Re: [R] Create matrix with subset from unlist

[David Winsemius]

On Feb 1, 2010, at 9:38 AM, Muhammad Rahiz wrote:
Hello all,

Thanks for all your replies.

Usually, when I make a post to the R-mailing list, I would keep on  
trying to get the solution myself rather than waiting for one. At  
times, I was able to derive my own solution. This would explain why  
my solution and that of Dennis's produces the same result - not that  
I totally ignore the method of his...

Anyway, I did manage to create the matrix I require. But there is  
still the recurring problem of " (list) object cannot be coerced to  
type 'double'  " in further analysis using the dataset. I thought I   
could resolve it by changing converting to matrix. Seems not.

Given, the following, is there any other way I can define y other  
that using list()? Seems that producing a list of matrices does not  
work.

I produced an array as an intermediate result in an earlier posting.  
If you defined the dimensions properly that would seem to be a  
sensible alternative.

> dput(x)
list(structure(list(V1 = c(-27.3, 29), V2 = c(14.4, -38.1)), .Names =  
c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")), structure(list(
   V1 = c(14.4, -38.1), V2 = c(29, -3.4)), .Names = c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")), structure(list(
   V1 = c(29, -3.4), V2 = c(-38.1, 55.1)), .Names = c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")))

> xx <- array( , dim=c(2,2,3))
> xx[,,1:3] <- sapply(x, data.matrix)

All you would need to do is change the 3's to 32.





y <-
for (i in 1:32){
y[[i]] <- matrix(xx[c(1:4)],2,2)
}


Muhammad




Dennis Murphy wrote:
Correct me if I'm wrong, but isn't this the solution I gave??

On Fri, Jan 29, 2010 at 9:43 AM, Muhammad Rahiz <muhammad.ra...@ouce.ox.ac.uk 
 <mailto:muhammad.ra...@ouce.ox.ac.uk>> wrote:

   Thanks David & Dennis,

   I may have found something.

   Given that the object xx is the product of unlist(x), to create a
   2x2 matrix with subsets, I could do,

   > y <- matrix(xx[c(1:4)], 2, 2).

First object named y...

   This returns,


       [,1]  [,2]
   [1,] -27.3  14.4
   [2,]  29.0 -38.1

   If I do,

   > y2 <- matrix(xx[c(5:8)],2,2)


second object named y2

   it returns,


       [,1] [,2]
   [1,]  14.4 29.0
   [2,] -38.1 -3.4


And I presume you want to do the same with the remaining 30 matrices,
assigning them to different objects. That is *precisely* what my  
solution
provided. Run it, observe the results and tell me what it is that  
differs from
what you want, because I don't see it.

Dennis

   The results are exactly what I want to achieve.

   The question is, how can I incorporate the increment in a for loop
   so that it becomes

   c(1:4)
   c(5:8)
   c(9:12) and so on

   How should I modify this code?

   y <-            # typeof ? for (i in 1:32){
   y[[i]] <- matrix(xx[c(1:4)],2,2)
   }


   Muhammad


   David Winsemius wrote:

       On Jan 29, 2010, at 9:45 AM, Dennis Murphy wrote:


           Hi:

           The problem, I'm guessing, is that you need to assign each
           of the  matrices
           to an object.
           There's undoubtedly a slick apply family solution for this
           (which I  want to
           see, BTW!),


       I don't have a method that would assign names but you could
       populate  an array of sufficient size and dimension. I
       populated a three-element  list with his data:

        > dput(x)
       list(structure(list(V1 = c(-27.3, 29), V2 = c(14.4, -38.1)),
       .Names =  c("V1",
       "V2"), class = "data.frame", row.names = c("1", "2")),
       structure(list(
           V1 = c(14.4, -38.1), V2 = c(29, -3.4)), .Names = c("V1",
       "V2"), class = "data.frame", row.names = c("1", "2")),
       structure(list(
           V1 = c(29, -3.4), V2 = c(-38.1, 55.1)), .Names = c("V1",
       "V2"), class = "data.frame", row.names = c("1", "2")))

        > xx <- array( , dim=c(2,2,3))

        > xx[,,1:3] <- sapply(x, data.matrix)
        > xx
       , , 1

             [,1]  [,2]
       [1,] -27.3  14.4
       [2,]  29.0 -38.1

       , , 2

             [,1] [,2]
       [1,]  14.4 29.0
       [2,] -38.1 -3.4

       , , 3

            [,1]  [,2]
       [1,] 29.0 -38.1
       [2,] -3.4  55.1

       Without the more complex structure ready to accept the 2x2
       arrays I  got this:

        > sapply(x, data.matrix)
             [,1]  [,2]  [,3]
       [1,] -27.3  14.4  29.0
       [2,]  29.0 -38.1  -3.4
       [3,]  14.4  29.0 -38.1
       [4,] -38.1  -3.4  55.1




David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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