Re: [R] Sweave for inclusion of p value in a sentence of a LaTeX document
Dear Julia,
my way to do that is to attribute the t.test to an object,
and then refer to its p.value with the function
\Sexpr
e.g.
\documentclass{article}
\usepackage{Sweave}
\begin{document}
<>=
x<-cbind(1,2,3)
y<-cbind(3,4,5)
t <- t.test(x,y)
@
The p value for my data was \Sexpr{ round(t$p.value, 3) }
which is not significant.
\end{document}
Best,
Matthieu
Matthieu Dubois
Post-doctoral researcher
Department of Psychology and Neural Science, NYU
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Re: [R] Data frame reordering to time series
In the real data the months are all complete, but the years can be missing.
So years can be missing up front, in the middle, at the end. but if a year
is present than every month has a value or NA.
To create regular R ts I had to plow through the data frame, collect a year
caluculate an index to put it into the final time series.
I had tried zoo out and it handled the irregular spaced data, but a large
data structure of zoo objects had stumped me. espcially since I need to do
matching and selecting
of the zoo objects.
In the real data, there are about 7000 time series of 1500 months and those
7000
get averaged and combined in different ways
On Sat, Aug 7, 2010 at 8:45 PM, Gabor Grothendieck
wrote:
> On Sat, Aug 7, 2010 at 9:18 PM, steven mosher
> wrote:
> > Very Slick.
> > Gabor this is a Huge speed up for me. Thanks. ha, Now I want to rewrite a
> > bunch of working code.
> >
> >
> >
> > Id<-c(rep(67543,4),rep(12345,3),rep(89765,5))
> > Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1))
> > Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14)
> > Values<-c(12,14,34,21,54,65,23,12,13,13,13,14)
> >
>
> Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values)
> > Data
> >Index Year Jan Feb Mar Apr Jun
> > 1 67543 1989 12 6.0 12 12 12
> > 2 67543 1990 14 7.0 NA NA 14
> > 3 67543 1991 34 17.0 34 34 34
> > 4 67543 1992 21 10.5 21 21 21
> > 5 12345 1991 54 27.0 NA NA 54
> > 6 12345 1993 65 32.5 65 65 65
> > 7 12345 1994 23 11.5 23 23 23
> > 8 89765 1991 12 6.0 NA NA 12
> > 9 89765 1992 13 6.5 13 13 13
> > 10 89765 1993 13 6.5 NA NA 13
> > 11 89765 1994 13 6.5 13 13 13
> > 12 89765 1995 14 7.0 14 14 14
> > # Gabor's solution
> > f <- function(x) ts(c(t(x[-(1:2)])), freq = 12, start = x$Year[1])
> > do.call(cbind, by(Data, Data$Index, f))
> > 12345 67543 89765
>
>
> The original data had consecutive months in each series (actually
> there was a missing 1992 in one case but I assumed that was an
> inadvertent omission and the actual data was complete); however, here
> we have missing 6 month chunks in addition. That makes the series
> non-consecutive so to solve that we could either apply this to the
> data (after putting the missing 1992 year back in):
>
> Data <- cbind(Data, NA, NA, NA, NA, NA, NA)
>
> or we could use a time series class that can handle irregularly spaced
> data:
>
> library(zoo)
> f <- function(x) {
>dat <- x[-(1:2)]
>tim <- as.yearmon(outer(x$Year, seq(0, length = ncol(dat))/12, "+"))
>zoo(c(as.matrix(dat)), tim)
> }
> do.call(cbind, by(Data, Data$Index, f))
>
> The last line is unchanged from before. This code will also handle
> the original situation correctly even if the missing 1992 is truly
> missing.
>
[[alternative HTML version deleted]]
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Re: [R] Data frame reordering to time series
On Sat, Aug 7, 2010 at 9:18 PM, steven mosher wrote:
> Very Slick.
> Gabor this is a Huge speed up for me. Thanks. ha, Now I want to rewrite a
> bunch of working code.
