similar to: when to use which apply function?

Friends I clearly donot understand how sapply and vapply work. What I have is a function that returns a matrix with an indeterminate number of rows (some times zero) but a constant number of columns. I cannot reliably use an apply function to assemble the matrices into a matrix. I am not sure it is possible. I can demonstrate the core of my confusion with this simple code. A.f <-

Not paying close attention to detail, I entered the equivalent of pstr<-c("b1=200", "b2=50", "b3=0.3") when what I wanted was pnum<-c(b1=200, b2=50, b3=0.3) There was a list thread in 2010 that shows how to deal with un-named vectors, but the same lapply solution doesn't seem to work here i.e., pnum<-lapply(pstr, as.numeric) or similar vapply

Hi, Both vapply() and sapply() support the 'USE.NAMES' argument. According to the man page: USE.NAMES: logical; if ?TRUE? and if ?X? is character, use ?X? as ?names? for the result unless it had names already. But if 'X' has names already and 'USE.NAMES' is FALSE, it's not clear what will happen to the names. Are they going to propagate to the result

Changing the vapply() behavior makes sense in principle. I analyzed the CRAN code base using the R parser and found 143 instances of calling vapply with USE.NAMES=FALSE. These would need to be inspected to understand the consequences of the change. For reference: /AzureML/R/datasets.R:226 /BBmisc/R/toRangeStr.R:33 /DBI/R/DBDriver.R:205 /Kmisc/R/str_rev.R:37 /Matrix/R/diagMatrix.R:98

Hi all, The function call series genericForPrimitive -> .findBasicFuns -> .findAll happens 4400 times while the GenomicRanges package is loading. Each time .findAll follows a chain of environments to determine that the methods namespace is the only one that holds a variable called .BasicFunsList. This accounts for ~10% of package loading time. I'm sure there is some history to that

I am trying to calculate the weighted mean for a of 10 deciles and I get an error: > decile <- tapply(X=mat$trt1m, INDEX=mat$Rank, FUN=weighted.mean, w=mat$mcap) Error in FUN(X[[1]], ...) : 'x' and 'w' must have the same length All three of my inputs have the same length, as shown below, and the weighted.mean calculation works by itself, just not in tapply() >

Doing it like this: genericForPrimitive <- function(f, where = topenv(parent.frame()), mustFind = TRUE) { ans = .BasicFunsList[[f]] ## this element may not exist (yet, during loading), dom't test null if(mustFind && identical(ans, FALSE)) stop(gettextf("methods may not be defined for primitive function %s in this version of R",

Hello, I've been dealing with a set of values that contain time stamps and part of my summary needs to look at just weekend data. In trying to limit the data I've found a large difference in performance in the way I index a data frame. I've constructed a minimal example here to try to explain my observation. is.weekend <- function(x) { tm <-

I am running into a limitation of the grid::grid.newpage function, for which I would like to overwrite this function with a slightly modified one. Hopefully this is a temporary working solution until the package gets updated. I found a way to overwrite the function in the package:grid namespace. However, lattice imports grid rather than depending on it. So I need a way to overwrite this imported

Dear list, I'm trying to create a matrix by combining the sites that species occur in a new matrix with species as rows and sites as columns. The main matrix is "mat": mat <- as.data.frame(cbind(sp1=c(rep(0, 5), rep(1, 5)),sp2=sample(c(rep(0, 6),rep(1, 4))), fac=c(rep("a", 3), rep("b", 3),rep("c", 4)))) The first two columns are species and the

Dear helpers in this forum, This is a clarified version of my previous questions in this forum. I really need your generous help on this issue. > Suppose I have the following data set: > > id x y > 023 1 2 > 023 2 5 > 023 4 6 > 023 5 7 > 412 2 5 > 412 3 4 > 412 4 6 > 412 7 9 > 220 5 7 > 220 4 8 > 220 9 8 > ...... > Now I want to compute the

Hi R community, I have a question concerning passing arguments between apply and lapply? Or maybe, once my problem is explained, the question is really about how to best transform my nested for loops into list/matrix operations; I am just beginning this transformation away from nested for loops, so I beg of you to have some lenience regarding my ignorance. Part I: I used a set of nested for

> > If you want to define an alias symbol "bar" to "foo" (which is an > extension you want to provide), one way is to create an object file that > defines "bar" and "__imp_bar" as aliases to "foo" and "__imp_foo", > respectively, and add that object file to the import library. As a result, > the import library file

Hello, I have a data.frame > names(popA) [1] "Year" "Series" "Age" "WM" "WF" "HM" "HF" "BM" [9] "BF" "IM" "IF" "AM" "AF" "Yr" how do i loop over a subset of variables in this frame to create a vector of

vapply <- function(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE) { FUN <- match.fun(FUN) if(!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(vapply(X, FUN, FUN.VALUE, USE.NAMES)) } This is an implementor question. Basically, what happened to the '...' args in the call to the .Internal? cf lapply:, where the ... is passed. lapply <- function (X, FUN, ...)

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,

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]]

The following code seems to contain an inconsistency in the behavior of vapply(). Am I missing something here? ## This function assumes v is a 3d vector, beta a scalar. f3d <- function(v,beta) { v+beta } ## This expression applies f3d to a vector of scalars, and ## specifies the template 'array(10,3)' for the return value. dat <- vapply(seq(0,1,length=10), function(beta) {

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan

FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut some corners compared to identical(): > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) Unit: nanoseconds expr min lq mean median uq max neval identical(FALSE, FALSE) 984 1138 1694.13 1218.0 1337.5 13584 100 isFALSE(FALSE) 713 761 1133.53 809.5 871.5