search for: uqs

It really is unclear what is claimed to be a bug here. But see https://stat.ethz.ch/pipermail/r-devel/2007-May/045592.html for why the bug is not in R: your old and new data do not match. Your fit is to a category. [The problem with the web interface to R-bugs was reported last week: it is being worked on.] On Mon, 30 Apr 2007, r.darnell at uq.edu.au wrote: > This is a multi-part

Hi R-users, I have a list mylist <- list(c(0.79, 0.92, 0.91, 0.86, 0.96, 0.96, 0.95, 0.94, 0.99), c(0.28, 0.45, 0.59, 0.69, 0.80, 0.87, 0.95, 0.94, 0.98), c(0.29, 0.39, 0.59, 0.69, 0.68, 0.80, 0.93, 0.95, 0.98)) Is there a way to find the index of the list element that contains the lowest value among all the other elements? As the lowest value in each element is the first, the

Hello R-users, I have a very large vector (a) containing elements consisting of numbers and letters, this is the i.e. a [1] "1.11.2a" "1.11.2d" "1.11.2e" "1.11.2f" "1.11.2x1" "1.11.2x1b" [7] "1.11.2x2" "1.11.2x2a" "1.11.2x2b" "1.11.2x3" "1.11.2x4"

Hi, S4 method dispatch can be very slow. Would it be reasonable to cache the most recent dispatch, anticipating the next invocation will be on the same type? This would be very helpful in loops. fun0 <- function(x) sapply(x, paste, collapse="+") fun1 <- function(x) { paste <- selectMethod(paste, class(x[[1]])) sapply(x, paste,

Dear All, I like to know how to simulate survival data using median (or mean) survival time. Any help will be greatly appreciated. Best wishes, Kere Kerenaftali Klein PhD| Biostatistician | Queensland Clinical Trials & Biostatistics Centre The University of Queensland | School of Population Health | Building 33, Level 1| Princess Alexandra Hospital |Ipswich Road | Woolloongabba QLD 4102 |

>>>>> Winston Chang <winstonchang1 at gmail.com> >>>>> on Fri, 23 Jan 2015 10:15:53 -0600 writes: > I think you can simplify a little by replacing this: > pkg %in% loadedNamespaces() > with this: > .getNamespace(pkg) almost: It would be !is.null(.getNamespace(pkg)) > Whereas getNamespace(pkg) will load the

I have fitted a logistic regression model > failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in > ddist time.long$Age time.long$task2 Low:effect 45

Maybe a useful addition to the predict functions would be to return the values of the predictor variables. It just (unless there are problems) requires an extra line. I have inserted an example below. "predict.glm" <- function (object, newdata = NULL, type = c("link", "response", "terms"), se.fit = FALSE,

Your example x = 5 exp = parse(text="f(uq(x)) + y +z") # expression: f(uq(x)) +y + z do_unquote(expr) # -> the language object f(5) + y + z could be done with the following wrapper for bquote my_do_unquote <- function(language, envir = parent.frame()) { if (is.expression(language)) { # bquote does not go into expressions, only calls

(Apologies if this is better suited for R-help.) On my system (macOS Sierra, late 2014 MacBook Pro; R 3.4.1, Homebrew build), I found that it is faster to construct a function using eval(call("function", ...)) than using as.function(list(...)). Example: make_fn_1 <- function(a, b) eval(call("function", a, b), env = parent.frame()) make_fn_2 <- function(a, b)

Could someone tell me where i can download the latest version of SSLeay. I have tried the following sites: * <ftp://ftp.psy.uq.oz.au/pub/Crypto/SSL/> * <ftp://ftp.uni-mainz.de/pub/internet/security/ssl> * <ftp://ftp.cert.dfn.de/pub/tools/crypt/sslapps> * <ftp://ftp.funet.fi/pub/crypt/mirrors/ftp.psy.uq.oz.au> * <ftp://ftp.sunet.se/ftp/pub/security/tools/crypt/ssleay>

...*BUT* the /tmp partition shrinking to a ridiculous size. /tmp only has the usual stuff on it, as I can now no longer create temporary files there: % du /tmp 4 /tmp/.X11-unix 0 /tmp/.XIM-unix 0 /tmp/.ICE-unix 0 /tmp/.font-unix 4 /tmp/ssh-tEgl0QxQHp 4 /tmp/ksocket-uqs 12 /tmp/kde-uqs 4 /tmp/fam-uqs 8 /tmp/.vbox-uqs-ipc 0 /tmp/worker-uqs 44 /tmp Anything I could try? Uli

Hi R-users, I have a function (myfun) that I want to apply to the rows of a matrix. Basically, "myfun" takes the values from the matrix ("exp.des"), which represent the different combinations of my experimental design, and pass them as arguments to some other functions (fun1 and fun2). As I want to replicate the results of fun1 and fun2 a certain number of time (e.g. 5), I

HLM question? Is there a minmum number of observations required for a category..I have individusals in work teams.I have incomplete data for all the teams ..sometimes I only have data for one person in a team.I assume that HLM can't work here! But what would be the mimimal.at the moment I have a sample of about 240 in about 100 teams with teamsizes form 2 to 5. Any advice? Thanks

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

Hi guys, I am working on some code for automatically translating recursive functions into looping functions to implemented tail-recursion optimisations. See https://github.com/mailund/tailr As a toy-example, consider the factorial function factorial <- function(n, acc = 1) { if (n <= 1) acc else factorial(n - 1, acc * n) } I can automatically translate this into the loop-version

Interestingly, the <<- operator is also a lot faster than using a namespace explicitly, and only slightly slower than using <- with local variables, see below. But, surely, both must at some point insert values in a given environment ? either the local one, for <-, or an enclosing one, for <<- ? so I guess I am asking if there is a more low-level assignment operation I can get my

Hi all, I think there's an unnecessary line in [.Date which has a considerable impact on performance when subsetting large dates: x <- Sys.Date() + 1:1e6 microbenchmark::microbenchmark(x[1]) #> Unit: microseconds #> expr min lq mean median uq max neval #> x[1] 920.651 1039.346 3624.833 2294.404 3786.881 41176.38 100 `[.Date` <- function(x, ...,

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 R-users, I have a problem running my R code on a Linux cluster. What I did was write a .pbs file to instruct the cluster on what to do and how: #!/bin/sh #PBS -m ae #PBS -M uqlcatta@uq.edu.au #PBS -A uq-CSER #PBS -N job1_lollo #PBS -l select=1:ncpus=1:NodeType=fast:mem=8GB #PBS -l walltime=999:00:00 cd $PBS_O_WORKDIR source /usr/share/modules/init/bash module