Filling a matrix without for-loops and ifs

Asked by Mauricio Garcia

Mauricio Garcia (view profile)

on 18 Sep 2019 at 22:21
Latest activity Edited by James Tursa

James Tursa (view profile)

on 18 Sep 2019 at 22:39
So i have this piece of code:
for i = 1:L
n1 = M(i,1);
n2 = M(i,2);
for j = 1:N
if j == n1
A(i,j) = 1;
elseif j == n2
A(i,j) = -1;
else
A(i,j) = 0;
end
end
end
M is a (L,2) matrix which tells me where does a line start and where it ends.
What i am looking to do is fill a matrix "A" , that is already filled with zeros, with a 1 where the line starts [M(:,1)], and a -1 where the line ends [M(:,2)], and all other elements with zeros, line per line.
For example if my M matrix is:
M =
1 2
2 3
1 3
The matrix A would be:
A =
1 -1 0
0 1 -1
1 0 -1
What i would like to know, is if i can fill this matrix "A" without using for's and if's, just make my code cleaner and maybe faster to run.
I was talking with a professor and told me to do it like this:
A(ind(:,1) = 1
A(ind(:,2) = -1
But i feel something is missing and cant find how to do it properly.
Sorry for the long post, thank you for your time.
Cheers

Release

R2019a

Answer by James Tursa

James Tursa (view profile)

on 18 Sep 2019 at 22:35
Edited by James Tursa

James Tursa (view profile)

on 18 Sep 2019 at 22:39

Check out the sub2ind( ) function:
In particular, your code would look something like:
A = zeros(_____); % <-- You need to figure out what the size of A should be
A(sub2ind(_____)) = 1; % <-- You need to figure out the proper inputs for sub2ind( )
A(sub2ind(_____)) = -1; % <-- You need to figure out the proper inputs for sub2ind( )