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numbers loop in R

4 messages · emj83, David Winsemius, jim holtman +1 more

emj83
# 
I would like to create a matrix in R that looks similar to this:

     [,1] [,2] [,3] [,4]
[1,]  NaN  1  2  3
[2,]  NaN  1  2  4
[3,]  NaN   1  2  5
[4,]  NaN  2  3  4
[5,]  NaN  2  3  5
[6,]  NaN    3    4    5

I have the loop below:

where A for example is 5

matrixx<-function(A){
B=matrix(NaN,nrow=(A+1),ncol=4)
        for(k in 1:(A+1)){
	        for(i in 1:(A-2)){
		     for(j in (i+2):A){		
		     }		
		}	
       }
B[k,]=c(NaN,i,(i+1),j)
print(B)
}

But it only prints the final line in:
[,1] [,2] [,3] [,4]
[1,]  NaN  NaN  NaN  NaN
[2,]  NaN  NaN  NaN  NaN
[3,]  NaN  NaN  NaN  NaN
[4,]  NaN  NaN  NaN  NaN
[5,]  NaN  NaN  NaN  NaN
[6,]  NaN    3    4    5

Could anyone give me a hand? Would be much appreciated.

Thanks Emma
David Winsemius
# 
I would have expected to see the assignment to B[k,] inside the loops.  
And to see some connection with the k index in the inner loops if you  
did not want all of the rows to be similar. Because the assignment is  
outside the loops, it happens only once.

-- David Winsemius
On Apr 17, 2009, at 11:11 AM, emj83 wrote:

            

      
      
      
    

        

      
      
    

        
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      jim holtman
    

    

      
      #
    

  


  

      
try this:

            

      
      
      
    

        
+     B=matrix(NaN,nrow=(A+1),ncol=4)
+     k <- 1
+     for (i in 3:A){
+         for (j in i:A) {
+             B[k,] <- c(NaN, i-2, i-1, j)
+             k <- k + 1
+         }
+     }
+     B
+ }

            

      
      
      
    

        
[,1] [,2] [,3] [,4]
[1,]  NaN    1    2    3
[2,]  NaN    1    2    4
[3,]  NaN    1    2    5
[4,]  NaN    2    3    4
[5,]  NaN    2    3    5
[6,]  NaN    3    4    5

            

      
      
      
    

        

        
On Fri, Apr 17, 2009 at 11:11 AM, emj83 <stp08emj at shef.ac.uk> wrote:

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Hadley Wickham
    

    

      
      #
    

  


  

      
On Fri, Apr 17, 2009 at 12:19 PM, jim holtman <jholtman at gmail.com> wrote:

            

      
      
      
    

        
Here's a solution without the loop.  I think it illustrates the intent
of the algorithm more clearly.

candidates <- t(combn(5,3))
firstdiff <- candidates[,2] - candidates[, 1]
cbind(NaN, candidates[firstdiff == 1, ])

Hadley