simple question on rbgsl vectors

[baptiste Auguié]

Hi,


I'm trying to do a simple thing : access elements of a vector
specified by their index. In Matlab/Octave i would do,

newVector=oldVector( [1, 3, 56, 72] )

How does this work with Ruby vectors?

oldVector.get(3,4,5) returns the first 3 elements, where I would
expect the 2nd, 3d and 4th.

I've had to do a twisted version of this using arrays and lists,

a2=[]
(0..200).step(10) {|i| a2 << a }

I find it very inefficient and annoying, plus I do need to use
vectors rather than arrays and lists and do not really know how to
convert between these three


Oh, and just a quick question: how can i do two instructions in one
block ? I have to repeat the same instructions over and over like
shown below...

a2=[]
b2=[]
(0..200).step(10) {|i| a2 << a }
(0..200).step(10) {|i| b2 << b }

Best regards,

baptiste

 

[Axel Etzold]

Dear Baptiste,

you can get reference for the gsl bindings for ruby
here:

http://rb-gsl.rubyforge.org/ref.html .

oldVector.get(3,4,5) returns the first 3 elements, where I would
expect the 2nd, 3d and 4th.

oldVector.get([3,4,5]) worked for me - I'm not quite sure
why this:

oldVector.get(3,4,5) returns the first 3 elements
happens...



I've had to do a twisted version of this using arrays and lists,

a2=[]
(0..200).step(10) {|i| a2 << a }

I find it very inefficient and annoying, plus I do need to use
vectors rather than arrays and lists and do not really know how to
convert between these three

You'll find info about that on the website given above.

Oh, and just a quick question: how can i do two instructions in one
block ? I have to repeat the same instructions over and over like
shown below...

You can use the same index several times:

(0..200).step(10) {|i| a2 << a ; b2<<b }

Best regards,

Axel