Hi Duncan,
I would not have changed the COMPILED_BY option unless I thought I have
to.
In my "C:\R-Project\Rtools\mingw_32\bin" I have
c++.exe
g++.exe
gcc.exe
i686-w64-mingw32-c++.exe
i686-w64-mingw32-g++.exe
i686-w64-mingw32-gcc-4.9.3.exe
i686-w64-mingw32-gcc.exe
In my "C:\R-Project\Rtools\m
Hi Doug,
To expand a bit on what Bert has written, all the the "best
subset/best model" procedures use random variation in the dataset to
produce a result. This means that you will almost certainly include
variables in your "best model" that cannot be replicated. Sometimes
you can see this as a var
I am trying to plot an expression of the form "p^*" --- a bold letter p
with the asterisk as a superscript.
I can get *something* with code of the form
plot(1:10)
text(7.5,2.5,expression(paste(bolditalic(p)^"*")))
but the asterisk that appears is *tiny*.
Is there any way to increase its siz
This is a statistics question, which is largely off topic on this
list. However, I'll give you a very brief OT response:
I would strongly suggest you consult a local statistician to explain
to why what you are doing is likely to result in complete nonsense
(best subset of 5 or 6 from 21 predictors
I appreciate the pointers.
I am using ggplot2 2.1.0, but I was looking at the wrong version of the web
page example documentation.
The easiest way is to define the function
binomial_smooth <- function(...) {
geom_smooth(method = "glm", method.args = list(family = "binomial"), ...)
}
And th
For all practical purposes, the differences are zero.
If you want them to also look like zero, try
round( m - mma , 3)
or
signif( m - mma , 3)
(or some number of digits other than three; I picked 3 rather arbitrarily)
For anticipating the sign of these minuscule differences, I doubt there
Hello,
I would like to add a counter column in a data frame based on a set of
identical rows. To do this, I tested:
DF = data.table(x=c("a","a","a","b","c","d","e","f","f"),
y=c(1,3,2,8,8,4,NA,NA,NA))
DF[ , Index := .GRP, by = c("y") ]
DF
However, the rows with NAs are considered to be ide
Hello All,
I am working on a linear regression model and trying to find the best subset of
variables for my dataset. I have 21 predictors, 1 response variable, and 79
observations. I need to find the best 5 or 6 predictors for my model. I've used
leaps for lm() and I'm now trying bestglm for glm
> On Jun 29, 2016, at 2:17 PM, Nathan Pace wrote:
>
> I want to add a logistic plot to data.
>
> My call to ggplot is:
>
>
> ggplot(data = SSI.dt, aes(x = elapsed, y = 1 - control)) + geom_point() +
> stat_smooth(method = 'glm', family = binomial) +
> xlab('Surgery Duration (min)') + ylab('Pr
On 30/06/16 02:08, Ulrik Stervbo wrote:
Then I don't understand what you want to achieve. If you want to combine
the two columns you can do
c(df1$col1, df2$col2)
It is indeed difficult to understand what the OP wants to do. Perhaps
he/she wants to *paste* the two columns together?
In whic
On 29/06/2016 4:49 PM, Christophe Dutang wrote:
Dear list,
How can I link a vignette of a package in a man page (Rd files)?
I try \link[=pkgname/doc/filename]{a name} without success.
The link gives the following error message: Only help files, NEWS, DESCRIPTION
and files under doc/ and demo/
I want to add a logistic plot to data.
My call to ggplot is:
ggplot(data = SSI.dt, aes(x = elapsed, y = 1 - control)) + geom_point() +
stat_smooth(method = 'glm', family = binomial) +
xlab('Surgery Duration (min)') + ylab('Probability SSI') +
labs(title = 'THA Surgical Site Infections')
ggsave(f
Please post this on R-package-devel, not here.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Jun 29, 2016 at 1:49 PM, Christophe Dutang
Dear Steven,
> -Original Message-
> From: Steven Yen [mailto:sye...@gmail.com]
> Sent: June 29, 2016 9:48 AM
> To: Fox, John
> Cc: R-help ; Sandy Weisberg (sa...@umn.edu)
>
> Subject: Re: [R] t-test for regression estimate
>
> Also,
> Is there a way to get the second command (hypothesis
Dear list,
How can I link a vignette of a package in a man page (Rd files)?
I try \link[=pkgname/doc/filename]{a name} without success.
The link gives the following error message: Only help files, NEWS, DESCRIPTION
and files under doc/ and demo/ in a package can be viewed
Thanks in advance
K
Dear Steven,
> -Original Message-
> From: Steven Yen [mailto:sye...@gmail.com]
> Sent: June 29, 2016 9:39 AM
> To: Fox, John
> Cc: R-help ; Sandy Weisberg (sa...@umn.edu)
>
> Subject: Re: [R] t-test for regression estimate
>
> Thanks John. Yes, by using verbose=T, I get the value of the
Hi,
Just to augment Bert's comments, I presume that you are aware of the relevant R
FAQ:
https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
That you had an expectation of the difference being 0 suggested to me that you
might not be, but my apo
Also,
Is there a way to get the second command (hypothesis defined with
externally scalars) below to work? Thanks.
linearHypothesis(U,"0.5*eq1_DQ+0.3*eq2_DQ",verbose=T)
w1<-0.5; w2<-0.3
linearHypothesis(U,"w1*eq1_DQ+w2*eq2_DQ",verbose=T) # does not work
On 6/29/2016 12:38 PM, Steven Yen wrote:
>
Thanks John. Yes, by using verbose=T, I get the value of the hypothesis.
But tell me again, how would I get the variance (standard error)?
On 6/29/2016 11:56 AM, Fox, John wrote:
> Dear Steven,
>
> OK -- that makes sense, and there was also a previous request for
> linearHypothesis() to return t
Dear Steven,
OK -- that makes sense, and there was also a previous request for
linearHypothesis() to return the value of the hypothesis and its covariance
matrix. In your case, where there's only 1 numerator df, that would be the
value and estimated sampling variance of the hypothesis.
I've no
On 29/06/2016 10:48 AM, g.maub...@weinwolf.de wrote:
Hi Duncan,
indeed, I did not see the other part of your message.
I did
BINPREF ?= C:/R-Project/Rtools/mingw_32/bin/
COMPILED_BY = g++ # instead of gcc-4.9.3
I wouldn't change the COMPILED_BY; some packages use it to configure
themselves f
I am certainly no expert, but I would assume that:
1. Roundoff errors depend on the exact numerical libraries and
versions that are used, and so general language comparisons are
impossible without that information;
2. Roundoff errors depend on the exact calculations being done and
machine precisi
Did you mean: strip.custom(factor.levels...) ?
(I know of no "panel.custom()" function)
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Jun 29,
Hi Duncan,
indeed, I did not see the other part of your message.
I did
BINPREF ?= C:/R-Project/Rtools/mingw_32/bin/
COMPILED_BY = g++ # instead of gcc-4.9.3
in "C:\R-Project\R-3.3.0\etc\i386\Makeconf"
and
BINPREF ?= C:/R-Project/Rtools/mingw_64/bin/
COMPILED_BY = g++ # instead of gcc-4.9.3
i
Hi Duncan,
many thanks for your reply.
I did insert die paths to the g++ compiler because I got the message about
the not existent compiler.
I took the directories for the compiler out again:
C:\R-Project\Rtools\bin;C:\ProgramData\Oracle\Java\javapath;C:\Program
Files\Python 3.5\Scripts\;C:\P
Your answer did not make it to the list...
Le 29/06/2016 à 16:06, Witold E Wolski a écrit :
If you do not understand than why do you reply?
On 29 June 2016 at 15:54, Ivan Calandra wrote:
Hi,
I don't really understand why you split every row... This makes it very
slow. Try with a more realis
On 29/06/2016 10:17 AM, g.maub...@weinwolf.de wrote:
Hi Duncan,
many thanks for your reply.
I did insert die paths to the g++ compiler because I got the message about
the not existent compiler.
I took the directories for the compiler out again:
C:\R-Project\Rtools\bin;C:\ProgramData\Oracle\Ja
I won't go into why splitting data.frames (or factors) uses time
proportional to the number of input rows times the number of
levels in the splitting factor, but you will get much better mileage
if you call split individually on each 'atomic' (numeric, character, ...)
variable and use mapply on the
It looks like License was imported as character data and converted to a factor.
