ents and
p-values for the 5 baseline variables, so I assumed
that it was due to the small number of levels (in fact, too few ). However,
when computing anova(model.rand, model.fix),
the output indicates a p-value < 0.001 in favour of the "model.rand". What's
happ
I just got a new Linux computer running Pop!_OS. If I download R from the
repository, which is basically he same as on Ubuntu, I get an outdated
version that can't run ggplot2. So I went to the R download page and
downloaded the newest version. It has make and config files but they
require an in
all( df1 == df3) # it works, but how can I understand that and how it is proven?
Regards
Frank Schwidom
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using anova.cca with margins?
I thank you very much in advance!
Frank
P.S.
Here are some outputs from the reduced to the full axes model:
Call: capscale(formula = gendist ~ scores(cur) + scores(ms) + scores(el) +
scores(lgm) + scores(lig) + scores(mis19) +
Condition(scores
Chris, thank you so much for your answer!!
Best,
Frank S.
De: Andrews, Chris
Enviado: martes, 3 de septiembre de 2019 14:14
Para: Frank S.
Cc: r-help@r-project.org
Asunto: Re: [R] Efficient way to update a survival model
library("survival")
set
Charles, thank you for your suggestion!
Frank S.
De: Berry, Charles
Enviado: s�bado, 31 de agosto de 2019 19:21
Para: Frank S.
Cc: Andrews, Chris ; r-help@r-project.org
Asunto: Re: Efficient way to update a survival model
The i^th model is included in the Cox
that, when computing Cox[[1]], the term
Cox[[k -1]]
does not exist. Is it possible to perform some "trick" (e.g. re-indexing) in
order to achieve this?
Best,
Frank
De: Andrews, Chris
Enviado: viernes, 30 de agosto de 2019 15:08
Para: Frank S. ; Vito
date(Cox0, substitute(. ~ . + Z[, 1:k]), data = pbc)
attr(Cox[[k]]$coefficients, "names")[2:(k+1)] <- paste0("sin(", 1:k, "* v)")
}
Cox
Best,
Frank
De: Frank S.
Enviado: jueves, 29 de agosto de 2019 12:38
Para: Vito Michele Rosa
hat OK?
Additionally, in my original question I wondered about the possibility of
reducing the
10 lines of code to one general expression or some loop. Is it possible?
Best,
Frank
De: Vito Michele Rosario Muggeo
Enviado: jueves, 29 de agosto de 2019 8:54
Para: Frank S.
in 1:10) {
Cox[[k + 1]] <- update(Cox[[k]], . ~ . + cos((k + 1) * v), data = pbc)
}
However, from Cox[[3]] onwards, the intermediate values of integer k are not
included here (for
instance, the model Cox[[10]] would only include the cosinus terms for
cos(1*v) and cos(10*v)).
Tha
s interested in it, can take a look:
https://github.com/schwidom/simplefs
Suggestions are welcome.
Kind regards
Frank Schwidom
On 2019-06-12 12:50:12, John via R-help wrote:
> On Wed, 5 Jun 2019 18:07:15 +0200
> Frank Schwidom wrote:
>
> In reading file names, names with spaces require
it can handle long vectors
of strings, booleans and also lists.
Kind regards,
Frank
On 2019-06-11 09:49:17, Gabriel Becker wrote:
>Hi Frank,
>I'm hesitant to be "that guy", but in case no one else has brought this up
>to you, having files with a tilde in th
Hi,
yes, I have seen this package and it has the same tilde expanding problem.
Please excuse me I will cc this answer to r-help and r-devel to keep the
discussion running.
Kind regards,
Frank Schwidom
On 2019-06-11 09:12:36, Gábor Csárdi wrote:
> Just in case, have you seen the fs pack
Hi,
to get rid of any possible filename modification I started a little project to
cover my usecase:
https://github.com/schwidom/simplefs
This is my first R package, suggestions and a review are welcome.
Thanks in advance
Frank Schwidom
On 2019-06-07 09:04:06, Richard O'Keefe wrote:
>
On 2019-06-05 20:32:07, Enrico Schumann wrote:
> >>>>> "FS" == Frank Schwidom writes:
>
> FS> Hi,
> FS> As I can see via path.expand a filename which contains a tilde
> anywhere gets automatically crippled.