>
>
>
> Id<-c(rep(67543,4),rep(12345,3),rep(89765,5))
> Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1))
> Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14)
> Values<-c(12,14,34,21,54,65,23,12,13,13,13,14)
> Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values)
> Data
> Index Year Jan Feb Mar Apr Jun
> 1 67543 1989 12 6.0 12 12 12
> 2 67543 1990 14 7.0 NA NA 14
> 3 67543 1991 34 17.0 34 34 34
> 4 67543 1992 21 10.5 21 21 21
> 5 12345 1991 54 27.0 NA NA 54
> 6 12345 1993 65 32.5 65 65 65
> 7 12345 1994 23 11.5 23 23 23
> 8 89765 1991 12 6.0 NA NA 12
> 9 89765 1992 13 6.5 13 13 13
> 10 89765 1993 13 6.5 NA NA 13
> 11 89765 1994 13 6.5 13 13 13
> 12 89765 1995 14 7.0 14 14 14
> # Gabor's solution
> f <- function(x) ts(c(t(x[-(1:2)])), freq = 12, start = x$Year[1])
> do.call(cbind, by(Data, Data$Index, f))
> 12345 67543 89765
The original data had consecutive months in each series (actually
there was a missing 1992 in one case but I assumed that was an
inadvertent omission and the actual data was complete); however, here
we have missing 6 month chunks in addition. That makes the series
non-consecutive so to solve that we could either apply this to the
data (after putting the missing 1992 year back in):
Data <- cbind(Data, NA, NA, NA, NA, NA, NA)
or we could use a time series class that can handle irregularly spaced data:
library(zoo)
f <- function(x) {
dat <- x[-(1:2)]
tim <- as.yearmon(outer(x$Year, seq(0, length = ncol(dat))/12, "+"))
zoo(c(as.matrix(dat)), tim)
}
do.call(cbind, by(Data, Data$Index, f))
The last line is unchanged from before. This code will also handle
the original situation correctly even if the missing 1992 is truly
missing.
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Re: [R] How to start R
Hi satimis, I also use Ubuntu 10.04 and R. I just wanted to say that it is very hard to get started in R. I'm guessing you know this, but I just wanted to add some encouragement and mention that if you find yourself getting frustrated at times or taking a very long time to do things that are extremely simple, do not give up! Once you struggle through it, you will be amazed at how much control you have over your data. Also, I would highly recommend getting a book. I'm not sure what the best ones are: R in a nutshell is quite good (and can be found at borders and barnes and noble). I think it's like 40 dollars but if you can spare it it's worth it in my opinion. The intro pdf is very well written but is too concise I think. Best of luck and enjoy the ride! -- View this message in context: http://r.789695.n4.nabble.com/How-to-start-R-tp2317188p2317539.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave for inclusion of p value in a sentence of a LaTeX document
Dear R Users,
I would like to include the p value in the results returned by the t.test
function in a sentence of a LaTeX document. For this purpose, I use the
following code (file.Rnw):
\documentclass{article}
\begin{document}
The p value for my data was
<>=
x<-cbind(1,2,3)
y<-cbind(3,4,5)
t.test(x,y)
@
which is not significant.
\end{document}
I use "R CMD Sweave file.Rnw" and "pdflatex file.tex" to create a PDF document
of it.
However, the all details of the t-test are included in my document and form a
new paragraph in another format than the rest of the original sentence.
The sentence should look like this: "The p value for my data was 0.2879 which
was not significant."
Thanks in advance.
Julia
Wassertemperaturen in Deutschland
Sommer, Sonne, Strand - wer braucht Abkühlung? Die aktuellen Wassertemperaturen
und Windgeschwindigkeiten für Deutschlands Badeseen gibt´s auf arcor.de.
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Re: [R] package for measurement error models
I believe there was a fairly recent exchange (within the last 6 months) about linear measurement error models/error-in-variable models/Deming regression/total least squares/orthogonal regression. Terry Therneau provided code for an R function that can estimate such models: http://www.mail-archive.com/r-help@r-project.org/msg85070.html If your knowledge of Fortran is up to the task, there is a Netlib package called ODRpack that can fit such models. Kind of surprising that no one has not written a R wrapper for this library (yet). Christos > Date: Sat, 7 Aug 2010 11:21:01 -0400 > From: carrieands...@gmail.com > To: R-help@r-project.org > Subject: [R] package for measurement error models > > Hi,all, > > I posted this question couple of days again, but haven't gotten any answers > back. I would like to post it again, and if you have any ideas, please let > me know. Any helps and suggestions are very much appreciated. > > The problem is about linear regression with both y and x have measurement, > and the variance of errors are heterogeneous. The estimated regression > coefficient and its variance are of interest. Is any R package doing this > task ? > > Thank you > > Carrie-- > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data frame reordering to time series
