从Octave 3x n矩阵的col 2中减去常数的最佳实践是什么

Question

Here, I subtract 2000 from column 2 and return the complete 3 column vector...

This "works"; but, isn't it processing the matrix 3 times?

xx = [X(:,1),X(:,2) .-2000,X(:,3)]

Best practice please... ;-0

Answer 1

The simplest way to do this operation is to simply:

X(:,2) -= 2000;

which is also a lot easier to read. This will modify the second column X "in place". If you want to make a copy of it where the second column is subtracted, then simply:

xx = X;
xx(:,2) -= 2000;

An example:

octave-cli-3.8.2> X = randi (9, 5, 3)
X =

   1   4   4
   1   2   6
   8   4   3
   7   7   1
   7   7   2

octave-cli-3.8.2> X(:,2) -= 10
X =

   1  -6   4
   1  -8   6
   8  -6   3
   7  -3   1
   7  -3   2