>
> FS> +> path
On 2019-06-05 20:32:07, Enrico Schumann wrote:
> >>>>> "FS" == Frank Schwidom writes:
>
> FS> Hi,
> FS> As I can see via path.expand a filename which contains a tilde
> anywhere gets automatically crippled.
>
> FS> +> path
ther its name contains any character unless
0.
The unix filesystem allow the creation of such files, it sould be possible to
open these.
How can I switch off any file crippling activity?
Kind regards,
Frank
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when explaining the
results of complex statistical models. We also have discussions during the
lunches in the dining hall next to the lecture room. Many of the topics I
cover are set by the participants during the week. And all of the course
handouts are freely available for everyone.
*Regression Modeling Strategies Short Course 2019*
Frank E. Harrell, Jr., Ph.D., Professor
Department of Biostatistics, Vanderbilt University School of Medicine
fharrell.com @f2harrell
*May 14-17, 2019* With Optional R Workshop May 13
9:00am - 4:00pm
Alumni Hall
Vanderbilt University
er_3.4.3 Matrix_1.2-16 tools_3.4.3 splines_3.4.3
[5] grid_3.4.3 lattice_0.20-38
Thank you for your help!
Frank
From: R-help on behalf of
r-help-requ...@r-project.org
Sent: Saturday, March 16, 2019 11:00 AM
To: r-help@r-project.org
Subject: R-help Di
er_3.4.3 Matrix_1.2-16 tools_3.4.3 splines_3.4.3
[5] grid_3.4.3 lattice_0.20-38
Thank you for your help!
Frank
From: R-help on behalf of
r-help-requ...@r-project.org
Sent: Saturday, March 16, 2019 11:00 AM
To: r-help@r-project.org
Subject: R-help Di
to where the problem may occur? Is it something within the C-code used within
the package (as the last error seems to indicate), or is it related to the
computer cluster?
Any help or insights is much appreciated.
Kind regards,
Frank
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on 4.1-1 (2018-01-03)) * describe: quit rounding values when =
20 distinct values no matter how far apart any two values are spaced.https ...
____
Frank E Harrell Jr Professor School of Medicine
Department of Biostatistics Vand
*RMS Short Course 2018*
Frank E. Harrell, Jr., Ph.D., Professor
Department of Biostatistics, Vanderbilt University School of Medicine
fharrell.com @f2harrell
*May 15-18, 2018* With Optional R Workshop May 14
9:00am - 4:00pm
Alumni Hall
Vanderbilt University
Nashville Tennessee USA
See http
Hi All,
I want to know how to run an R file on my computer in R-Fiddle?
I tried source("filename.r"), but not working.
thanks,
Frank
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le in http://biostat.mc.vanderbilt.edu/tmp/bbr.pdf
Use lrm, not lrm.fit for this. Adding maxit=20 will probably make it work
on the small dataset but still not clear on why you are using this dataset.
Frank
--
Frank E Harrell Jr Professor School of Medicine
Dep
ou can use options(prType='html') to get special html output for RMarkdown
html reports and html notebooks.
There are small changes in Hmisc, e.g., leave off file="" for
latex(describe()).
Frank
--
Frank E Harrell Jr Professor and Ch
wo
group bariables because ggplot2 geom_ribbon doesn't support multiple
aesthetics
* predictrms: set contrasts for ordered factors to contr.treatment instead
of contr.poly
* val.prob: changed line of identity to use wide grayscale line instead of
dashed line
--
do.call(
what = rbind,
args = lapply(
X = paste0("dt_sp_", 1:length(sp)),
FUN = get,
envir = environment()
)
)
____
De: Frank S.
Envi
ce of the
statistic was appreciable; added option to compute posterior mean
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *Biostatistics* *Vanderbilt University*
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now a reasonable explanation for this?
Thanks.
Kind regards,
Frank
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Many thanks Ista and Bert for your nice solutions!
As Ista commented in a previous mail, the 0.87 value in my example is not
fixed, but for each subject
it depends on the difference "2007-01-01 - fini". However, both of your
solutions take into account this
fact.
lt;- 1 # 1st case
exposure[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30")]
<- difftime(as.Date("2007-01-01"), fini, units="days")/365.25 # 2nd case
exposure[fini >= as.Date("2006-07-01") & fini <= as.Dat
should I change to avoid this message?
Any help would be appreciated!
Best,
Frank
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This happens when you have not strat variables in the model.
