search for: aperm

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...

...d 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 columns. However, I then got an error from aperm(): Error in aperm(X, c(s.call, s.ans)) : aperm: invalid second argument, must be a vector even though the second argument tested as a vector. This happened both within the apply() function and calling aperm() separately, even if I explicitly used a numeric vector for the second argument. I then...

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, c("fr...

Hi everyone. I'm playing with aperm(): a <- 1:24 dim(a) <- c(2,3,2,2) permutation <- c(1,2,4,3) b <- aperm(a,permutation) So if my understanding is right, a[1,3,2,1] == b[c(1,3,2,1)[permutation] ] but this isn't what I want because the RHS evaluates to a vector, and I am trying to identify a single element of b...

...d now, printing a prints these dimension labels nicely: > A to from [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 I don't know exactly what these labels are called, but they can be handy sometimes. For example, they could be handy when using aperm. Instead of writing aperm(A, c(2,1) ) It would be nice to be able to write: aperm(A, c("from","to") ) This way, I don't have to remember what is the order of the dims of A. It is easy to extend the functionality of aperm, so that the above code works: --- my.aperm=fun...

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 elem...

...80% of the seeds produced by a plant will fell into the parent cell, the 15% in the first ring according to the king movement (in chess), and a 5% in the second ring defined by the queen2 matrix. Someone said me the functions that order the position of the seed dispersed, they are: neighb <- aperm(apply(aperm(outer(king,IJ,'+'),c(2,4,1,3)),3:4,diag),c(2,1,3)) #Seed going back into source cell??? for(i in 1:nrow(IJ)) SB[matrix(neighb[,,i],nc=2)]<- SR1[matrix(IJ[i,],1,2)]/8 + SB[neighb[,,i]] #Seed going to anuli 1 neighbor cells from all possible sources neighb <- aperm(apply...

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:...

...ion 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)) all.equal(aperm(a, c(3,1,2)), iapply(a, 3, force)) I have also generalised apply so that the function can return an array of any dimension and it will be slotted in orthogonally to the existing dimensions: iapply(a, 1, min) iapply(a, 2, min) iapply(a, 3, min) ia...

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...

...ng sd=0 for the parameters -- thus, every matrix will be the same. The correct matrix would be: ???? [,1] [,2] [1,]? 0.0 0.96 [2,]? 0.5 0.80 In fact, this is what is returned by the mapply approach, while the array approach returns the transpose. I gather the 'missing step' is to use aperm, but haven't figured out how to get that to work...yet. On 9/28/2017 5:11 AM, Duncan Murdoch wrote: > ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2)) [[alternative HTML version deleted]]

...ter) if (no.na && no.nb) nms <- NULL else { if (no.na) na <- vector("list", length(dima)) else if (no.nb) nb <- vector("list", length(dimb)) nms <- c(na[kpa], nb[kpb]) } A <- matrix(aperm(A, c(kpa, da)), ncol = prod(dima[da])) B <- matrix(aperm(B, c(db, kpb)), nrow = prod(dimb[db])) array(A %*% B, c(dima[kpa], dimb[kpb]), dimnames = nms) } # examples A <- array(1:6, 2:3, list(LETTERS[1:2], LETTERS[3:5])) B <- 1:3 tensor(A, B, 2, 1) # same as drop(A %*% B) A <...

...ure out the formula interface, and think there's a bug, but not sure how to find it, yet alone fix it. It seems to only work with independence models: > data(HairEyeColor) > mosaicplot(~ Hair + Eye + Sex, data=HairEyeColor) > mosaicplot(~ Hair * Eye + Sex, data=HairEyeColor) Error in aperm(a, perm, resize) : `perm' is of wrong length here's a traceback: > debug(mosaicplot) > mosaicplot(~Hair + Eye + Sex + Hair:Eye, data=HairEyeColor) debugging in: mosaicplot(~Hair + Eye + Sex + Hair:Eye, data = HairEyeColor) debug: UseMethod("mosaicplot") Browse[1]> Erro...

Hi Abby, Thanks a lot for your paraphrasing and your suggestion! The problem of wrapping the list into a S3/S4 object, i.e. subclassing array or matrix, is that one also has to define a bunch of methods for subsetting, joining, etc, in order to make it behave like a list array. The reason is that most R functions for subsetting, joining, etc. do not preserve class attributes of the input, which

I've dug around but not been able to find anything, am probably missing something obvious. How can I fill a three-dimensional (or higher dimension) array by rows instead of columns. eg new1 <- array(c(1:125), c(5,5,5)) works fine for me but fills it by columns and new2 <- array(c(1:125), c(5,5,5), byrow=TRUE) throws an error. Am I missing something obvious? I also tried

...23,24, 25,26,29,30,33,34, 27,28,31,32,35,36 )),6,6) So identical(x[,,1] , y[1:2,1:2]) identical(x[,,2] , y[3:4,1:2]) [snip] identical(x[,,9] , y[5:6,5:6]) all return TRUE. OBattempts: (i) dim(x) <- c(4,9);x <- t(x) ; dim(x) <- c(6,6) ; y <- x (ii) y <- aperm(x,c(2,1,3)) ; dim(y) <- c(6,6) Any genius out there with some ideas? -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre SO14 3ZH tel +44(0)23-8059-7743 initialDOTsurname at soc.soton.ac.uk (edit in obvious way; spam precaution)

...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 problem: "B" has dimensions (<No name>, "x2", "x1"). How can I give (back) the name "y" to the dimension with no name in the matrix "B"? (Unless I misunderstand something, dimnames() won...

...tion multipliers (i x k) ARRAY!! # target: restriction value vector/scalar (k) htest.lm <- function(object, resmat, target = 0, level = 0.95) { tsu <- summary(object, corr=FALSE) cfs <- as.array(object$coefficients) dfr1 <- nrow(resmat) dfr2 <- object$df.residual S <- aperm(resmat %*% aperm(cfs) - target) %*% solve(resmat %*% tsu$cov.unscaled %*% aperm(resmat)) %*% (resmat %*% aperm(cfs) - target) / (nrow(resmat) * tsu$sigma^2) cat(" F(",dfr1,",",dfr2,") =",qf(level,dfr1,dfr2), " vs. S =",S," [ p = ",df(S,dfr1,df...

...particular, how do I get \usage{m1 %eq% m2} and \examples{..} to work nicely? best wishes rksh > > > >>What has this to do with the original subject line? Replacement functions >>are not intended to be used directly, and certainly not in do.call. >> >>See ?aperm, as in >> >>xx <- x >>dim(xx) <- c(2,2,3,3) >>xx <- aperm(xx, c(1,3,2,4)) >>dim(xx) <- c(6, 6) >>xx >> >>as required. >> >>BTW, you have a broken package `magic' on CRAN: please do us the courtesy >>of fixing it so we...

there are really two related problems here I have a 2D matrix A <- matrix(1:100,nrow=20,ncol =5) S <- matrix(1:10,nrow=2,ncol =5) #I want to subtract S from A. so that S would be subtracted from the first 2 rows of #A, then the next two rows and so on. #I have a the same problem with a 3D array # where I want to subtract Q for every layer (1-10) in Z # I thought I solved this one