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Merging data in arrays

8 messages · arun, Ray Cheung, Jeff Newmiller +1 more

Original
arun
# 
Hi,

I didn't fully understand the logic.
You could get the result by:
?list1<-lapply(mapply(cbind,lapply(1:2,function(i) E[,,i]),lapply(c(1,3),function(i) C[,i,]),SIMPLIFY=FALSE),function(x) x[,c(TRUE,apply(matrix(!x[,-1]%in% x[,1],nrow=5),2,all))])
Fnew<-array(unlist(list1),dim=c(dim(list1[[1]]),length(list1)))

?identical(F,Fnew)
#[1] TRUE
A.K.


----- Original Message -----
From: Ray Cheung <ray1728 at gmail.com>
To: r-help at r-project.org
Cc: 
Sent: Wednesday, February 6, 2013 10:16 PM
Subject: [R] Merging data in arrays

Dear All,

Here is a hypothetical sample (sorry for the clumsy code):

A1 <- matrix(1:5, nrow=5, ncol=1)
A2 <- matrix(6:10, nrow=5, ncol=1)
A3 <- matrix(11:15, nrow=5, ncol=1)
A4 <- matrix(16:20, nrow=5, ncol=1)
A5 <- matrix(21:25, nrow=5, ncol=1)
A6 <- matrix(26:30, nrow=5, ncol=1)

B1 <- matrix(c(A1, A2, A3), nrow=5, ncol=3)
B2 <- matrix(c(A2, A3, A4), nrow=5, ncol=3)
B3 <- matrix(c(A3, A4, A5), nrow=5, ncol=3)

C <- array(c(B1, B2, B3), dim = c(5,3,3))

D1 <- matrix(c(A1, A4, A5), nrow=5, ncol=3)
D2 <- matrix(c(A3, A5, A6), nrow=5, ncol=3)

E <- array(c(D1, D2), dim = c(5,3,2))

In the above example, I want to merge array C to array E by matching the
column 1. That is, the resultant array F should look like this:

F1 <- matrix(c(A1, A4, A5, A2, A3), nrow=5, ncol=5)
F2 <- matrix(c(A3, A5, A6, A4, A5), nrow=5, ncol=5)
F <- array(c(F1, F2), dim = c(5,5,2))

Would you please advise on the codes? Thank you very much for any help.

Best Regards,
Ray

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13 days later
Jeff Newmiller
# 
I think this specification is insufficient to respond accurately to. Please make a reproducible subset of your data (or simulated data) and provide it in dput form, and describe your desired result data set more clearly.

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

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Ray Cheung <ray1728 at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        
      

    

    

      
      

        


  

        

      Rui Barradas
    

    

      
      #
    

  


  

      
Hello,

I bet there are simpler solutions but I'm not thinking of anything else, 
right now.

fun <- function(x, y){
	f <- function(a, b){
		a <- as.data.frame(a)
		b <- as.data.frame(b)
		names(a)[1] <- names(b)[1] <- "V1"
		res <- merge(b, a, by = "V1")
		if(nrow(res) > 0) as.matrix(res) else NULL
	}
	dx <- dim(x)[3]
	dy <- dim(y)[3]
	res <- list()
	ires <- 0
	for(idx in seq_len(dx)){
		for(idy in seq_len(dy)){
			tmp <- f(x[, , idx], y[, , idy])
			if(!is.null(tmp)){
				ires <- ires + 1
				res[[ires]] <- tmp
			}
		}
	}
	k <- nrow(res[[1]])
	m <- ncol(res[[1]])
	n <- length(res)
	array(unlist(res), dim = c(k, m, n))
}

G <- fun(C, E)
identical(F, G)  #TRUE



Hope this helps,

Rui Barradas

Em 22-02-2013 03:40, Ray Cheung escreveu:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    
1 day later

    

      
      

        


  

        

      Jeff Newmiller
    

    

      
      #
    

  


  

      
There is a reason why merge is not designed to work directly with arrays: the matching of rows can easily yield a result with a different number of rows than the initial data had. This plays havoc with the nature of arrays. I would recommend that if a merge is really what you want then unrolling your original data into data frames and leaving them that way as much as possible will be a wise move to make sooner rather than later.
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DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live Go...
                                      Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
--------------------------------------------------------------------------- 
Sent from my phone. Please excuse my brevity.

        
Rui Barradas <ruipbarradas at sapo.pt> wrote: