search for: aaply

...), 1:(C-1)] } Now, I want to extend this to higher-way arrays, using apply-like methods over the strata (all but the first two dimensions), and returning an array in which the last dimensions correspond to strata. That is, I want to define something like the following using an a*ply method, but aaply gives a result in which the applied .margin(s) do not appear last in the result, contrary to the documentation for ?aaply. I think this is a bug, either in the function or the documentation, but perhaps there's something I misunderstand for this case. fun <- function(f, stratum=NULL) {...

Dear helpers, I'm using plyr to process a large matrix for the first time. My code is set up to work with matrixes, since I learned the hard way that dataframes are considerably slower to process. I started using aaply(), but the data was rearranged from a flat matrix to a [, , 4] array for larger input matrixes. I'm sure something clever is happening that I'm just not seeing - anyone have any insight? I can provide the code for index.frames() if you like, but it's pretty turgid stuff. For now I'...

...able 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't do it --...

Oh, this seemed so simple (and I'm sure the answer will be, as usual, so thanks in advance for enlightening me). I need to sort each row of a matrix independent of the others. For example, > test <- matrix(c(8,7,1,2,6,5,9,4,3),nrow=3) > test [,1] [,2] [,3] [1,] 8 2 9 [2,] 7 6 4 [3,] 1 5 3 I can get each row sorted well enough. > sort(test[1,]) [1]

Dear list, I was curious how to subtract a vector from matrix? Say, I have mat <- matrix(1:40, nrow=20, ncol=2) x <-c(1,2) I want, x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements of x against column elements in mat. But x-mat won't do it. Thanks, Mike [[alternative HTML version deleted]]

. HI there, I'd like to show demonstrate how the chi-squared distribution works, so I've come up with a sample data frame of two categorical variables y<-data.frame(gender=sample(c('Male', 'Female'), size=100000, replace=TRUE, c(0.5, 0.5)), tea=sample(c('Yes', 'No'), size=100000, replace=TRUE, c(0.5, 0.5))) And I'd like to create a list of 100

Hi. I have 2 tables, with same dimensions (8000 x 5). Something like: tab1: V1 V2 V3 V4 V5 14.23 1.71 2.43 15.6 127 13.20 1.78 2.14 11.2 100 13.16 2.36 2.67 18.6 101 14.37 1.95 2.50 16.8 113 13.24 2.59 2.87 21.0 118 tab2: V1 V2 V3 V4 V5 1.23 1.1 2.3 1.6 17 1.20 1.8 2.4 1.2 10 1.16 2.6 2.7 1.6 11 1.37 1.5 2.0 1.8 13 1.24 2.9 2.7 2.0 18 I need generate a table of averages, the

...ype of object they input (first letter) and output (second letter): * llply = from a list to a list * alply = from an array (or vector, or matrix) to a list * ldply = from a list to a data.frame * d_ply = from a data.frame, ignore output * and so on for llply, laply, ldply, l_ply, alply, aaply, adply, a_ply, dlply, daply, dply, d_ply plyr also provides: * m*ply which works in a similar way to mapply * r*ply which works in a similar way to replicate You can find out more at http://had.co.nz/plyr/, including a 20 page introductory guide, http://had.co.nz/plyr/plyr-intro.pdf. Regard...

Folks, I've got a matrix x as follows: > x <- matrix(c(1,2,3,5,3,4,3,2,1), ncol = 3, byrow = TRUE) > x [,1] [,2] [,3] [1,] 1 2 3 [2,] 5 3 4 [3,] 3 2 1 In each row of x, I want to replace the minimum value by -1, the maximum value by +1 and all other values by 0. So in the above case I want to end up as follows: [,1] [,2] [,3] [1,] -1 0

...ype of object they input (first letter) and output (second letter): * llply = from a list to a list * alply = from an array (or vector, or matrix) to a list * ldply = from a list to a data.frame * d_ply = from a data.frame, ignore output * and so on for llply, laply, ldply, l_ply, alply, aaply, adply, a_ply, dlply, daply, dply, d_ply plyr also provides: * m*ply which works in a similar way to mapply * r*ply which works in a similar way to replicate You can find out more at http://had.co.nz/plyr/, including a 20 page introductory guide, http://had.co.nz/plyr/plyr-intro.pdf. Regard...