Very Slick. Gabor this is a Huge speed up for me. Thanks. ha, Now I want to rewrite a bunch of working code. Id<-c(rep(67543,4),rep(12345,3),rep(89765,5)) Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1)) Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14) Values<-c(12,14,34,21,54,65,23,12,13,13,13,14) Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values) Data Index Year Jan Feb Mar Apr Jun 1 67543 1989 12 6.0 12 12 12 2 67543 1990 14 7.0 NA NA 14 3 67543 1991 34 17.0 34 34 34 4 67543 1992 21 10.5 21 21 21 5 12345 1991 54 27.0 NA NA 54 6 12345 1993 65 32.5 65 65 65 7 12345 1994 23 11.5 23 23 23 8 89765 1991 12 6.0 NA NA 12 9 89765 1992 13 6.5 13 13 13 10 89765 1993 13 6.5 NA NA 13 11 89765 1994 13 6.5 13 13 13 12 89765 1995 14 7.0 14 14 14 # Gabor's solution f <- function(x) ts(c(t(x[-(1:2)])), freq = 12, start = x$Year[1]) do.call(cbind, by(Data, Data$Index, f)) 12345 67543 89765 Jan 1989NA 12.0NA Feb 1989NA 6.0NA Mar 1989NA 12.0NA Apr 1989NA 12.0NA May 1989NA 12.0NA Jun 1989NA 14.0NA Jul 1989NA 7.0NA Aug 1989NANANA Sep 1989NANANA Oct 1989NA 14.0NA Nov 1989NA 34.0NA Dec 1989NA 17.0NA Jan 1990NA 34.0NA Feb 1990NA 34.0NA Mar 1990NA 34.0NA Apr 1990NA 21.0NA May 1990NA 10.5NA Jun 1990NA 21.0NA Jul 1990NA 21.0NA Aug 1990NA 21.0NA Sep 1990NANANA Oct 1990NANANA Nov 1990NANANA Dec 1990NANANA Jan 1991 54.0NA 12.0 Feb 1991 27.0NA 6.0 ... On Sat, Aug 7, 2010 at 5:09 PM, steven mosher wrote: > Thanks Gabor, I probably should have done an example with fewer columns. > > i will rework the example and post it up so the next guys who has this > issue can have a > clear example with a solution. > > > > On Sat, Aug 7, 2010 at 5:04 PM, Gabor Grothendieck < > ggrothendi...@gmail.com> wrote: > >> On Sat, Aug 7, 2010 at 4:49 PM, steven mosher >> wrote: >> > Given a data frame, or it could be a matrix if I choose to. >> > The data consists of an ID, a year, and data for all 12 months. >> > Missing values are a factor AND missing years. >> > >> > Id<-c(rep(67543,4),rep(12345,3),rep(89765,5)) >> > Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1)) >> > Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14) >> > Values<-c(12,14,34,21,54,65,23,12,13,13,13,14) >> > >> >> Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values,July=Values/3,Aug=Values2,Sep=Values, >> > + Oct=Values,Nov=Values,Dec=Values2) >> > Data >> > Index Year Jan Feb Mar Apr Jun July Aug Sep Oct Nov Dec >> > 1 67543 1989 12 6.0 12 12 12 4.00 12 12 12 12 12 >> > 2 67543 1990 14 7.0 NA NA 14 4.67 NA 14 14 14 NA >> > 3 67543 1991 34 17.0 34 34 34 11.33 34 34 34 34 34 >> > 4 67543 1992 21 10.5 21 21 21 7.00 21 21 21 21 21 >> > 5 12345 1991 54 27.0 NA NA 54 18.00 NA 54 54 54 NA >> > 6 12345 1993 65 32.5 65 65 65 21.67 65 65 65 65 65 >> > 7 12345 1994 23 11.5 23 23 23 7.67 23 23 23 23 23 >> > 8 89765 1991 12 6.0 NA NA 12 4.00 NA 12 12 12 NA >> > 9 89765 1992 13 6.5 13 13 13 4.33 13 13 13 13 13 >> > 10 89765 1993 13 6.5 NA NA 13 4.33 NA 13 13 13 NA >> > 11 89765 1994 13 6.5 13 13 13 4.33 13 13 13 13 13 >> > 12 89765 1995 14 7.0 14 14 14 4.67 14 14 14 14 14 >> > >> > >> > The Goal is to return a Time series object for each ID. Alternatively >> one >> > could return a matrix that I can turn into a Time series. >> > The final structure would be something like this ( done in matrix form >> for >> > illustration) >> > 1989.0 1989.083 >> >1991 ..19921993. 1994 1995 >> > 67543 12 6.0 12 12 12 4.00 12 12 12 12 12... >> > .34...21.. NA.NANA >> > 12345 NA, NA, >> > NA,.54 27 >> > >> > Basically the time series will have patches at the front, middle and end >> > where you may have years of NA >> > The must be column ordered by time and aligned so that averages for all >> > series can be computed per month. >> > >> > Now I have looping code to do this, where I loop through all the IDs and >> map >> > the row of data into the correct >> > column. and create column names based on the data and row names based on >> the >> > ID, but it's painfully >> > slow. Any wizardry would help. >> >> Your email came out a bit garbled so its not clear what you want to >> get out but this code will produce a multivariate ts series, i.e. an >> mts series, with one column for each series: >> >> f <- function(x) ts(c(t(x[-(1:2
Re: [R] Data frame reordering to time series
