Add a diagonal of zeros to a matrix in MATLAB - matlab

Question

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 ];
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];
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;
Could you suggest an efficient way to vectorise it?

Answer 1

You can do this with upper and lower triangular parts of the matrix (triu and tril).
Then it's a 1 line solution:
B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
Edit: benchmark
This is a comparison of the loop method, the 2 methods in Sardar's answer, and my method above.
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
Results:
Sardar method 1: 2.83 secs
Sardar method 2: 1.82 secs
My method: 1.45 secs
Looping method: 3.80 secs (!)
Conclusions:
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.

Answer 2

Generate a matrix with zeros at diagonal and ones at non-diagonal indices. Replace the non-diagonal elements with the transpose of A (since MATLAB is column major). Transpose again to get the correct order.
B = double(~eye(N)); %Converting to double since we want to replace with double entries
B(find(B)) = A.'; %Replacing the entries
B = B.'; %Transposing again to get the matrix in the correct order
Edit:
As suggested by Wolfie for the same algorithm, you can get rid of conversion to double and the use of find with:
B = 1-eye(N);
B(logical(B)) = A.';
B = B.';

Answer 3

If you want to insert any vector on a diagonal of a matrix, one can use plain indexing. The following snippet gives you the indices of the desired diagonal, given the size of the square matrix n (matrix is n by n), and the number of the diagonal k, where k=0 corresponds to the main diagonal, positive numbers of k to upper diagonals and negative numbers of k to lower diagonals. ixd finally gives you the 2D indices.
function [idx] = diagidx(n,k)
% n size of square matrix
% k number of diagonal
if k==0 % identity
idx = [(1:n).' (1:n).']; % [row col]
elseif k>0 % Upper diagonal
idx = [(1:n-k).' (1+k:n).'];
elseif k<0 % lower diagonal
idx = [(1+abs(k):n).' (1:n-abs(k)).'];
end
end
Usage:
n=10;
k=3;
A = rand(n);
idx = diagidx(n,k);
A(idx) = 1:(n-k);