similar to: apply() returning a list?

Hello, I have been experiencing an odd error with the RGL package that I cannot figure out. I have searched without luck for fixes to this problem on Rseek, Google, and the R-help-list. RGL used to work on my machine about a month ago. It was a very useful tool. The graphics device will not initialize now, and I cannot figure out why. When I try to load the package using ... >

Hi Yeah I ended up figuring it out My usage for the tool will be to gather hardware specs of new servers that we receive and servers we have in storage I will then see if I can script something to phrase the dumps to update our inventory tool. Our main issue is that gathering information manually by opening the server and having to locate each parts is time consuming and being humans we are

The Bioconductor build for a package DirichletMultinomial on R Under development (unstable) (2013-02-26 r62077) -- "Unsuffered Consequences" at http://bioconductor.org/checkResults/devel/bioc-LATEST/DirichletMultinomial/george2-buildsrc.html shows * creating vignettes ... ERROR ... Error: processing vignette ?DirichletMultinomial.Rnw? failed with diagnostics: chunk 21 (label

Hello all I would like to ask your help use mapply. I have a function called findCell that takes two arguments(x,sr) where x is a vector of size two (e.g x<-c(2,3) and sr is a matrix. I would like to call many times the findCell function (thus I need mapply) for different x inputs but always for the same sr. as x is a vector of size two (two cells) I want to pass inside inside the following

I need to develop a simple list manipulation. Although it seems easier to do it in matrix form, but I need it in list form. I have a matrix x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2) for list form example, the conversion is x.list <- lapply(seq_len(nrow(x)), function(i) x[i,]) ### list version calcnorm=function(a, b){ diff <- mapply("-", a, b)

dear all, Given a matrix A, say, I would like to apply a bivariate function to each combination of its colums. That is if myfun<-function(x,y)cor(x,y) #computes simple correlation of two vectors x and y then the results should be something similar to cor(A). I tried with mapply, outer,...but without success Can anybody help me? many thanks in advance, vito

Hello, I'm not quite sure I understand how do you use HDT for dumping over tftp server. The wiki on dumping does not offer much details on how to set this up. >From my understanding is you need to run pxelinux somewhere but being that HDT is a bootable cd ISO, I don't see how I can get HDT up from a TFTP server without the need to install something on that TFTP server. Is there a more

Hello! I have noticed that mapply() drops names in R 2.3.1 as well as in r-devel. Here is a simple example: l <- list(a=1, b=2) k <- list(1) mapply(FUN="+", l, k) [1] 2 3 mapply(FUN="+", l, k, SIMPLIFY=FALSE) [[1]] [1] 2 [[2]] [1] 3 Help page does not indicate that this should happen. Argument USE.NAMES does not have any effect here as it used only in a bit special

I know this is something simple that I cannot do because I do not yet "think" in R. I have a data frame has a variable participation (a factor), and several other factors. I want a chisq test (no contingency tables) for participation vs all of the other factors. In SPSS I would do: CROSSTABS /TABLES= (my other factors) BY participation /FORMAT=NOTABLES /STATISTICS=CHISQ

Dear R community, I'm running "a mlogit" function with a very large data set. Since the data size is so large, I got the Error: Error: cannot allocate vector of size 5.4 Mb In addition: Warning messages: 1: In mapply("*", X, P, SIMPLIFY = FALSE) : Reached total allocation of 16340Mb: see help(memory.size) 2: In mapply("*", X, P, SIMPLIFY = FALSE) : Reached

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

Here are several ways to get there, but your original loop is fine once it is corrected: > for (i in 1:2) smean[i] <- mean(toy$diam[toy$group==i][1:nsel[i]]) > smean [1] 0.271489 1.117015 Using sapply() to hide the loop: > smean <- sapply(1:2, function(x) mean((toy$diam[toy$group==x])[1:nsel[x]])) > smean [1] 0.271489 1.117015 Or use head() > smean <- sapply(1:2,

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 > > 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),

Hello. Please forgive me if this problem has already been posted (and solved) by someone else ... I can't find it anywhere though it seems so very basic. Here it is: I have a list comprised of several matrices, each of which has two columns. > list [[1]] [,1] [,2] [1,] 1 3 [2,] 2 4 [[2]] [,1] [,2] [1,] 5 7 [2,] 6 8 [[3]] [,1] [,2]

Does any know of a way to sync the hardware clock without a reboot? Graham Johnston Senior Network Analyst Westman Communications Group 204.725.4300 ext 382 johnstong at westmancom.com -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.centos.org/pipermail/centos/attachments/20060719/70ef23ce/attachment-0002.html>

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

Hello, I would like to use mapply to avoid using a loop but for some reason, I can't seem to get it to work. I've included copies of my code below. The first set of code uses a loop (and it works fine), and the second set of code attempts to use mapply but I get a "subscript out of bounds" error. Any guidance would be greatly appreciated. Xj, Yj, and Wj are also lists, and s2,

Dear all, #Last week, I asked about merge x and y as list. #Now I have a dataset with list of list like: x <- list(list(matrix(1:20, 5, 4),matrix(1:20, 5, 4)), list(matrix(1:20, 5, 4),matrix(1:20, 5, 4))) y <- list(list(c(1, -1, -1, 1, 1),c(1, 1, -1, -1, -1)), list(c(1, 1, 1, 1, 1),c(1, 1, -1, 1, -1))) x y #I need merge x and y, I have tried with list.uni <-

Hi, Starting with ordinary vectors, so we know what to expect: > mapply(function(x, y) {x * y}, 101:106, rep(1:3, 2)) [1] 101 204 309 104 210 318 > mapply(function(x, y) {x * y}, 101:106, 1:3) [1] 101 204 309 104 210 318 Now with an S4 object: setClass("A", representation(aa="integer")) a <- new("A", aa=101:106) > length(a)

Vectorize is defined to return a function that acts as if 'mapply' was called. So we have: > mapply(dput,1:2) # mapply form 1L # calls dput on each element of 1:2 2L [1] 1 2 > Vectorize(dput)(1:2) # Vectorize form 1L # same behavior 2L [1] 1 2 Same thing with a named argument: >