Thanks Gabor, I probably should have done an example with fewer columns. i will rework the example and post it up so the next guys who has this issue can have a clear example with a solution. On Sat, Aug 7, 2010 at 5:04 PM, Gabor Grothendieck wrote: > On Sat, Aug 7, 2010 at 4:49 PM, steven mosher > wrote: > > Given a data frame, or it could be a matrix if I choose to. > > The data consists of an ID, a year, and data for all 12 months. > > Missing values are a factor AND missing years. > > > > Id<-c(rep(67543,4),rep(12345,3),rep(89765,5)) > > Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1)) > > Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14) > > Values<-c(12,14,34,21,54,65,23,12,13,13,13,14) > > > > Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values,July=Values/3,Aug=Values2,Sep=Values, > > + Oct=Values,Nov=Values,Dec=Values2) > > Data > > Index Year Jan Feb Mar Apr Jun July Aug Sep Oct Nov Dec > > 1 67543 1989 12 6.0 12 12 12 4.00 12 12 12 12 12 > > 2 67543 1990 14 7.0 NA NA 14 4.67 NA 14 14 14 NA > > 3 67543 1991 34 17.0 34 34 34 11.33 34 34 34 34 34 > > 4 67543 1992 21 10.5 21 21 21 7.00 21 21 21 21 21 > > 5 12345 1991 54 27.0 NA NA 54 18.00 NA 54 54 54 NA > > 6 12345 1993 65 32.5 65 65 65 21.67 65 65 65 65 65 > > 7 12345 1994 23 11.5 23 23 23 7.67 23 23 23 23 23 > > 8 89765 1991 12 6.0 NA NA 12 4.00 NA 12 12 12 NA > > 9 89765 1992 13 6.5 13 13 13 4.33 13 13 13 13 13 > > 10 89765 1993 13 6.5 NA NA 13 4.33 NA 13 13 13 NA > > 11 89765 1994 13 6.5 13 13 13 4.33 13 13 13 13 13 > > 12 89765 1995 14 7.0 14 14 14 4.67 14 14 14 14 14 > > > > > > The Goal is to return a Time series object for each ID. Alternatively one > > could return a matrix that I can turn into a Time series. > > The final structure would be something like this ( done in matrix form > for > > illustration) > > 1989.0 1989.083 > >1991 ..19921993. 1994 1995 > > 67543 12 6.0 12 12 12 4.00 12 12 12 12 12... > > .34...21.. NA.NANA > > 12345 NA, NA, > > NA,.54 27 > > > > Basically the time series will have patches at the front, middle and end > > where you may have years of NA > > The must be column ordered by time and aligned so that averages for all > > series can be computed per month. > > > > Now I have looping code to do this, where I loop through all the IDs and > map > > the row of data into the correct > > column. and create column names based on the data and row names based on > the > > ID, but it's painfully > > slow. Any wizardry would help. > > Your email came out a bit garbled so its not clear what you want to > get out but this code will produce a multivariate ts series, i.e. an > mts series, with one column for each series: > > f <- function(x) ts(c(t(x[-(1:2)])), freq = 12, start = x$Year[1]) > do.call(cbind, by(Data, Data$Index, f)) > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data frame reordering to time series
On Sat, Aug 7, 2010 at 4:49 PM, steven mosher wrote: > Given a data frame, or it could be a matrix if I choose to. > The data consists of an ID, a year, and data for all 12 months. > Missing values are a factor AND missing years. > > Id<-c(rep(67543,4),rep(12345,3),rep(89765,5)) > Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1)) > Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14) > Values<-c(12,14,34,21,54,65,23,12,13,13,13,14) > Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values,July=Values/3,Aug=Values2,Sep=Values, > + Oct=Values,Nov=Values,Dec=Values2) > Data > Index Year Jan Feb Mar Apr Jun July Aug Sep Oct Nov Dec > 1 67543 1989 12 6.0 12 12 12 4.00 12 12 12 12 12 > 2 67543 1990 14 7.0 NA NA 14 4.67 NA 14 14 14 NA > 3 67543 1991 34 17.0 34 34 34 11.33 34 34 34 34 34 > 4 67543 1992 21 10.5 21 21 21 7.00 21 21 21 21 21 > 5 12345 1991 54 27.0 NA NA 54 18.00 NA 54 54 54 NA > 6 12345 1993 65 32.5 65 65 65 21.67 65 65 65 65 65 > 7 12345 1994 23 11.5 23 23 23 7.67 23 23 23 23 23 > 8 89765 1991 12 6.0 NA NA 12 4.00 NA 12 12 12 NA > 9 89765 1992 13 6.5 13 13 13 4.33 13 13 13 13 13 > 10 89765 1993 13 6.5 NA NA 13 4.33 NA 13 13 13 NA > 11 89765 1994 13 6.5 13 13 13 4.33 13 13 13 13 13 > 12 89765 1995 14 7.0 14 14 14 4.67 14 14 14 14 14 > > > The Goal is to return a Time series object for each ID. Alternatively one > could return a matrix that I can turn into a Time series. > The final structure would be something like this ( done in matrix form for > illustration) > 1989.0 1989.083 > 1991 ..19921993. 1994 1995 > 67543 12 6.0 12 12 12 4.00 12 12 12 12 12... > .34...21.. NA.NANA > 12345 NA, NA, > NA,.54 27 > > Basically the time series will have patches at the front, middle and end > where you may have years of NA > The must be column ordered by time and aligned so that averages for all > series can be computed per month. > > Now I have looping code to do this, where I loop through all the IDs and map > the row of data into the correct > column. and create column names based on the data and row names based on the > ID, but it's painfully > slow. Any wizardry would help. Your email came out a bit garbled so its not clear what you want to get out but this code will produce a multivariate ts series, i.e. an mts series, with one column for each series: f <- function(x) ts(c(t(x[-(1:2)])), freq = 12, start = x$Year[1]) do.call(cbind, by(Data, Data$Index, f)) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] basic question about t-test with adjusted p value
ANOVA will give you an adjusted t-test. For example:
fit<-lm(height~sex+age)
summary(fit)
Will give an age-adjusted t-test of the heights of men and women.
John
John Sorkin
Chief Biostatistics and Informatics
Univ. of Maryland School of Medicine
Division of Gerontology and Geriatric Medicine
jsor...@grecc.umaryland.edu
-Original Message-
From: Erik Iverson
To:
Cc:
Sent: 8/7/2010 4:24:35 PM
Subject: Re: [R] basic question about t-test with adjusted p value
On 08/07/2010 03:08 PM, josef.kar...@phila.gov wrote:
> I have read the R manual and help archives, sorry but I'm still stuck.
>
> How would I do a t-test with an adjusted p-value?
Please be more specific about what you mean by 'adjusted p-value'. See below...
>
> Suppose that I use t.test ( ) , with the function argument alternative =
> "two.sided", and data such that degrees of freedom = 20. The function
> calculates a t-statistic of 2.086, and p-value =0.05
>
> How do I then adjust the p-value? My thought is to do
> p.adjust (pt(2.086, df=20),"BH")
> but that doesn't change anything (returns 0.975)
A couple things here.
1) You can get the p-value returned from t.test by inspecting the object with
?str, and noting that it's stored in an element called p-value.
So,
p.val <- t.test(extra ~ group, data = sleep)$p.value
assigns the p-value to an object called p.val.
2) You're calling p.adjust with a single p-value, so what kind of adjustment
did
you have in mind? You would normally be adjusting p-values because of a
multiple
comparison issue (i.e., multiple tests performed). You'd pass p.adjust the
*vector* of p-values, and a method to adjust them by. Your vector is of length
1, so in p.adjust's opinion, there is nothing to adjust for.
So, what do *you* mean by 'adjusted p-value'? `Why are you looking to adjust?`
is another way of stating that.
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Re: [R] plot the dependent variable against one of the predictors with other predictors as constant
There are many ways to do this. Here is one.
install.packages('rms')
require(rms)
dd <- datadist(x, y); options(datadist='dd')
f <- ols(z ~ x + y)
plot(Predict(f))# plot all partial effects
plot(Predict(f, x)) # plot only the effect of x
plot(Predict(f, y)) # plot only the effect of y
f <- ols(z ~ pol(x,2)*pol(y,2) # repeat, not assuming linearity
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Sat, 7 Aug 2010, Yi wrote:
Hi, folks,
Happy work in weekends >_<
My question is how to plot the dependent variable against one of the
predictors with other predictors as constant. Not for the original data, but
after prediction. It means y is the predicted value of the dependent
variables. The constane value of the other predictors may be the average or
some fixed value.
###
y=1:10
x=10:1
z=2:11
lin_model=lm(z~x+y)
x_new=11:2
###
How to plot predicted value of z from the regression model with x takes
x_new and y as a constant (let's say y=1)
I am thinking about using 'predict' command to generate the prediction of z
with the new data.frame but there should be a better way.
Thanks all.
