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comp.lang.ruby

simple question on rbgsl vectors

baptiste Auguié

7/4/2007 12:48:00 PM

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] }

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

Best regards,

baptiste
1 Answer

Axel Etzold

7/4/2007 2:04:00 PM

0

Dear Baptiste,

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

http://rb-gsl.rubyforge.or... .


> 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] }
>
> 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[i] ; b2<<b[i] }

Best regards,

Axel

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