similar to: Add a vector to the values in a specific dimension of an array

I'm attempting to sort a 3 dimensional array that looks like this > x , , 1 [,1] [,2] [1,] 9 9 [2,] 7 9 , , 2 [,1] [,2] [1,] 6 5 [2,] 4 6 , , 3 [,1] [,2] [1,] 2 1 [2,] 3 2 Such that it ends up like this .... > y , , 1 [,1] [,2] [1,] 2 1 [2,] 3 2 , , 2 [,1] [,2] [1,] 6 5 [2,] 4 6 , , 3 [,1] [,2]

I have a multidimensional array - in this case with 4 dimensions of x,y,z and time. I'd like to take the time derivative of this matrix, i.e. perform the diff operator along dimension number 4. In Matlab, there is a simple option to specify which dimension the difference is to be taken across., but I can't see this in R. I realise I can use aperm to rotate my array so that time is in

I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, "workers", sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks

Hello all, I'm working with a matrix that will have varying dimensions. It will populate an array such that the number of matrix columns will determine the number of 3rd dimension levels of the array. Rows will be the same for both. For this example lets say the array will have 2 columns, but that's not fixed. dim(arr)<-c(dim(mat)[1],2,dim(mat)[2]) I wish to repeat each matrix

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

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

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

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

Dear r-help, I have an array a1 with dimensions [1:660,1:65,1:25] I would like the first dimension to be the last one. That is I want and array [1:65,1:25,1:660] The only way to do this, I know, is tmp.a<-array(dim=dim(a1)[c(2,3,1)]) for(i in 1:dim(a1)[1]) tmp.a[,,i]<-a1[i,,] a1<-tmp.a rm(tmp.a) Is it possible to avoid 'for' loop here? Thank you! ---

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

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

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

Dear all, When I coerce a vector into a multi dimensional array, I would like R to start filling the array along the last dimension, then the 2nd last etc. Let's jump straight into an example. x <- 1 : 24 y <- array(dim=c(2,2,6)) I would like to have: y[1,1,1] = 1 y[1,1,2] = 2 ... y[1,1,6] = 6 y[1,2,1] = 7 y[1,2,2] = 8 ... y[2,1,1] = 13 ... y[2,2,1] = 19 if I do y<- array(x,

Hi, I would like to write a one-line R code to create a 3-dim array, B, of dimension (n,n,m) from a matrix, A, of dimension (m,n) such that the i-th element of the 3-dim array, B[, , i] is the outer product of the i-th row of A. Thanks for any help, Ravi. [[alternative HTML version deleted]]

Hello! Is it posible to apply /cumsum()/ along the 3rd dimension of 3D array? Something like matrlab function - /cumsum (*A*,dim)/ which returns the cumulative sum of the elements along the dimension of *A* specified by scalar dim. Thanks in advance ?eljka -- View this message in context: http://r.789695.n4.nabble.com/cumsum-in-3d-arrays-tp4110470p4110470.html Sent from the R help mailing

Dear R-users, I would like to apply a function (more precisely sd()) over the third dimension of a three-dimension array. The function apply would be interesting but the chosen function can only be applied on the rows and columns of the array according to the help file. I can use a loop to cut the array in matrices and then use apply for each replication, but it's not very nice. A small

I am trying to multiply a 3D array of 4x4x4 by the 4 3D arrays of a 4D array with dimensions 4x4x4x4 (the last dimension being the one that I want to split by). (4x4x4 array) > hiaAry , , a1 i1 i2 i3 i4 h1 9.5936098 6.001040 0.08772 0.3138600 h2 1.2003500 1.454570 2.79248 0.0000000 h3 0.1346500 0.201220 0.39256 0.5464000 h4 0.0109000 0.012270 0.16417 0.2766900 ,

Hi everyone, I created a four dimensional vector (dim (128,128,1,8)). This third dimension is necessary for another function somewhere. Now I'd like to perform a t-test on every vector of length 8 in my array on the fourth dimension. I'd like to obtain a new array of three dimensions with dimensions 128x128x1 with all these test statistics. I tried this with a double loop: A <-

If I have a two column data frame like: > dat <- cbind("x"=c(1:100),"y"=c(100:1)) How can I create an array that splits every ten rows of that data frame into a third dimension of an array so that: > newarray[,,1] ,,1 x y 1 100 2 99 3 98 ... ... 10 91 ,,2 x y 11 90 12 89 ... ... ... Thanks. [[alternative HTML version deleted]]

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