I have got a general problem when applying a function to a dataframe using the function; Apply(df,1,myfunct) Where myfunct has an IF statement in it along the lines of; If (z == 0) X = 1 If (z ==0) Z = 1 I.e. Two If statements Is there something I am missing or have a just formed the if statements wrong ­ just checking there is not some trick you have to use when using apply with functions

I have a list of data frames like the following: set.seed(123) a<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10)) b<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10)) L<- list(a,b) All data frames in the list have the same dimensions. I need to calculate the sample means for x and y. The real data are lists of several thousand quite large dataframes, so I need

Dear all, I have a question regarding the possibility of parallel computation in plyr version 1.7. The help files of the following functions mention the argument '.parallel': ddply, aaply, llply, daply, adply, dlply, alply, ldply, laply However, the help files of the following functions do not mention this argument: ?d_ply, ?aply, ?lply Is it because parallel computation is not supported for these latter functions? Or is it just because the documentation was not updated for these...

Hi. On Sun, Feb 13, 2011 at 10:18 AM, TakeoKatsuki <takeo.katsuki at gmail.com> wrote: > > Hi Henrik, > > It would be nice if functions of the matrixStats package can handle array > data. > For example, rowSums() of the base package sums along the third axis of an > array by rowSums(x, dim=2). That is a good idea. This was indeed the initial objective before starting

Hey all, I have a list of matrices. I'd like to calculate the median across all those matrices for each element. What I'd like to end up with is a matrix containing the median of all [1,1] [1,2] etc. elements across all matrices. Is there a concise way of doing that? Thanks!

...* massive speed ups for splitting large arrays * fixed typo that was causing a 50% speed penalty for d*ply * rewritten rbind.fill is considerably (> 4x) faster for many data frames * colwise about twice as fast Bug fixes: * daply: now works when the data frame is split by multiple variables * aaply: now works with vectors * ddply: first variable now varies slowest as you'd expect plyr 0.1.5 (2008-02-23) --------------------------------------------------- * colwise now accepts a quoted list as its second argument. This allows you to specify the names of columns to work on: colwise(mean...

...* massive speed ups for splitting large arrays * fixed typo that was causing a 50% speed penalty for d*ply * rewritten rbind.fill is considerably (> 4x) faster for many data frames * colwise about twice as fast Bug fixes: * daply: now works when the data frame is split by multiple variables * aaply: now works with vectors * ddply: first variable now varies slowest as you'd expect plyr 0.1.5 (2008-02-23) --------------------------------------------------- * colwise now accepts a quoted list as its second argument. This allows you to specify the names of columns to work on: colwise(mean...

Dear list, I'm working with a large data set, where I grouped several species in one group (guild). Then I reshaped my data as shown below. Now, I just want to have "Rep" only as 1 or 0. I'm not being able to change the values of rep>=1 to 1... tried many things and I'm not being successful! > melting=melt(occ.data,id.var=c("guild", "Site",

Dear all, I have a dataset that looks like this: x <- read.table(textConnection("col1 col2 3 1 2 2 4 7 8 6 5 10"), header=TRUE) I want to rewrite it as below: var1 var2 var3 var4 var5 var6 var7 var8 var9 var10 1 0 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 1 0 0

Can you please help on the issue? I using the apply command on a matrix below the example: Create a vector x =c(5, 3, 2:4, NA, 7, 3, 9, 2, 1, 5) create a matrix of 2 rows by 6 columns b=matrix(x, 2,6) print(b) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 5 2 4 7 9 1 [2,] 3 3 NA 3 2 5 using the command apply print(apply(b, 1, function(y) sort(y, na.last=F))) the