search for: simplify2array

Should simplify2array(higher=TRUE) treat 1 by 1 matrices differently than others? I expected a 3-dimensional array from all of the following 3 examples, not just the last 2. > str(simplify2array(list(array(11,c(1,1)), array(21,c(1,1))), higher=TRUE)) num [1:2] 11 21 > str(simplify2array(list(array(11:13,c(...

Reading the doc for ?simplify2array, I got the impression that with the 'higher = T' argument the function returns an array of dimension greater than 2 when it makes sense (the doc says "when appropriate", which is rather vague). I would expect a <- list( list(list(1, 2), list(3, 4)), list(list(5, 6), list(7...

Jean-Claude: Well, here's my "explanation". Caveat emptor! Note that: "simplify2array() is the utility called from sapply() when simplify is not false" and > sapply(a, I, simplify = "array") [,1] [,2] [1,] list,2 list,2 [2,] list,2 list,2 So it seems that simplify2array() is not intended to operate in the way that you expected, i.e. that recursive simplif...

...()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > library(parallel) > simplify2array(mclapply(rep(4, 5), rnorm)) *** caught segfault *** address 0x7fee2a7229d0, cause 'memory not mapped' Traceback: 1: mcexit(0L) 2: FUN(1:2[[1L]], ...) 3: lapply(seq_len(cores), inner.do) 4: mclapply(rep(4, 5), rnorm) 5: lapply(x, length) 6: unlist(lapply(x, length)) 7: unique(unlis...

i'm somehow embarrassed to even ask this, but is there any built-in method for doing this: my_list <- list() my_list[[1]] <- matrix(1:20, ncol = 5) my_list[[2]] <- matrix(20:1, ncol = 5) now, knowing that these matrices are identical in dimension, i'd like to unfold the list to a 2x4x5 (or some other permutation of the dim sizes) array. i know i can initialize the array, then

...\"j\")" [3] "c(\"k\", \"l\")" > as.vector(x) [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] "f" "g" "h" "i" "j" [[3]] [1] "k" "l" > simplify2array(x) [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] "f" "g" "h" "i" "j" [[3]] [1] "k" "l" What I would need to get instead is: > letters[1:12] [1] "a" "b" &quot...

Hi, Suppose you created a dataframe like this: set.seed(28) ?dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 5 obs. of? 3 variables: # $ V1: chr? "a" "b" "c" "d" ... # $ V2: chr? "1" "2" "10" "18" ... # $ V3: chr?...

I am curious if anyone knows of R code where the "{" function is redefined in a useful way. Or "(" for that matter. Thanks Mick

...t;- function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE){ FUN <- match.fun(FUN) answer <- lapply(X = X, FUN = FUN, ...) if (USE.NAMES && is.character(X) && is.null(names(answer))) names(answer) <- X if (simplify && length(answer)) simplify2array(answer, higher = (simplify == "array")) else answer } > microbenchmark(sapply(myList, length), times = 10000L) Unit: microseconds expr min lq mean median uq max neval sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000 &g...

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")

Hi experts. I have a tibble? with a column containing a nested list (<list<list<double>>>? data type to be specific). Looks something like the following (but in R/Arrow? format): ID Nestedvals 001 [[1]](1,0.1)[[2]](2,0.2)[[3]](3,0.3)[[4]](4,0.4)[[5]](5,0.5) 002 [[1]](1,0.1)[[2]](2,0.2)[[3]](3,0.3)[[4]](4,0.4) 003 [[1]](1,0.1)[[2]](2,0.2)[[3]](3,0.3) 004 [[1]](1,0.1)[[2]](2,0.2)

...= TRUE, USE.NAMES = TRUE){ > FUN <- match.fun(FUN) > answer <- lapply(X = X, FUN = FUN, ...) > if (USE.NAMES && is.character(X) && is.null(names(answer))) > names(answer) <- X > if (simplify && length(answer)) > simplify2array(answer, higher = (simplify == "array")) > else answer > } > > > > microbenchmark(sapply(myList, length), times = 10000L) > Unit: microseconds > expr min lq mean median uq max neval > sapply(myList, length) 14.156 15.572...

Hello, I spent a lot of a time on a weird bug, and I just managed to narrow it down. In parallel code (here with parallel::mclappy, but I got it doMC/multicore too), if the library(tcltk) is loaded, R hangs when trying to open a DB connection. I got the same behaviour on two different computers, one dual-core, and one 2 xeon quad-core. Here's the code: library(parallel) library(RSQLite)

...unction (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE){ FUN <- match.fun(FUN) answer <- lapply(X = X, FUN = FUN, ...) if (USE.NAMES && is.character(X) && is.null(names(answer))) names(answer) <- X if (simplify && length(answer)) simplify2array(answer, higher = (simplify == "array")) else answer } > microbenchmark(sapply(myList, length), times = 10000L) Unit: microseconds expr min lq mean median uq max neval sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000 &g...

...fy = TRUE, USE.NAMES = TRUE){ > FUN <- match.fun(FUN) > answer <- lapply(X = X, FUN = FUN, ...) > if (USE.NAMES && is.character(X) && is.null(names(answer))) > names(answer) <- X > if (simplify && length(answer)) > simplify2array(answer, higher = (simplify == "array")) > else answer > } > > >> microbenchmark(sapply(myList, length), times = 10000L) > Unit: microseconds > expr min lq mean median uq max neval > sapply(myList, length) 14.156 15....

...fy = TRUE, USE.NAMES = TRUE){ > FUN <- match.fun(FUN) > answer <- lapply(X = X, FUN = FUN, ...) > if (USE.NAMES && is.character(X) && is.null(names(answer))) > names(answer) <- X > if (simplify && length(answer)) > simplify2array(answer, higher = (simplify == "array")) > else answer > } > > >> microbenchmark(sapply(myList, length), times = 10000L) > Unit: microseconds > expr min lq mean median uq max neval > sapply(myList, length) 14.156 15....

...) { DF <- as.data.frame(unclass(as.POSIXlt( dt )), stringsAsFactors=stringsAsFactors) `names<-`(cbind(dt, DF, deparse.level=0L), c(dName, names(DF))) } dt2df(.leap.seconds) # date+time dt2df(Sys.Date() + 0:9) # date ##' Even simpler: Date -> Matrix: d2mat <- function(x) simplify2array(unclass(as.POSIXlt(x))) d2mat(seq(as.Date("2000-02-02"), by=1, length.out=30)) # has R 1.0.0's release date ------------------------------------------------------------ In the distant past / one of the last times I touched on people using (base) R's Date / Time-Date objects, I...

Hello, I'm trying to figure out how to find the index of the second occurrence of "/" in a string (which happens to represent a date) within a data frame column. I've used the following code successfully to find the first instance of "/". dframe <- data.frame(date=c("5/14/2011", "4/7/2011")) dframe$x1 <- regexpr("/", dframe[, 1])

Hi: Summary: I am trying to determine the 90th percentile of ambulance response times for groups of data. Background: A fire chief would like to look at emergency response times at the 90th percentile for 1 kilometer grids in Cape Coral, Florida. I have mapped out ambulance response times on a GIS map. Then I superimpose a regularly-spaced grid over the response times and spatially join the

say I have a matrix and lists like x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2) x.list <- lapply(seq_len(nrow(x)), function(i) x[i,]) if I want a column of the matrix x, I write x[, 2] for example. But how can I do something similar for a set of lists, x.list, above? > x.list [[1]] [1] 12.1 0.1 12.0 1.1 [[2]] [1] 3.44 3.00 33.10 23.00 unlist(x.list)[,2] does