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *Biostatistics* *Vanderbilt University*
On Thu, Jun 2, 2016 at 10:55 AM, Steve Lianoglou
wrote:
> Hello f
y = c(1, 2,3), Predator = 2)
times <- seq(0, 200, by = 1)
out <- ode(yini, times, LVmod, pars)
summary(out)
Thanks
Frank
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PLEASE do re
Hi Dalila,
is your function a part of a packacke or is it self-written? If you
have the code it would be helpful to provide it.
Frank
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Should I not try to update recommended packages in
/usr/local/lib/R/site-library on my Ubuntu system but specify
/usr/lib/R/library instead?
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *B
This does seem to be a good situation for ordinal regression. The R rms
package's orm function allows for thousands of categories in Y. But it
doesn't handle censoring.
This discussion would be better for stats.stackexchange.
Many thanks Petr!
Best,
Frank
> From: petr.pi...@precheza.cz
> To: f_j_...@hotmail.com; r-help@r-project.org
> Subject: RE: [R] Overlapping subject-specific histograms
> Date: Thu, 14 Jan 2016 12:03:16 +
>
> Hi
>
> change
>
> points( factor(names(tab)
"sim" variable.
Thanks in advance for suggestions!!
Frank
data <- data.frame(id = rep(c(1,3,4,7), c(9,5,3,3)),
count = c(0, 10, 15, 0, 16, 7, 14, 11, 12, 1, 8, 17, 19, 0, 9, 10, 14, 2,
3, 10),
sims = c(1, 9, 15, 1, 14, 5, 12, 10, 12, 2, 6, 15, 18, 1,
s have not succeeded.
Any help on how to approach this is kindly appreciated.
Kind regards,
Frank van Berkum
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Many thanks to David L Carlson, Ben Gunter and David Winsemius for your quick
and very elegant solutions!!
With your list answers I am learning sa lot of things that will help me in the
future to program.
Best,
Frank S.
> Subject: Re: [R] Random selection of a fixed number of values
LSE, breaks =
0:ceiling(max(data$value
and the size vector:
size <- c(10, 7, 5, 5, 3)
But I'm not able to get it by using sample function. Does anyone have some idea?
Thank you very much for any suggestions!!
Frank S.
Rolf I believe \textsf is the correct font to use for the symbol R but
not necessarily for the names of R variables and functions. I'd like
more discussion about that.
Frank
--
Frank E Harrell Jr Professo
Thank you very much that substr works great on my data. For future users, I
did in the following code.
z<-matrix(c(substr(tmp,5,8),substr(tmp,1,4)),5,2)
dataU16 <- c(t(z))
On Sun, Nov 22, 2015 at 8:38 AM, Duncan Murdoch
wrote:
> On 22/11/2015 2:04 AM, Frank Wang wrote:
>
>&
Hi,
I am new user on R. I want to split a vector of hex data such as "00ff8020"
"02d0" "001e0240" "00010096" "00010033"
into 16 bits such as:
"00ff", "8020", "02d0","","001e", "024
correction:
On 2015-10-10 16:08:39, Frank Schwidom wrote:
> On 2015-10-09 20:15:16, liqunhan--- via R-help wrote:
>
> > for (k in 1 : 5) {
> > xlist <- list(x5 = dailyrecord$a[k], x6 = dailyrecord$e[k], x7 =
> > dailyrecord$f[k])
> > result_forlo
On 2015-10-09 20:15:16, liqunhan--- via R-help wrote:
> for (k in 1 : 5) {
> xlist <- list(x5 = dailyrecord$a[k], x6 = dailyrecord$e[k], x7 =
> dailyrecord$f[k])
> result_forloop[k] <- fun3(list1, list2, xlist)
> }
result_forloop <- lapply( 1 : 5, function( k) {
tmpRow <- dailyreco
Please note that Kirsten is cross-posting to stats.stackexchange.com
creating extra work for everyone.
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *Biostatistics
the model and options(contrasts) is not set properly
* rms transformation functions: made more robust by checking !
length instead of is.null
--
----
Frank E Harrell Jr Professor and Chairman School of Medicine
test1 <- (rbind(c(0.1,0.2),0.3,0.1))
rownames(test1)=c('y1','y2','y3')
colnames(test1) = c('x1','x2');
test2 <- (rbind(c(0.8,0.9,0.5),c(0.5,0.1,0.6)))
rownames(test2) = c('y2','y5')
colnames(test2) = c('x1','x3','x2')
lTest12 <- list( test1, test2)
namesRow <- unique( unlist( lapply( lTest12, ro
And if we want to use the approach of William Dunlap for sequence.optimization
, then we can write:
rev( xr[ seq_len(sum(vec)) - rep.int(cumsum(c(0L, vec[-length(vec)])), vec)] -
rep.int( xr[ -1], vec))
Regards.
On 2015-09-22 23:43:10, Frank Schwidom wrote:
> Hi,
>
> xr <- rev( x
Hi,
xr <- rev( x)
vec <- 1:(length( x) - 1)
rev( xr[ sequence( vec)] - rep.int( xr[ -1], vec))
On 2015-09-21 14:17:40, Dan D wrote:
> I need an efficient way to build a new n x (n-1)/2 vector from an n-vector x
> as:
>
> c(x[-1]-x[1], x[-(1:2)]-x[2], ... , x[-(1:(n-1)] - x[n-1])
>
> x is incre
Hi
I have to correct myself, this last solution is not universally valid
here a better one:
tmp1 <- ( 1 - outer( max( x):1, x, FUN='-'))
tmp1[ tmp1 > 0]
On 2015-09-17 21:06:30, Frank Schwidom wrote:
>
> how abount a more complicated one?
>
> outer( 1:5
better ( if year is an vector of more than 1 element):
df[ df$Year %in% outer(as.numeric( as.character( year)), -1:1, FUN='+'), ]
Year Amount
2 2002120
3 2003175
4 2004160
On Mon, Sep 21, 2015 at 10:49:34PM +0200, Frank Schwidom wrote:
>
> year <- df$Year[ w
year <- df$Year[ which.max( df$Amount)]
df[ df$Year %in% (as.numeric( as.character( year)) + -1:1), ]
Year Amount
2 2002120
3 2003175
4 2004160
On Mon, Sep 21, 2015 at 04:52:46PM +0200, Nico Gutierrez wrote:
> Hi All,
>
> I need to do the following operation from data.frame:
>
>
Hi,
when you can plot this graph using the rgl-package,
then you can use "rgl::writeWebGL" to create an 3D-View
in the Browser.
Regards
On Sat, Sep 19, 2015 at 04:42:47PM +0200, bgnumis bgnum wrote:
> Hi al,
>
> I want to put a graph in a html5 webpage plotted with R (I want to get dar
> from
how abount a more complicated one?
outer( 1:5, 1:5, '-')[ outer( 1:5, 1:5, '>')]
[1] 1 2 3 4 1 2 3 1 2 1
On Thu, Sep 17, 2015 at 11:52:27AM -0700, David Winsemius wrote:
> You can add this to the list of options to be tested, although my bet would
> be placed on `sequence(5:1)`:
>
> > Reduce
sequence( 5:1)
Regards.
On Thu, Sep 17, 2015 at 11:19:05AM -0700, Dan D wrote:
> Can anyone think of a slick way to create an array that looks like c(1:n,
> 1:(n-1), 1:(n-2), ... , 1)?
>
> The following works, but it's inefficient and a little hard to follow:
> n<-5
> junk<-array(1:n,dim=c(n,n)
Hi
where can i find 'melt' and 'dcast' ?