Yi
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[R] plot the dependent variable against one of the predictors with other predictors as constant
Hi, folks, Happy work in weekends >_< My question is how to plot the dependent variable against one of the predictors with other predictors as constant. Not for the original data, but after prediction. It means y is the predicted value of the dependent variables. The constane value of the other predictors may be the average or some fixed value. ### y=1:10 x=10:1 z=2:11 lin_model=lm(z~x+y) x_new=11:2 ### How to plot predicted value of z from the regression model with x takes x_new and y as a constant (let's say y=1) I am thinking about using 'predict' command to generate the prediction of z with the new data.frame but there should be a better way. Thanks all. Yi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data frame reordering to time series
Given a data frame, or it could be a matrix if I choose to. The data consists of an ID, a year, and data for all 12 months. Missing values are a factor AND missing years. Id<-c(rep(67543,4),rep(12345,3),rep(89765,5)) Years<-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1)) Values2<-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14) Values<-c(12,14,34,21,54,65,23,12,13,13,13,14) Data<-data.frame(Index=Id,Year=Years,Jan=Values,Feb=Values/2,Mar=Values2,Apr=Values2,Jun=Values,July=Values/3,Aug=Values2,Sep=Values, + Oct=Values,Nov=Values,Dec=Values2) Data Index Year Jan Feb Mar Apr Jun July Aug Sep Oct Nov Dec 1 67543 1989 12 6.0 12 12 12 4.00 12 12 12 12 12 2 67543 1990 14 7.0 NA NA 14 4.67 NA 14 14 14 NA 3 67543 1991 34 17.0 34 34 34 11.33 34 34 34 34 34 4 67543 1992 21 10.5 21 21 21 7.00 21 21 21 21 21 5 12345 1991 54 27.0 NA NA 54 18.00 NA 54 54 54 NA 6 12345 1993 65 32.5 65 65 65 21.67 65 65 65 65 65 7 12345 1994 23 11.5 23 23 23 7.67 23 23 23 23 23 8 89765 1991 12 6.0 NA NA 12 4.00 NA 12 12 12 NA 9 89765 1992 13 6.5 13 13 13 4.33 13 13 13 13 13 10 89765 1993 13 6.5 NA NA 13 4.33 NA 13 13 13 NA 11 89765 1994 13 6.5 13 13 13 4.33 13 13 13 13 13 12 89765 1995 14 7.0 14 14 14 4.67 14 14 14 14 14 The Goal is to return a Time series object for each ID. Alternatively one could return a matrix that I can turn into a Time series. The final structure would be something like this ( done in matrix form for illustration) 1989.0 1989.083 1991 ..19921993. 1994 1995 67543 12 6.0 12 12 12 4.00 12 12 12 12 12... .34...21.. NA.NANA 12345 NA, NA, NA,.54 27 Basically the time series will have patches at the front, middle and end where you may have years of NA The must be column ordered by time and aligned so that averages for all series can be computed per month. Now I have looping code to do this, where I loop through all the IDs and map the row of data into the correct column. and create column names based on the data and row names based on the ID, but it's painfully slow. Any wizardry would help. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] basic question about t-test with adjusted p value
On 08/07/2010 03:08 PM, josef.kar...@phila.gov wrote: I have read the R manual and help archives, sorry but I'm still stuck. How would I do a t-test with an adjusted p-value? Please be more specific about what you mean by 'adjusted p-value'. See below... Suppose that I use t.test ( ) , with the function argument alternative = "two.sided", and data such that degrees of freedom = 20. The function calculates a t-statistic of 2.086, and p-value =0.05 How do I then adjust the p-value? My thought is to do p.adjust (pt(2.086, df=20),"BH") but that doesn't change anything (returns 0.975) A couple things here. 1) You can get the p-value returned from t.test by inspecting the object with ?str, and noting that it's stored in an element called p-value. So, p.val <- t.test(extra ~ group, data = sleep)$p.value assigns the p-value to an object called p.val. 2) You're calling p.adjust with a single p-value, so what kind of adjustment did you have in mind? You would normally be adjusting p-values because of a multiple comparison issue (i.e., multiple tests performed). You'd pass p.adjust the *vector* of p-values, and a method to adjust them by. Your vector is of length 1, so in p.adjust's opinion, there is nothing to adjust for. So, what do *you* mean by 'adjusted p-value'? `Why are you looking to adjust?` is another way of stating that. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] basic question about t-test with adjusted p value
On Sat, Aug 07, 2010 at 04:08:40PM -0400, josef.kar...@phila.gov wrote: > I have read the R manual and help archives, sorry but I'm still stuck. > > How would I do a t-test with an adjusted p-value? > > Suppose that I use t.test ( ) , with the function argument alternative = > "two.sided", and data such that degrees of freedom = 20. The function > calculates a t-statistic of 2.086, and p-value =0.05 > > How do I then adjust the p-value? My thought is to do > p.adjust (pt(2.086, df=20),"BH") > but that doesn't change anything (returns 0.975) > > what is the procedure? I'm sorry if there is a basic concept that I am > missing here... I'm confused - what result where you expecting? p.adjust will need to know the number of test you are trying to adjust for - either by giving explicitly giving the number or by handing a vector of p-values to the function. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan Maximus-von-Imhof-Forum 3 85354 Freising, Germany http://webclu.bio.wzw.tum.de/~pagel/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] basic question about t-test with adjusted p value
On Sat, Aug 7, 2010 at 1:08 PM, wrote: > I have read the R manual and help archives, sorry but I'm still stuck. > > How would I do a t-test with an adjusted p-value? > > Suppose that I use t.test ( ) , with the function argument alternative = > "two.sided", and data such that degrees of freedom = 20. The function > calculates a t-statistic of 2.086, and p-value =0.05 > > How do I then adjust the p-value? My thought is to do > p.adjust (pt(2.086, df=20),"BH") > but that doesn't change anything (returns 0.975) > > what is the procedure? I'm sorry if there is a basic concept that I am > missing here... These adjustments are generally designed to control a family wise or experiment wise error rate. If you only have one p-value, then the adjustment is equivalent to the unadjusted. The argument n defaults to length(p) (which is the length of the first argument). Here are some examples: > p.adjust(p = 0.975, method = "BH") [1] 0.975 > p.adjust(p = 0.975, method = "BH", n = 1) [1] 0.975 > p.adjust(p = 0.975, method = "BH", n = 5) [1] 1 HTH, Josh > > > > > > > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] basic question about t-test with adjusted p value
I have read the R manual and help archives, sorry but I'm still stuck. How would I do a t-test with an adjusted p-value? Suppose that I use t.test ( ) , with the function argument alternative = "two.sided", and data such that degrees of freedom = 20. The function calculates a t-statistic of 2.086, and p-value =0.05 How do I then adjust the p-value? My thought is to do p.adjust (pt(2.086, df=20),"BH") but that doesn't change anything (returns 0.975) what is the procedure? I'm sorry if there is a basic concept that I am missing here... [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Q] a dummy variable used with sapply
On 08/07/2010 01:00 PM, GMail (KU) wrote:
Hello,
While learning how to manipulate data with R, I found one example that I
could not understand.