I'm not sure I understand what you want, but to make them into a single column,
you need to convert the factor back to numeric using
as.numeric(as.character(License)), but you use the command levels(License) to
look
Then I don't understand what you want to achieve. If you want to combine
the two columns you can do
c(df1$col1, df2$col2)
Ulrik
schrieb am Mi., 29. Juni 2016 15:45:
> Hi Ulrik,
> Thnaks for your reply, merge() does not give me one column, it returns two
> columns
>
>
> On Wednesday, June 29,
Hi,
I don't really understand why you split every row... This makes it very
slow. Try with a more realistic example (with a factor to split).
Ivan
--
Ivan Calandra, PhD
Scientific Mediator
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, Fr
It looks like the function you are searching for is merge()
HTH
Ulrik
On Wed, 29 Jun 2016 at 15:11 ch.elahe via R-help
wrote:
> Hi all,
> I have this column as a part of df:
>
> $License : Factor W/384 levels
> "41005","41006","41034","41097","41200",...
> and I have other column which is
Hi,
Here is an complete example which shows the the complexity of split or
by is O(n^2)
nrows <- c(1e3,5e3, 1e4 ,5e4, 1e5 ,2e5)
res<-list()
for(i in nrows){
dum <- data.frame(x = runif(i,1,1000), y=runif(i,1,1000))
res[[length(res)+1]]<-(system.time(x<- split(dum, 1:nrow(dum
}
res <- do.
Hi all,
I have this column as a part of df:
$License : Factor W/384 levels "41005","41006","41034","41097","41200",...
and I have other column which is a part of other df lets say df2:
$Diff: int 41166 41202 41290 41353 41503 41507 41548
these two columns df$License and df$Diff
I don't think this is an RStudio issue. The plot() method for "rpart"
objects draws no labels. The text() method has to be called additionally:
library("rpart")
rp <- rpart(Species ~ ., data = iris)
plot(rp)
text(rp)
As these plots produced by rpart itself are not very appealing, there are
var
Patrick
Have a look at
https://stat.ethz.ch/pipermail/r-help/2006-August/110621.html
and
https://stat.ethz.ch/pipermail/r-help/2008-June/165279.html
I remember working it out with an example but I cannot remember any of the
details
Regards
Duncan
-Original Message-
From: R-help [mail
What happens if you run the code in a terminal rather than RStudio? My
experience is that very, very occasionally RStudio does something a bit funny
with plots.
And while this may sound funny just shut down RStudio, reload it and try again.
John Kane
Kingston ON Canada
> -Original Mess
On 29/06/2016 5:49 AM, g.maub...@weinwolf.de wrote:
Hi All,
I would like to install R packages from source on Windows 7 64-Bit.
Currently my settings are:
-- cut --
sessionInfo()
R version 3.3.0 (2016-05-03)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) S
Hi,
May be it is a basic thing but I would like to know if we can anticipate
round-off errors sign.
Here is an example :
# numerical matrix
m <- matrix(data=cbind(rnorm(10, 0), rnorm(10, 2), rnorm(10, 5)), nrow=10,
ncol=3)
> m
[,1] [,2] [,3]
[1,] 0.4816247 1
On 29/06/16 21:16, Witold E Wolski wrote:
It's the inverse problem to merging a list of data.frames into a large
data.frame just discussed in the "performance of do.call("rbind")"
thread
I would like to split a data.frame into a list of data.frames
according to first column.
This SEEMS to be eas
Hi All,
I would like to install R packages from source on Windows 7 64-Bit.
Currently my settings are:
-- cut --
> sessionInfo()
R version 3.3.0 (2016-05-03)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
locale:
[1] LC_COLLATE=German_Germany.
It's the inverse problem to merging a list of data.frames into a large
data.frame just discussed in the "performance of do.call("rbind")"
thread
I would like to split a data.frame into a list of data.frames
according to first column.
This SEEMS to be easily possible with the function base::by. How
On Mon, 27-Jun-2016 at 10:17PM -0700, Bert Gunter wrote:
[...]
|>
|> You seem to be making this way more difficult than you should.
Though I didn't get any closer to an understanding of which.panel, the
question I asked was simply answered by
panel.custom(factor.levels = )
Thanks to Duncan Ma
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