Regards
On Thu, Sep 17, 2015 at 08:22:10AM +, PIKAL Petr wrote:
> Hi
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Kai Mx
> > Sent: Wednesday, September 16, 2015 10:43 PM
> > To: r-help mailing
Hi
res <- sapply( df1[ , -1], function( x) table(x)[as.character( 0:5)])
rownames( res) <- paste( sep='', 'result', 0:5)
res[ is.na( res)] <- 0
res
item1 item2 item3 item4 item5
result0 1 0 1 1 0
result1 1 2 0 0 0
result2 1 2 1 1
On Mon, Sep 14, 2015 at 04:11:57PM +0100, JORGE COLACO wrote:
> > X-mean
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,] -0.8 -0.2 0.8 -1.2 0.8 -0.2
> [2,] -0.2 0.2 0.8 0.8 0.8 -1.2
> [3,] -1.2 -0.2 -0.8 0.8 -0.2 0.8
> [4,] 0.8 0.8 -1.2 -0.8 -0.2 -0.2
> [5,] -0.2 -0.2 0.8 0.8 -0.8 0.8
> >
df <- data.frame( V1= 1, V2= c( 2, 3, 2, 1), V3= c( 1, 2, 1, 1))
dfO <- df[ do.call( order, df), ]
dfOD <- duplicated( dfO)
dfODTrigger <- ! c( dfOD[-1], FALSE)
dfOCounts <- diff( c( 0, which( dfODTrigger)))
cbind( dfO[ dfODTrigger, ], dfOCounts)
V1 V2 V3 dfOCounts
4 1 1 1 1
3 1 2
Just for fun:
> colSums( outer( VAS, VAS, '<'))
[1] 3 3 0 3 7 8 8 0 3 0
On Sat, Sep 05, 2015 at 02:14:18PM -0700, Dan D wrote:
> # your data
> VAS<-c("Green","Green","Black","Green","White","Yellow","Yellow","Black","Green","Black")
>
> # declare the new vector
> New_Vector<-numeric(length(V
c( as.factor( VAS))
On Sat, Sep 05, 2015 at 02:14:18PM -0700, Dan D wrote:
> # your data
> VAS<-c("Green","Green","Black","Green","White","Yellow","Yellow","Black","Green","Black")
>
> # declare the new vector
> New_Vector<-numeric(length(VAS))
>
> # brute force:
> New_Vector[VAS=="White"]<-1
>
> rawToChar( charToRaw( str)[ c( TRUE, FALSE)])
[1] "ACEG"
Regards
On Sat, Sep 05, 2015 at 04:59:54PM -0400, Evan Cooch wrote:
> Suppose I had the following string, which has length of integer multiple of
> some value n. So, say n=2, and the example string has a length of (2x4) = 8
> characters
# my last one:
xtfrm( VAS)
On Tue, Sep 08, 2015 at 11:55:51AM -0700, Dan D wrote:
> Great!
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Help-with-vectors-tp4711801p4712023.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
On Sat, Sep 05, 2015 at 02:14:18PM -0700, Dan D wrote:
> # your data
> VAS<-c("Green","Green","Black","Green","White","Yellow","Yellow","Black","Green","Black")
>
> # declare the new vector
> New_Vector<-numeric(length(VAS))
>
> # brute force:
> New_Vector[VAS=="White"]<-1
> New_Vector[VAS=="Yell
It's implemented in the R rms package's lrm and orm functions.
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Nagelkerke-Pseudo-R-squared-tp4710014p4710031.html
Sent from the R help mailing list
Thank you for all your observations and comments!!
As you suggest, the option x <- x*c(rep(1,19), -1) is a more elegant and a fast
way!
Frank S.
> From: dwinsem...@comcast.net
> Date: Thu, 18 Jun 2015 21:33:57 -0700
> To: marc_schwa...@me.com
> CC: r-help@r-project.org
&g
he desired result. II
wonder wether there is a cool way to do so, that is, for example with apply or
sign function.
Thans in advanced for your help!
Frank S.
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Thank you very much Terry. I'm still puzzled at why this worked a year
ago. What changed? I'd very much like to reverse the change by setting
an argument somewhere or manipulating the terms object.
I echo your sentiments about the general approach.
Frank
On 06/15/2015 09:05 AM
should not be there
(aa1:strat(b)b2).
This does seem to be a change in R. Any help appreciated.
Note that after subsetting out strat terms using Terms[ - temp$terms],
Terms attributes factor and term.labels are:
attr(,"factors")
a a:strat(b)
y0 0
a
herneau uses more complex logic to construct the
design matrix reliably. I'd like to avoid that logic because it creates
an overly wide design matrix before removing the unneeded columns.
Thanks for any assistance,
Frank
--
Thank you very much Bill,
Your answer to my question is exactly what I was trying to do in my R code.
Best regards.
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rse. Does anybody can help me? Thanks a bunch!! Frank S.
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Interesting solutions. Thanks guys!
On Wed, May 27, 2015 at 9:27 AM, William Michels
wrote:
> Hi Frank!
>
> Ok, bind columns together in a state-wise fashion, allowing for state
> duplicates. Below (maybe cheesy) uses the state abbreviations
> "state.abb" in the datase
.
While this is clunky I can just write a function to do it all at once.