In the following example, the function definition of "maxcor" has an argument
named "i" and I don't understand why. Could someone explain why the maxcor
function definition needs to have this argument?
maxcor = function(i, n = 10, m = 5) { mat = matrix(rnorm(n*m),n,m) corr =
cor(mat) diag(corr) = NA max(corr,na.rm=TRUE) }
maxcors = sapply(1:1000, maxcor, n = 100)
Well, the short answer is that you're instinct is correct, it doesn't need it.
The `i` parameter is just being used as a counter. `sapply` is passing the
numbers from 1 to 1000 to this function each time it is invoked, but the
function definition does not actually use this number.
I'm guessing it is there because whoever wrote maxcor did not know about, or
understand, the ?replicate function, and just used the sapply function. (Note
`replicate` is essentially automatically doing what the above code does.)
The following is untested, but will probably work...
maxcor <- function(n = 10, m = 5)
{
mat <- matrix(rnorm(n*m), n, m)
corr <- cor(mat)
diag(corr) <- NA
max(corr, na.rm = TRUE)
}
maxcors <- replicate(n = 1000, maxcor(n = 100))
where the first n represents the number of replicates to perform,
and the second n is the value passed to maxcor. It's just a coincidence that
they are named the same thing.
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[R] [Q] a dummy variable used with sapply
Hello,
While learning how to manipulate data with R, I found one example that I could
not understand.
In the following example, the function definition of "maxcor" has an argument
named "i" and I don't understand why.
Could someone explain why the maxcor function definition needs to have this
argument?
maxcor = function(i, n = 10, m = 5)
{
mat = matrix(rnorm(n*m),n,m)
corr = cor(mat)
diag(corr) = NA
max(corr,na.rm=TRUE)
}
> maxcors = sapply(1:1000, maxcor, n = 100)
Thanks in advance.
Young-Jin Lee
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[R] Fwd: quantmod Example-google data download-problems
-- Forwarded message --
From: Velappan Periasamy
Date: Sat, Aug 7, 2010 at 11:20 PM
Subject: quantmod Example-google data download-problems
To: r-sig-fina...@stat.math.ethz.ch
getSymbols("YHOO",src="google") is working
getSymbols("NSE:RCOM",src="google") is not working.
then how to download the stock data for the above symbol?.
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Re: [R] Colours on conditional levelplots in package "lattice"
gordon.morrison wrote:
>
> I am having difficulty in getting the colours on a conditional levelplot
> in
> the package lattice to work as I want them to - I wonder if someone could
> help me.
>
I did not understand exactly what you wanted because the example was a bit
complext, but the following comes closer to your requests:
require(lattice)
#col1 <-
c(1.37,2.35,0.20,1.85,0.274,1.56,-0.76,-0.03,-1.39,-0.95,-0.50,0.91,1.58,0.17,-0.34,-0.48,1.50,-0.68,1.05)
col2 <-
c("A","B","C","F","G","H","B","C","E","G","H","A","B","C","D","E","F","G","H")
col3 <-
c("X","Y","X","Y","X","Y","Y","X","X","X","Y","X","Y","X","Y","X","Y","X","Y")
col4 <-
c("1","1","1","1","1","1","2","2","2","2","2","3","3","3","3","3","3","3","3")
dat1 <- data.frame(X1=col1, var1=col2, var2=col3, var3=col4)
dat1$X1 = 1:nrow(dat1)-nrow(dat1)/2
maxAbsValue <- max(abs(as.vector(dat1[, "X1"])))
colorkeyAt <- pretty(c(-maxAbsValue, maxAbsValue), n = 10)
boxColours <- c("#8B", "#A83500", "#C46B00", "#E2A100", "#FFD700",
"#ADD8E6", "#81A2CF", "#566CB8", "#2B36A1", "#8B")
levelplot(X1 ~ var1 * var2 | var3, data = dat1,
col.regions = boxColours,
colorkey = list(at = colorkeyAt,
labels = list(labels = colorkeyAt, at = colorkeyAt,
cex =0.6)),
at = pretty(c(-maxAbsValue, maxAbsValue), n = 10),
panel = function(x, y, z, ...) {
panel.fill(col = "darkgrey")
panel.levelplot(x, y, z,...)
