Extracting columns from large matrices

You probably won't notice any direct speed difference with manually specifying the indices as opposed to using subvec. However, your options are narrowed depending on your use-case, as well as the elegance of the code involved.

Using a basic representation of x:

>> x = reshape(seqa(1,1,25), 5, 5);
>> x

       1.0000000        2.0000000        3.0000000        4.0000000        5.0000000 
       6.0000000        7.0000000        8.0000000        9.0000000        10.000000 
       11.000000        12.000000        13.000000        14.000000        15.000000 
       16.000000        17.000000        18.000000        19.000000        20.000000 
       21.000000        22.000000        23.000000        24.000000        25.000000 

1) If you are extracting the same column for every row:

>> x[.,2]

       2.0000000 
       7.0000000 
       12.000000 
       17.000000 
       22.000000 

is the most straight-forward

If you plan on extracting multiple columns in this manner, you could gain a significant speed advantage by transposing the matrix first.

2) Extracting all rows with a specific column for each row.
Note that this method requires the column vector to equal the rows of the input matrix.

>> subvec(x, seqa(1,1,rows(x)));

       1.0000000 
       7.0000000 
       13.000000 
       19.000000 
       25.000000 

3) Extracting specific columns (not all) for specific row(s):

>> c = {1,3,5};
>> x[.,c]

       1.0000000        3.0000000        5.0000000 
       6.0000000        8.0000000        10.000000 
       11.000000        13.000000        15.000000 
       16.000000        18.000000        20.000000 
       21.000000        23.000000        25.000000