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?
D = diag( v ) returns a square diagonal matrix with the elements of vector v on the main diagonal. D = diag( v , k ) places the elements of vector v on the k th diagonal. k=0 represents the main diagonal, k>0 is above the main diagonal, and k<0 is below the main diagonal.
Calculate the sum of diagonal elements of a matrix Let us understand the logic before going to a solution. (i=j). and condition for the right diagonal element is ( i+j=N-1). where i and j are position coordinates and N is the size of Matrix.
A diagonal matrix is defined as a square matrix in which all off-diagonal entries are zero. (Note that a diagonal matrix is necessarily symmetric.) Entries on the main diagonal may or may not be zero.
A hollow matrix may be a square matrix whose diagonal elements are all equal to zero.
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:
Conclusions:
find
for large matrices is important, saving ~35% processing time between Sardar's methods. If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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