similar to: lapply version with [ subseting - a suggestion

Dear all, I am looking for some procedure to apply 'ifelse' condition on function. I have created an alternative to lapply() function with exactly same set of arguments named lapply1(), however with different internal codes. Therefore I want something like, if (some condition) then call lapply1() otherwise lapply() function. Ofcourse I can create a complete if()... else()... like of

Hey guys, I noticed something curious in the lapply call. I'll copy+paste the function call here because it's short enough: lapply <- function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(lapply(X, FUN)) } Notice that lapply coerces X to a list if the !is.vector || is.object(X) check passes. Curiously,

Hello and thank you both for your answers! Dennis, I tried to simply run lm(a ~ b) after re-importing "a" as a matrix, but I get the following error message: Error in model.frame.default(formula = a ~ b, drop.unused.levels = TRUE) : invalid type (list) for variable 'a' so maybe I have to specify something in the arguments? What do you think? David, I tried your syntax as

You could replace your 'depth' argument with one that shows where in the original data you are at: leaf.func <- function(data, where) { if(is.null(data)) stop("Null data at ", deparse(where)) return(mean(data)) } visit.level <- function(data, where = integer()) { if (length(where) == 2) { return(leaf.func(data, where)) } else

I have a dataset of property transactions that includes the transaction ID (TranID), property ID (UnitID), and transaction date (TranDt). I need to create a data frame (or data table) that includes the previous transaction date, if one exists. This is an easy problem in SQL, where I just run a sub-query, but I'm trying to make R my one-stop-shopping program. ?The following code works on a

Hi, by replacing 'll' with 'wh' in the source code for base::which() one gets ~20% speed up for *named logical vectors*. CURRENT CODE: which <- function(x, arr.ind = FALSE) { if(!is.logical(x)) stop("argument to 'which' is not logical") wh <- seq_along(x)[ll <- x & !is.na(x)] m <- length(wh) dl <- dim(x) if (is.null(dl)

This might be just beyond of my understanding of how assignment works in R, but the documentation does not say anything about: > tv <- c(a="dsf", b="sss") > tl <- list(232) > `$<-`(tl, tv[[1]], "sdfdsfdsfsd") Error: invalid subscript type 'language' This happens even before the method is dispatched. I can not handle the "name"

I'm trying to understand the behavior of seq_along in the following example: x <- 1:5; sum(x) y <- 6:10; sum(y) data <- c(x,y) S <- sum( data[seq_along(x)] ) S T <- sum( data[seq_along(y)] ) T Why is T != sum(y) ?

I am moving this from r-help to r-devel. Based on offline communications with Jim, suppose dat is defined as follows: set.seed(123) dat <- data.frame(ID= c(rep(1,2),rep(2,3), rep(3,3), rep(4,4), rep(5,5)), var1 =rnorm(17, 35,2), var2=runif(17,0,1)) # Then this ave call works as expected: ave(dat$ID, dat$ID, FUN = function(x) seq_along(x)) # but this apparently identical calculation

Using the IRanges package from Bioconductor and somewhat recent R-2.9.1. ov = IRanges(1:3, 4:6) length(ov) # 3 seq(along = ov) # 1 2 3 as wanted seq_along(ov) # 1! I had expected that the last line would yield 1:3. My guess is that somehow seq_along don't utilize that ov is an S4 class with a length method. The last line of the *Details* section of ?seq has a typeo. Currently it is

Hello, I am forwarding this from ESS mailing list, as it's a failure of internal R completion system: This fails: utils:::.assignLinebuffer('iris[iris$Spec') utils:::.assignEnd(15) utils:::.guessTokenFromLine() utils:::.completeToken() utils:::.retrieveCompletions() ## -> [1] "iris[iris$Spec" This works utils:::.assignLinebuffer('iris[ iris$Spec') # note the

Hi! I have a 3d array: > dim(ib5km15.dbc) [1] 190 241 19 and a set of positions to extract: > ib5km.lincol.random[1:3,] [,1] [,2] [1,] 78 70 [2,] 29 213 [3,] 180 22 Geting the values of a 2D array for that set of positions would be: > ima <- ib5km15.dbc[,,1] > ima[ib5km.lincol.random[1:10,]] but don't find the way for the case of the 3D array: >

Hi I see a changed behaviour in xts indexed on class Date in the latest versions, versus 2. It seems to be related to changes to/from daylight savings time, happens those weekends. Is it not intended that class Date be used like this, or is this new behaviour incorrect? Giles Example: > a<-as.Date(15423:15426) > x<-xts(seq_along(a),a) > print(x) [,1] 2012-03-24

Hi all, A couple of ideas for improving seq_along: * It would be really useful to have a second argument dim: seq_along(mtcars, 1) seq_along(mtcars, 2) # equivalent to seq_len(dim(mtcars)[1]) seq_len(dim(mtcars)[2]) I often find myself wanting to iterate over the rows or column of a data frame, and there isn't a particularly nice idiom if you want to avoid problems

In R-3.6.0 autoprinting was changed so that print methods for the storage modes are not called when there is no explicit class attribute. E.g., % R-3.6.0 --vanilla --quiet > print.function <- function(x, ...) { cat("Function with argument list "); cat(sep="\n ", head(deparse(args(x)), -1)); invisible(x) } > f <- function(x, ...) { sum( x * seq_along(x) ) }

Dear R Helpers, I am missing something very elementary here, and I don't seem to get it from the help pages of the ave, seq and seq_along functions, so I wonder if you could offer a quick help. To use an example from an earlier post on this list, I have a dataframe of this kind: dat = data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4))) dat$freq =

Dear list! ? I have question of?'correct function formation'. Which function (fun1 or fun2; see below) is written more correctly? Using ''structure'' as output or creating empty ''data.frame'' and then transform it as output? (fun1 and fun1 is just for illustration). ? Thanks a lot, OV ? code: input <- data.frame(x1 = rnorm(20), x2 = rnorm(20), x3 =

Dear useRs,while trying to plot the yearly curves of 1000 stations and overlapping each set of curves with mean curve and then saving it automatically in a pdf file, i tried the following commands >Path = "C:\\R\\003.pdf">pdf(file=Path) for (i in seq(1:1000) >a<-lapply(seq_along(tcp), function(x) tcp[[x]][,-1]) >b<-lapply(seq_along(a), function(a)

It also is a problem with storage.modes "integer" and "complex": 3.6.0> print.integer <- function(x,...) "integer vector" 3.6.0> 1:10 [1] 1 2 3 4 5 6 7 8 9 10 3.6.0> print(1:10) [1] "integer vector" 3.6.0> 3.6.0> print.complex <- function(x, ...) "complex vector" 3.6.0> 1+2i [1] 1+2i 3.6.0>

Below is a version of [.data.frame that is faster for subscripting rows of large data frames; it avoids calling duplicated(rows) if there is no need to check for duplicate row names, when: i is logical attr(x, "dup.row.names") is not NULL (S+ compatibility) i is numeric and negative i is strictly increasing "[.data.frame" <- function (x, i, j,