similar to: mclapply Segmentation Fault for Ubuntu

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(3,1)), array(21:23,c(3,1))), higher=TRUE)) int [1:3, 1, 1:2] 11 12 13 21 22

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, 8)) ) simplify2array(a, higher = T) to return the same

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

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

Dear all How does one convert a "non-symmetric" list to a vector? See below: > x <- list() > x[[1]] <- letters[1:5] > x[[2]] <- letters[6:10] > x[[3]] <- letters[11:12] > x [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] "f" "g" "h" "i" "j" [[3]] [1] "k"

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

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")

While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function if (!identical(simplify, FALSE) && length(answer)) This seems superfluous to me, in particular this part: !identical(simplify, FALSE) The preceding

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)

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?

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)

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])

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

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

Martin In terms of context of the actual problem, sapply is called millions of times because the work involves scoring individual students who took a test. A score for student A is generated and then student B and such and there are millions of students. The psychometric process of scoring students is complex and our code makes use of sapply many times for each student. The toy example used

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

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

This is mostly a RFC [but *not* about the many extra packages, please..]: Noticing to my chagrin how my students work in a project, googling for R code and cut'n'pasting stuff together, accumulating this and that package on the way all just for simple daily time series (though with partly missing parts), using chron, zoo, lubridate, ... all for things that are very easy in base R *IF*

On 03/13/2018 09:23 AM, Doran, Harold wrote: > While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function > > if (!identical(simplify, FALSE) && length(answer)) > > This seems superfluous to me,

Hello, I am very new to R (as my Subject probably indicates). I want to do something that should, I think, be very simple. I have five vectors in a list and I want to construct a covariance matrix out of them. Given a 5X5 matrix cvm1, and the list of vectors, cvm1_list, I thought the following would work (sorry cannot find code tags): for(i in 1:5){ for(j in 1:5){ cvm1[i,j] <-