Displaying 20 results from an estimated 90 matches similar to: "Hard crash of lme4 in R-devel"
2010 Dec 10
1
survreg vs. aftreg (eha) - the relationship between fitted coefficients?
Dear R-users,
I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different
2008 Jul 03
1
subset function within a function
Hi,
I am using this subset statement and it works
outside a function.
LIS[[i]]<- lapply(LI, subset, select=cov[[i]])
However, wrapped inside a function this statement
produces the same values for every LIS[[1]] which
is only the first subset of LI.
Does anyone know why is not working correctly inside
a function?
ff = factor(covariate)
nLev <- nlevels(ff)
cov <-
2009 Apr 17
1
cast function in package reshape
Hello R useRs,
I have a function which returns a list of functions :
freq1 <- function(x) {
lev <- unique(x[!is.na(x)])
nlev <- length(lev)
args <- alist(x=)
if (nlev == 1) {
body <- c("{", "sum(!is.na(x))", "}")
f <- function() {}
formals(f) <- as.pairlist(args)
body(f) <- parse(text = body)
namef <-
2012 Nov 05
1
Another code to drop factor levels
I apologize if this is not appropriate for this mailing list.
In R, there is already functionality to drop unused factor levels. However, I am proposing the code below that I wrote. In some occasions, it was faster than applying function 'factor'. In any case, there is no restriction for anyone to use the code below.
droplevels2 <- function(x) {
if (is.null(levels(x)))
stop("no
2007 Jan 28
1
plot.lm (PR#9474)
Full_Name: Robert Kushler
Version: 2.4.1
OS: Windows XP
Submission from: (NULL) (69.245.71.40)
In the constant leverage case, plot #5 is not correctly produced.
The labels on the x-axis are sorted correctly by magnitude of the
fitted value, but the data are plotted in the original factor order.
I changed
facval[ord] <- facval
xx <- facval
2003 Aug 04
1
coxph and frailty
Hi:
I have a few clarification questions about the elements returned by
the coxph function used in conjuction with a frailty term.
I create the following group variable:
group <- NULL
group[id<50] <- 1
group[id>=50 & id<100] <- 2
group[id>=100 & id<150] <- 3
group[id>=150 & id<200] <- 4
group[id>=200 & id<250] <- 5
group[id>=250
2002 May 17
1
Strange R CMD check \usage parse error
In running R CMD check I get an error I can't debug. Would someone please let me know if they spot a syntax error in the code below or if there is a workaround for the parse error? Thanks -Frank
Error in parse(file, n, text, prompt) : parse error
Error in codoc(package = "Hmisc") : cannot source usages in documentation object 'plsmo'
Execution halted
* checking for
2009 Mar 06
2
sm.options
Hi,
I am doing kernel density plots, and am trying to make the lines thicker. I
comparing three groups, in sm.density.compare. I tried changing lwd to make
the line sthicker right on the density compare call, but was not able to do
it. There is not an option in sm.options to specify line thickness, as well
as cex.ylab or cex.xlab- I tried it and it does not change the thickness of
the lines.
2018 Mar 24
1
Function 'factor' issues
I am trying once again.
By just changing
f <- match(xlevs[f], nlevs)
to
f <- match(xlevs, nlevs)[f]
, function 'factor' in R devel could be made more consistent and back-compatible. Why not picking it?
--------------------------------------------
On Sat, 25/11/17, Suharto Anggono Suharto Anggono <suharto_anggono at yahoo.com> wrote:
Subject: Re: [Rd] Function
2017 Nov 25
0
Function 'factor' issues
>From commits to R devel, I saw attempts to speed up subsetting and 'match', and to cache results of conversion of small nonnegative integers to character string. That's good.
I am sorry for pushing, still.
Is the partial new behavior of function 'factor' with respect to NA really worthy?
match(xlevs, nlevs)[f] looks nice, too.
- Using
f <- match(xlevs, nlevs)[f]
2024 Jun 15
1
Hard crash of lme4 in R-devel
? Sat, 15 Jun 2024 02:04:31 +0000
"Therneau, Terry M., Ph.D. via R-devel" <r-devel at r-project.org> ?????:
> other attached packages:
> [1] lme4_1.1-35.1 Matrix_1.7-0
I see you have a new Matrix (1.7-0 from 2024-04-26 with a new ABI) but
an older lme4 (1.1-35.1 from 2023-11-05).
