similar to: aperm() and as.list() args to "["

OK guys another problem. I have a 3D array "x" with dim(x)=c(a,a,b^2) and I want to rearrange the elements of x to make a matrix "y" with dimensions c(a*b,a*b). Neither a nor b is known in advance. I want the "n-th" a*a submatrix of y to be x[,,n] (where 1 <= n <= b^2). Needless to say, this has gotta be vectorized! Toy example with a=2, b=3 follows:

aperm() was designed for multidimensional arrays, but is also useful for table objects, particularly with the lattice, vcd and vcdExtra packages. But aperm() was designed and implemented before other related object classes were conceived, and I propose a small tune-up to make it more generally useful. The problem is that aperm() always returns an object of class 'array', which

R version 1.0.1 OS RedHat Linux 6.1 In attempting to test for numeric vectors in a data frame, I tried: apply(dataframe,2,is.numeric) and found that it returned FALSE for all vectors whether they were numeric or not. I tracked this to the fact that as.array() was converting the data frame to character vectors, and thought I could solve it by using array(), which preserved the mode of the

I think that the "dimname names" of tables and arrays could make aperm() and apply() (and probably some other functions) easier to use. (dimname names are, for example, created by table() ) The use would be something like: -- x <-table( from=sample(3,100,rep=T), to=sample(5,100,rep=T)) trans <- x / apply(x,"from",sum) y <- aperm( trans,

Dear all, I have been working on an idempotent version of apply, such that applying a function f(x) = x (ie. force) returns the same array (or a permutation of it depending on the order of the margins): a <- array(1:27, c(2,3,4)) all.equal(a, iapply(a, 1, force)) all.equal(a, iapply(a, 1:2, force)) all.equal(a, iapply(a, 1:3, force)) all.equal(aperm(a, c(2,1,3)), iapply(a, 2, force))

Hi, I have noticed that aperm is very slow, and I wondered if there was a way of speeding it up. Let me tell you a bit about the context of my problem, because perhaps I shouldn't be using aperm at all. The context is probabilistic inference in graphical models. One of the most fundamental operations is two compute an element-wise multiplication of two arrays of different sizes, say A and B.

Professor Ripley thanks for this. Very much appreciated. The original subject line reflected my late-night conviction that the answer might involve passing a strange list to do.call(). Anyway, package magic is broken (only in R-devel, I might add) because I have a function called "%eq%". R-2.0.0 CMD check is stopping (I think) because it interprets the "%" as a

Hi everyone I've been playing with do.call() but I'm having problems understanding it. I have a list of "n" elements, each one of which is "d" dimensional [actually an n-by-n-by ... by-n array]. Neither n nor d is known in advance. I want to bind the elements together in a higher-dimensional array. Toy example follows with d=n=3. f <-

Hi I have a little vector function that takes a vector A of strictly positive integers and outputs a matrix M each of whose columns is the vector, modified in a complicated combinatorical way. Now I want to generalize the function so that A can include zeroes. Given A, I want to strip out the zeroes, pass it to my function, and pad M with rows at positions corresponding to the zeroes

i'm somehow embarrassed to even ask this, but is there any built-in method for doing this: my_list <- list() my_list[[1]] <- matrix(1:20, ncol = 5) my_list[[2]] <- matrix(20:1, ncol = 5) now, knowing that these matrices are identical in dimension, i'd like to unfold the list to a 2x4x5 (or some other permutation of the dim sizes) array. i know i can initialize the array, then

Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as opposed to really thinking hard about how it works ;-) On 9/28/2017 11:55 AM, Duncan Murdoch wrote: > On 28/09/2017 9:10 AM, Evan Cooch wrote: >> Thanks for both the mapply and array approaches! However, although >> intended to generate the same result, they don't: >> >> # mapply approach

The use of aperm is unnecessary if you call array() properly. ms <- array(c(rep(0, 5),so,sa*m,sa), c(5, 2, 2)) -- Sent from my phone. Please excuse my brevity. On September 28, 2017 9:10:26 AM PDT, Evan Cooch <evan.cooch at gmail.com> wrote: >Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as >opposed to really thinking hard about how it works ;-) > >On

I have been tinkering with mosaicplot() and friends as a way of learning R. As part of this, I've written a pair.table() method for mosaic matrices, and would like to extend mosaicplot to work with loglin and logln (MASS) objects. I'm using R 1.4.0 on Win 98. I've been trying to figure out the formula interface, and think there's a bug, but not sure how to find it, yet alone fix

Hi Everyone, To complete the outer() and kronecker() functions in the base, may I suggest the following tensor() function, which allows the multiplication of arrays through sets of conformable dimensions. I am happy to write a help page if required. The code also needs a setdiff() function which prompts me to ask: what about simple set functions? I expect many of us have written our own

Hi I'm using kronecker() with a matrix and a vector. I'm interested in the column names that kronecker() returns: > a <- matrix(1:9,3,3) > rownames(a) <- letters[1:3] > colnames(a) <- LETTERS[1:3] > b <- c(x=1,y=2) > kronecker(a,b,make.dimnames=TRUE) A: B: C: a:x 1 4 7 a:y 2 8 14 b:x 2 5 8 b:y 4 10 16 c:x 3 6 9 c:y 6 12 18 > The

Hi, xtabs() creates a table of counts. I want a table of proportions -- that is, I want to divide every vector (along a particular dimension) by its sum. The tiny example below does that. The call to xtabs() creates a matrix "A" with dimensions ("x1","x2","y"). I transform "A" using aperm() and aaply() to get the matrix "B". The

Hi I had a hard-to-find bug in some of my code the other day, which I eventually traced to my misusing of sweep(). I would expect sweep() to give me a warning if the elements don't recycle nicely, but X <- matrix(1:36,6,6) sweep(X,1,1:5,"+") [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2 9 16 23 30 32 [2,] 4 11 18 25 27 34 [3,] 6 13 20 22

Thanks for both the mapply and array approaches! However, although intended to generate the same result, they don't: # mapply approach n = 3 sa <- rnorm(n,0.8,0.1) so <- rnorm(n,0.5,0.1) m <- rnorm(n,1.2,0.1) mats = mapply(function(sa1, so1, m1) matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE) print(mats) [[1]] ????????? [,1]????? [,2] [1,] 0.0000000

Hi again everyone I have an arbitrarily dimensional array "a" and a list "jj" of length length(dim(a)). The elements of jj are vectors of indexes. How do I use do.call() to extract a[ jj[[1]], jj[[2]], jj[[3]], ...] ? Toy example follows: a <- matrix(1:30,5,6) jj <- list(5:1,6:1) I want the following a[ jj[[1]],jj[[2]] ] How do I do this? OBAttempts:

Hi listers, I am having some trouble in a matrix multiplication... I have already checked some posts, but I didn't find my problem... I have the following code... But I am not getting the right multiplication... I checked the dimension and they are fine... id_y <- array(1:10,dim=c(2,1,5)) id_yt<-aperm(id_y,c(2,1,3)) m_id<-array(dim=c(dim(id_y)[1],dim(id_y)[1],dim(id_y)[3])) for (i in