similar to: Problem with {

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 <-

I did not realize that ++ was available. Is there a comprehensive list somewhere of which operators are available for definition? I searched the R Language manual for ++ but that only came up with a reference to C++ . On 17 Nov 2006 11:38:43 +0100, Peter Dalgaard <p.dalgaard at biostat.ku.dk> wrote: > rfrancois at mango-solutions.com writes: > > > Hi, > > > > You

Within a function I'm assigning dynamic variable names and values to them using the "assign" function. I want to pass back the results but am uncertain how to do this. Basically, my function reads a number of data files and uses the filename of each file as the variable name for a list-to-become-dataframe. I want then to pass all these lists back, but again, the names of the

Dear all, I am working with a list of objects each of which contains two POSIXct objects (say, $Start and $End) and a number of different data in addition to that. Now an easy way to extract Start times of all object could be sapply(x, "[", "Start") but this converts them all to numeric, and so does sapply(x, "[[", "Start"). lapply preserves the class but

Le 19/04/2020 ? 20:46, Gabor Grothendieck a ?crit?: > You can get pretty close to that already using fn$ in the gsubfn package: >> library(gsubfn) fn$sapply(split(mtcars, mtcars$cyl), x ~ >> summary(lm(mpg ~ wt, x))$r.squared) > 4 6 8 0.5086326 0.4645102 0.4229655 Right, I thought about similar syntax but this implementation has similar flaws pointed by Simon, i.e. it reduces

Hi Consider the following example: f <- function(expr) g(expr) g <- function(expr) { ? h(expr) } h <- function(expr) { ? expr # evaluation happens here ? i(expr) } i <- function(expr) { ? expr # already evaluated, no costs here ? invisible() } rprof <- tempfile() Rprof(rprof) f(replicate(1e2, sample.int(1e4))) Rprof(NULL) cat(readLines(rprof), sep = "\n") #>

> On Feb 25, 2015, at 5:35 PM, Benjamin Tyner <btyner at gmail.com> wrote: > > Actually, it depends on the number of cores: Under current semantics, yes. Each 'stream' of function calls is lazily capturing the last value of `i` on that core. Under Luke's proposed semantics (IIUC), the result would be the same (2,4,6,8) for both parallel and serial execution. This is

Hi, I bet this one has be asked before, but doing sapply(x, FUN=as.character) where 'x' is a vector, then the result "should [] be simplified to a vector" according to ?sapply, correct? However, > x <- 1:10 > sapply(x, FUN=as.character) [1] "1" "2" "3" "4" "5" "6" "7" "8"

Yesterday I posted the following question (my apologies for not putting a subject line): =================question====================== Hello -- I would like to know of a more efficient way of writing the following piece of code. Thanks. options(stringsAsFactors=FALSE) orig <- c(rep('11111111',100000),rep('22222222',200000),rep('33333333'

> On Feb 24, 2015, at 10:50 AM, <luke-tierney at uiowa.edu> wrote: > > The documentation is not specific enough on the indented semantics in > this situation to consider this a bug. The original R-level > implementation of lapply was > > lapply <- function(X, FUN, ...) { > FUN <- match.fun(FUN) > if (!is.list(X)) > X <-

I've been trough the R documentation for about half an hour and it's not clear to me how to do this: I need to format to character a series of integers from 1 to 1000, and I like them to look like "0001" "0002", "0059", "0123" and so on. Padded with zeroes to have four digits. Cheers! Mario. r-help-request at r-project.org wrote: > Send

Hello everybody, I have problem with a lapply command, which rather proves that I don't fully understand it. I want to extract from a list that consists of dataframes, the length of the first sequences from a given variable (its part of a simulation exercises). Below is code which does the job, but I think it should be possible to make it more compact. ### Example Data dat <-list()

Hello everyone, I cannot seem to find information about objects of class "matrix" and mode "list", and how to handle them (apart from flattening the list). I get this type of object from using sapply(). Sorry for the long example, but the code below illustrates how I get this type of object. Is anyone aware of documentation regarding this object? Thanks very much, Stephen

Hello, I have a data manipulation problem that I can easily resolve by using perl or python to pre-process the data, but I would prefer to do it directly in R. Given, for example: month length ratio monthly1 monthly2 1 Jan 23 0.1 9 6 2 Jan 45 0.2 9 6 3 Jan 16 0.3 9 6 4 Feb 14 0.2 1 9 5 Mar 98 0.4 2 2 6 Mar 02 0.6

Dear List: I am running into a memory issue that I haven't noticed before. I am running a simulation with all of the code used below. I have increased my memory to 712mb and have a total of 1 gb on my machine. What appears to be happening is I run a simulation where I create 1,000 datasets with a sample size of 100. I then run each dataset through a gls and obtain some estimates. This works

Hi All, I am confused about the following code. I thought that the problem stemmed from lazy evaluation and the fact that 'i' is never evaluated within the first lapply. However, I am then confused as to why it gets bound to the final element of the lapply. The environments of the returned functions are indeed different in both cases and each environment has a local binding for

Hi everybody, with the following code I generate a list of functions. Each function reflects a "condition". When I evaluate this list of functions by another lapply/sapply, I get an unexpected result: all values coincide. However, when I uncomment the print(), it works as expected. Is this a bug or a feature? conditions <- 1:4 test <- lapply(conditions, function(mycondition){

From: Daniel Kaschek <daniel.kaschek at physik.uni-freiburg.de> > ... When I evaluate this list of functions by > another lapply/sapply, I get an unexpected result: all values coincide. > However, when I uncomment the print(), it works as expected. Is this a > bug or a feature? > > conditions <- 1:4 > test <- lapply(conditions, function(mycondition){ >

Hi, There are quiet a few different 'apply' functions, such as lapply, sapply and many more. I'm very familiar with the 'Apply' function in Mathematica. Can somebody point me a summary of all the 'apply' functions in R. Also, I'm curious that what 'l' and 's' (and other prefixes) stand for in 'lapply' and 'sapply' Regards, Peng

Wouldn't that change how simplify='array' is handled? > str(sapply(1:3, function(x)diag(x,5,2), simplify="array")) int [1:5, 1:2, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=TRUE)) int [1:10, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=FALSE)) List of 3 $ : int [1:5, 1:2] 1 0 0 0 0 0 1 0 0