Skip to content

MAtrix addressing

4 messages · Alaios, David Winsemius

Alaios
# 
Hello
I would like to ask you if it is possible In R Cran to change the default way of addressing a matrix.
for example
matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row numbering) # not having R at this pc

will create something like the following
1 2
3 4

the way R address this matrix is from top left corner moving to bottom right.
The cell numbers in that way are 
1 2
3 4

IS it possible to change this default addresing number to something that goes bottom left to top right? In this simple case I want to have
3 4
1 2

Would that be possible?

I would like to thank y for your help
Regards
Alex
David Winsemius
#
On Jan 25, 2011, at 4:50 PM, Alaios wrote:

            

      
      
      
    

        
Yes. it's possible but ... why?

            

      
      
      
    

        

      
      
    

        
David Winsemius, MD
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      Alaios
    

    

      
      #
    

  


  

      
The reason is the following image
http://img545.imageshack.us/i/maptoregion.jpg/
In the picture above you will find the indexes for each cell.

Also you will see that I place that matrix inside a x,y region that spans from -1 to 1. I am trying to write one function that will get as argument a (x,y) value x e[-1,1] y e[-1,1] and will return the indexes of that cell tha x,y value correspond to.

I really do not have a clue how I should try to approach that to solve it. So based on some version I had for 1-d vector I tried to extend it for 2-d. I used findInterval as a core to get results.
Unfortunately my code fails to produce accurate results as my approach 'assumes' (this is something inhereted by the find Interval function) that the numbering starts bottom left and goes high top right.
You will find my code below




sr.map <- function(sr){
# This function converts the s(x,y) matrix into a function x that spans #from -1 to 1 and y spans from -1 to 1.
# Input: sr a x,y matrix containing the shadowing values of a Region
     breaksX <- seq(from=-1, to = 1, length = nrow(sr) +1L )
     breaksY <- seq(from=-1, to = 1, length = ncol(sr) + 1L)
     function(x,y){ # SPAGGETI CODE FOR EVER
         indx <- findInterval(x, breaksX,rightmost.closed=TRUE)
	 indy <- findInterval(y, breaksY,rightmost.closed=TRUE)
	 c(indx,indy)
     }
 }



sr<-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)
f.sr.map<-sr.map((sr))
f.sr.map(-0.1,-0.1)
f.sr.map(0.1,0.1)



Best Regards
Alex

        
--- On Wed, 1/26/11, David Winsemius <dwinsemius at comcast.net> wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      David Winsemius
    

    

      
      #
    

  


  

      
On Jan 26, 2011, at 2:47 AM, Alaios wrote:

        

            

      
      
      
    

        
If one wants to take an ordinary r matrix and reorder it in the manner  
you describe:

mtx2 <- mtx[ nrow(mtx):1, ]

Whether that is an efficient way to get at the sokution your you  
programming task I cannot say. It sounds as though it has gotten too  
convoluted. I was not able to comprehend the overall goal from your  
problem description.

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        
David Winsemius, MD
West Hartford, CT