On Wed, May 27, 2015 at 6:20 AM, Frank Burbrink
wrote:
> Thanks Bill,
>
> However, unique(merge(x, y, by = 1, all=T)) is giving me:
>
>state locus.x locus.y
> 1 AR 5 2
> 2 AR
1
9 LA 2 NA
10LA 2 NA
11MS 3 NA
12MS 4 NA
13TN NA 3
14TN NA 4
On Wed, May 27, 2015 at 3:53 AM, William Michels
wrote:
> Hi Frank,
>
> It looks like you're very close. I think you want:
>
2 NA
10LA 2 NA
11MS 3 NA
12MS 4 NA
13TN NA 3
14TN NA 4
Any help would be much appreciated.
Thanks!
Frank
--
*
*Frank T. Burbrink, Ph.D.*
*Professor*
*Biology Department*
*6S
Group; similar for hoeffd
--
----
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *Biostatistics* *Vanderbilt University*
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r-pac
*RMS Short Course 2015*
Frank E. Harrell, Jr., Ph.D., Professor and Chair
Department of Biostatistics, Vanderbilt University School of Medicine
*March 3, 4, 5 & 6, 2015* With Optional R Workshop March 2
9:00am - 4:00pm
Student Life Center Board of Trust Room
Vanderbilt University
Nashv
Subject: Regression Modeling Strategies 4-Day Short Course March 2015
*RMS Short Course 2015*
Frank E. Harrell, Jr., Ph.D., Professor and Chair
Department of Biostatistics, Vanderbilt University School of Medicine
*March 3, 4, 5 & 6, 2015* With Optional R Workshop March 2
9:00am - 4:00pm
Stu
Hi,
I think I got it!
The clue: Function split!
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Hi everybody,
I have (as an example) the following
two data tables:
all <-
data.table(ID = c(rep(c(100:105),c(3,2,2,3,3,3))),
value =
c(100,120,110,90,45,35,270,50,65,40,25,55,75,30,95,70))
DT <-
data.table(ID = 100:105, code=c(2,1,3,2,3,1))
My aim is to construct as many sub
Jim, Thanks for the comment about else!
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Dear Berend and Petr,
I do apologise for the disorderly code I posted. I have tried to solve it in a
new mail.
Frank S.
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Hi to all members of R list,
I�m working with data.table package, and with 6
variables:
ID: Subject identifier
born: Birthdate
start: Starting date
register: date of measurement
value: Value of measurement
end: date of expiration of the measurements.
So, the natural order of d
Hi to all members of R list,
I�m working with data.table package, and with 6
variables: "ID" (Identifier), "born" (Birthdate), "start" (Starting date),
"register" (date of measurement), "value"and "end" (date of expiration). So,
the natural order of dates would be: born
=< start =< register =
Thanks Richard!
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classes created by survfit.formula
* logLik.Gls: added. Makes AIC(Gls object) work.
* NAMESPACE: several changes
See http://biostat.mc.vanderbilt.edu/rms and
https://github.com/harrelfe/rms for more information.
Frank Harrell
Vanderbilt University
handle Inf. Thanks: Benjamin Tyner
* DESCRIPTION, NAMESPACE: multiple function changes to work in R-devel
See http://biostat.mc.vanderbilt.edu/Hmisc,
https://github.com/harrelfe/Hmisc for more information.
Frank Harrell
Vanderbilt University
___
Hi to all members of the list,
I have a data frame with subjects who can get into a certain study from
2010-01-01 onwards. Small example:
DF <- data.frame(id=as.factor(1:3), born=as.Date(c("1939/10/28", "1946/02/23",
"1948/02/29")))
id born
1 1 1939-10-28
2 2 1946-02-23
3 3 1948
Many thanks for you help Uwe!
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Hi everyone, After trying to find the solution during days, I decided to write
in this help list in order to ask if anyone can help me.I would want to
construct an R function, with "initial", "final" and "specific" dates as 3
arguments (for example, becauseI'm not really sure it is the best wa
Greg I just re-copied the latest subplot and its help file from
TeachingDemos to Hmisc for the next release. Thanks for pointing this out.
Frank
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When using cph in the rms package there is a function Mean that operates
on cph objects to produce an R function for computing the mean or
restricted mean life time.
Frank
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Please try hard to create a simple reproducible example, then I'll work
on it.
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s and transitioned to R around 2000) and have never looked back.
Lately what has really made R powerful is its ability to interface
with other languages and especially the way it works in a reproducible
analysis/dynamic report document context.
Frank
--
Frank E Harrell Jr Professor and Cha
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