I reproduced the crash and the giant backtrace by first installing
latest lme4 and then
2006 Jun 29
2
Biobass, SAGx, and Jonckheere-Terpstra test
Hi list,
I tried to load the package SAGx and failed because it complains it's
looking for the Biobass which is not there. Then I looked up the package
list and Biobass is not found.
I'm trying to run the Jonckheere-Terpstra test and from what I see in
the R archive, SAGx is the only place it's been implemented.
> library(SAGx)
Loading required package: multtest
Loading required
2017 Oct 21
0
Function 'factor' issues
My idea (like in https://bugs.r-project.org/bugzilla/attachment.cgi?id=1540 ):
- For remapping, use
f <- match(xlevs, nlevs)[f]
instead of
f <- match(xlevs[f], nlevs)
(I have mentioned it).
- Remap only if length(nlevs) differs from length(xlevs) .
On use of 'order' in function 'factor' in R devel, factor.Rd still says 'sort.list' in "Details" section.
My
2012 Jan 19
1
Legend problem in line charts
Hi all,
Small problem in generating the line charts.
Question: Legend for the first graph is coming wrong., for second graph correctly. Please fix the legend postion at the down of graph.
Plesae give me the solution.
Thank you
Devarayalu
Orange1 <- structure(list(REFID = c(7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8,
8, 8, 8, 8, 9, 9, 9, 9), ARM = c(1, 1, 1, 1, 2, 2, 2, 2, 1, 1,
1, 1, 2, 2, 2,
2009 Jan 09
2
recursive relevel
Dear list,
I'm having second thoughts after solving a very trivial problem: I
want to extend the relevel() function to reorder an arbitrary number
of levels of a factor in one go. I could not find a trivial way of
using the code obtained by getS3method("relevel","factor"). Instead, I
thought of solving the problem in a recursive manner (possibly after
reading
2017 Oct 18
0
Function 'factor' issues
>>>>> Suharto Anggono Suharto Anggono via R-devel <r-devel at r-project.org>
>>>>> on Sun, 15 Oct 2017 16:03:48 +0000 writes:
> In R devel, function 'factor' has been changed, allowing and merging duplicated 'labels'.
Indeed. That had been asked for and discussed a bit on this
list from June 14 to June 23, starting at
2008 May 20
1
contr.treatments query
Hi Folks,
I'm a bit puzzled by the following (example):
N<-factor(sample(c(1,2,3),1000,replace=TRUE))
unique(N)
# [1] 3 2 1
# Levels: 1 2 3
So far so good. Now:
contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE)
contrasts(N)
# 1 2
# 1 1 0
# 2 0 1
# 3 0 0
whereas:
contr.treatment(3, base=1, contrasts=FALSE)
# 1 2 3
# 1 1 0 0
# 2 0 1 0
# 3 0 0 1
contr.treatment(3, base=1,
2017 Oct 15
2
Function 'factor' issues
In R devel, function 'factor' has been changed, allowing and merging duplicated 'labels'.
Issue 1: Handling of specified 'labels' without duplicates is slower than before.
Example:
x <- rep(1:26, 40000)
system.time(factor(x, levels=1:26, labels=letters))
Function 'factor' is already rather slow because of conversion to character. Please don't add slowdown.
2010 Dec 11
0
is there a packge or code to generate markov chains in R
Hi,
if i have data in the following time series format:
time, amount, state
1 2222 A
1 333 B
2 45 A
2 77 B
where states could be n and time periods t is there a package in R that would calculate the transition probabilities in a markov chain.
for each t except t=0 to generate
A B
A
B
perhaps the best structure might
2009 Jun 16
4
confusion on levels() function, and how to assign a wanted order to factor levels, intentionally?
Dear R-helpers,
I want to make a series of boxplots on several numeric univariates with two
group variables (species and population, population nested in species, and
with population as the X-axis). In order to get a proper order of the
individual populations in X-axis, I need to assign a wanted order to the
factor (population). I used the levels() function to do this assignment, but
it seemed