在MATLAB中将零的对角线添加到矩阵 [英] Add a diagonal of zeros to a matrix in MATLAB

问题描述

假设我在MATLAB中有一个尺寸为Nx(N-1)的矩阵A，例如

Suppose I have a matrix A of dimension Nx(N-1) in MATLAB, e.g.

N=5;
A=[1  2  3  4;
   5  6  7  8;
   9  10 11 12;
   13 14 15 16;
   17 18 19 20 ];

我想通过添加零对角线(即

I want to transform A into an NxN matrix B, just by adding a zero diagonal, i.e.,

B=[ 0  1   2   3   4;
    5  0   6   7   8;
    9  10  0   11  12;
    13 14  15  0   16;
    17 18  19  20  0];

---

这段代码可以满足我的要求:

---

This code does what I want:

B_temp = zeros(N,N); 
B_temp(1,:) = [0 A(1,:)];
B_temp(N,:) = [A(N,:) 0];
for j=2:N-1
    B_temp(j,:)= [A(j,1:j-1) 0 A(j,j:end)];
end
B = B_temp; 

您能建议一种有效的矢量化方法吗?

Could you suggest an efficient way to vectorise it?

推荐答案

您可以对矩阵的上下三角形部分(triu和tril)进行此操作.

You can do this with upper and lower triangular parts of the matrix (triu and tril).

那么这是1行解决方案:

Then it's a 1 line solution:

B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];

---

修改:基准测试

这是循环方法， Sardar的答案和我上面的方法中的2种方法的比较.

This is a comparison of the loop method, the 2 methods in Sardar's answer, and my method above.

基准代码，使用timeit进行计时，并直接从问题和答案中提取代码:

Benchmark code, using timeit for timing and directly lifting code from question and answers:

function benchie()
    N = 1e4; A = rand(N,N-1); % Initialise large matrix
    % Set up anonymous functions for input to timeit
    s1 = @() sardar1(A,N); s2 = @() sardar2(A,N); 
    w =  @() wolfie(A,N); u = @() user3285148(A,N);
    % timings
    timeit(s1), timeit(s2), timeit(w), timeit(u)
end
function sardar1(A, N) % using eye as an indexing matrix
    B=double(~eye(N)); B(find(B))=A.'; B=B.';
end
function sardar2(A,N) % similar to sardar1, but avoiding slow operations
    B=1-eye(N); B(logical(B))=A.'; B=B.';
end
function wolfie(A,N) % using triangular parts of the matrix
    B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
end
function user3285148(A, N) % original looping method
    B = zeros(N,N); B(1,:) = [0 A(1,:)]; B(N,:) = [A(N,:) 0];
    for j=2:N-1; B(j,:)= [A(j,1:j-1) 0 A(j,j:end)]; end
end

结果:

• Sardar方法1:2.83秒
• Sardar方法2:1.82秒
• 我的方法:1.45秒
• 循环方法:3.80秒(！)

结论:

• 您希望对此进行矢量化处理的理由很充分，循环比其他方法要慢得多.
• 避免大型矩阵的数据转换和find很重要，在Sardar方法之间可节省约35％的处理时间.
• 通过避免一起编制索引，可以节省20％的处理时间.

• Your desire to vectorise this was well founded, looping is way slower than other methods.
• Avoiding data conversions and find for large matrices is important, saving ~35% processing time between Sardar's methods.
• By avoiding indexing all together you can save a further 20% processing time.

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