     4    14    15   101

如果您只想访问对角元素的子集，则需要创建一个对角索引列表：

subsetIdx = [1 3];
diagonalIdx = (subsetIdx-1) * (nRows + 1) + 1;
A(diagonalIdx) = 203
A =
   203     2     3    13
     5   101    10     8
     9     7   203    12
     4    14    15   101

或者，您可以使用 diag 创建逻辑索引数组（仅适用于方形数组）

diagonalIdx = false(nRows,1);
diagonalIdx(subsetIdx) = true;
A(diag(diagonalIdx)) = -1
A =
    -1     2     3    13
     5   101    10     8
     9     7    -1    12
     4    14    15   101

I usually use EYE for that:

A = magic(4)
A(logical(eye(size(A)))) = 99

A =
    99     2     3    13
     5    99    10     8
     9     7    99    12
     4    14    15    99

Alternatively, you can just create the list of linear indices, since from one diagonal element to the next, it takes nRows+1 steps:

[nRows,nCols] = size(A);
A(1:(nRows+1):nRows*nCols) = 101
A =
   101     2     3    13
     5   101    10     8
     9     7   101    12
     4    14    15   101

If you only want to access a subset of diagonal elements, you need to create a list of diagonal indices:

subsetIdx = [1 3];
diagonalIdx = (subsetIdx-1) * (nRows + 1) + 1;
A(diagonalIdx) = 203
A =
   203     2     3    13
     5   101    10     8
     9     7   203    12
     4    14    15   101

Alternatively, you can create a logical index array using diag (works only for square arrays)

diagonalIdx = false(nRows,1);
diagonalIdx(subsetIdx) = true;
A(diag(diagonalIdx)) = -1
A =
    -1     2     3    13
     5   101    10     8
     9     7    -1    12
     4    14    15   101

回复收藏 0 原文

ㄟ。诗瑗 2024-10-05 04:35:40

>> tt = zeros(5,5)
tt =
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
>> tt(1:6:end) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     3     0     0
     0     0     0     3     0
     0     0     0     0     3

更一般地说：

>> V=[1 2 5]; N=5;
>> tt = zeros(N,N);
>> tt((N+1)*(V-1)+1) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     3

这是基于矩阵可以作为一维数组（向量）访问的事实，其中 2 个索引 (m,n) 被线性映射 m*N+n 替换。

>> tt = zeros(5,5)
tt =
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
>> tt(1:6:end) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     3     0     0
     0     0     0     3     0
     0     0     0     0     3

and more general:

>> V=[1 2 5]; N=5;
>> tt = zeros(N,N);
>> tt((N+1)*(V-1)+1) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     3

This is based on the fact that matrices can be accessed as one-dimensional arrays (vectors), where the 2 indices (m,n) are replaced by a linear mapping m*N+n.

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听不够的曲调 2024-10-05 04:35:40

>> B=[0,4,4;4,0,4;4,4,0]

B =

     0     4     4
     4     0     4
     4     4     0

>> v=[1,2,3]

v =

     1     2     3

>> B(eye(size(B))==1)=v
%insert values from v to eye positions in B

B =

     1     4     4
     4     2     4
     4     4     3

>> B=[0,4,4;4,0,4;4,4,0]

B =

     0     4     4
     4     0     4
     4     4     0

>> v=[1,2,3]

v =

     1     2     3

>> B(eye(size(B))==1)=v
%insert values from v to eye positions in B

B =

     1     4     4
     4     2     4
     4     4     3

回复收藏 0 原文

弥繁 2024-10-05 04:35:40

A = zeros(7,6);
V = [1 3 5];

[n m] = size(A);
diagIdx = 1:n+1:n*m;
A( diagIdx(V) ) = 1

A =
     1     0     0     0     0     0
     0     0     0     0     0     0
     0     0     1     0     0     0
     0     0     0     0     0     0
     0     0     0     0     1     0
     0     0     0     0     0     0
     0     0     0     0     0     0

A = zeros(7,6);
V = [1 3 5];

[n m] = size(A);
diagIdx = 1:n+1:n*m;
A( diagIdx(V) ) = 1

A =
     1     0     0     0     0     0
     0     0     0     0     0     0
     0     0     1     0     0     0
     0     0     0     0     0     0
     0     0     0     0     1     0
     0     0     0     0     0     0
     0     0     0     0     0     0

回复收藏 0 原文

无人接听 2024-10-05 04:35:40

我将使用 sub2ind 并将对角线索引作为 x 和 y 参数传递：

A = zeros(4)
V=[2 4]

idx = sub2ind(size(A), V,V)
% idx = [6, 16]

A(idx) = 1

% A =
% 0     0     0     0
% 0     1     0     0
% 0     0     0     0
% 0     0     0     1

I'd use sub2ind and pass the diagonal indices as both x and y parameters:

A = zeros(4)
V=[2 4]

idx = sub2ind(size(A), V,V)
% idx = [6, 16]

A(idx) = 1

% A =
% 0     0     0     0
% 0     1     0     0
% 0     0     0     0
% 0     0     0     1

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红墙和绿瓦 2024-10-05 04:35:40

假设 K 是值。该命令

A=A-diag(K-diag(A))

可能会快一点，

>> A=randn(10000,10000);

>> tic;A(logical(eye(size(A))))=12;toc

经过的时间是 0.517575 秒。

>> tic;A=A+diag((99-diag(A)));toc

经过的时间是 0.353408 秒。

但它会消耗更多的内存。

Suppose K is the value. The command

A=A-diag(K-diag(A))

may be a bit faster

>> A=randn(10000,10000);

>> tic;A(logical(eye(size(A))))=12;toc

Elapsed time is 0.517575 seconds.

>> tic;A=A+diag((99-diag(A)));toc

Elapsed time is 0.353408 seconds.

But it consumes more memory.

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烟花肆意 2024-10-05 04:35:40

我在有限差分代码中使用这个小内联函数。

A=zeros(6,3);
range=@(A,i)[1-min(i,0):size(A,1)-max(i+size(A,1)-size(A,2),0 ) ];
Diag=@(A,i) sub2ind(size(A), range(A,i),range(A,i)+i );
A(Diag(A, 0))= 10; %set diagonal 
A(Diag(A, 1))= 20; %equivelent to diag(A,1)=20;
A(Diag(A,-1))=-20; %equivelent to diag(A,-1)=-20;

通过更改函数范围，可以轻松修改它以在对角线的子范围上工作。

I use this small inline function in finite difference code.

A=zeros(6,3);
range=@(A,i)[1-min(i,0):size(A,1)-max(i+size(A,1)-size(A,2),0 ) ];
Diag=@(A,i) sub2ind(size(A), range(A,i),range(A,i)+i );
A(Diag(A, 0))= 10; %set diagonal 
A(Diag(A, 1))= 20; %equivelent to diag(A,1)=20;
A(Diag(A,-1))=-20; %equivelent to diag(A,-1)=-20;

It can be easily modified to work on a sub-range of the diagonal by changing the function range.

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