})
--
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Re: [R] apply family functions
Sapply is not significantly faster than a for loop. Vectorization
generally is, but you pay for it in RAM.
> dat.oc <- oc[dat$Class,]
> dat$Flag <- ifelse(with(dat.oc,(Open<=dat$Close_date &
dat$Close_date<=Close) | (Open1<=dat$Close_date &
dat$Close_date<=Close1)),"Valid","Invalid")
If you are really hurting for RAM, you might take a rather less
computationally efficient approach:
> dat$Flag <- sapply(1:length(dat$Class), function(idx,ddat,toc){cl <-
ddat[idx,"Class"]
cld <- ddat[idx,"Close_date"]
if ( (toc[cl,"Open"]<=cld && cld<=toc[cl,"Close"]) ||
(toc[cl,"Open1"]<=cld && cld<=toc[cl,"Close1"])) {result <- "Valid"}
else {result <- "Invalid"}
c(Flag=result) }, ddat=dat, toc=oc )
Steven Kang wrote:
ini <- as.Date("2010/1/1", "%Y/%m/%d")
# Generate arbitrary data frame consisting of date values
oc <- data.frame(Open = seq(ini, ini + 6, 1), Close = seq(ini + 365, ini +
365 + 6, 1), Open1 = seq(ini + 365*2, ini + 365*2 + 6, 1), Close1 = seq(ini
+ 365*3, ini + 365*3 + 6, 1), Open2 = seq(ini + 365*4, ini + 365*4 + 6, 1),
Close2 = seq(ini + 365*5, ini + 365*5 + 6, 1))
rownames(oc) <- c("AAA", "C", "AA", "A", "CC", "BB", "B")
oc
Open Close Open1Close1
Open2Close2
AAA 2010-01-01 2011-01-01 2012-01-01 2012-12-31 2013-12-31 2014-12-31
C 2010-01-02 2011-01-02 2012-01-02 2013-01-01 2014-01-01
2015-01-01
AA2010-01-03 2011-01-03 2012-01-03 2013-01-02 2014-01-02 2015-01-02
A 2010-01-04 2011-01-04 2012-01-04 2013-01-03 2014-01-03
2015-01-03
CC2010-01-05 2011-01-05 2012-01-05 2013-01-04 2014-01-04 2015-01-04
BB2010-01-06 2011-01-06 2012-01-06 2013-01-05 2014-01-05 2015-01-05
B 2010-01-07 2011-01-07 2012-01-07 2013-01-06 2014-01-06
2015-01-06
dat <- data.frame(Class = c("AAA", "C", "CC", "BB", "B", "A"), Close_date =
c(ini, ini, ini, ini+109, ini+39, ini+24), stringsAsFactors = FALSE)
ind <- sapply(dat$Class, function(x) match(x, rownames(oc)))
for (i in length(ind)) {
dat[["Flag"]] <- sapply(dat[["Close_date"]], function(x) ifelse((x >=
oc[ind[[i]], 1] & x < oc[ind[[i]], 2]) | (x >= oc[ind[[i]], 3] & x <
oc[ind[[i]], 4]) | (x >= oc[ind[[i]], 5] & x < oc[ind[[i]], 6]),
"Valid", "Invalid"))
}
dat
Class Close_dateFlag
*1 AAA2010-01-01 Invalid*
2 C 2010-01-01 Invalid
3CC2010-01-01Invalid
4BB2010-04-20Valid
5 B 2010-02-09Valid
6 A 2010-01-25Valid
The first record (highlighted in yellow) is flagged as "Invalid" where it
should really be "Valid".
Any suggestions on resolving this would be great.
Many thanks.
[[alternative HTML version deleted]]
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Re: [R] package for measurement error models
On Sat, 7 Aug 2010, Carrie Li wrote:
Hi,all,
I posted this question couple of days again, but haven't gotten any answers
back. I would like to post it again, and if you have any ideas, please let
me know. Any helps and suggestions are very much appreciated.
Start by reading the Posting Guide.
It will tell you to
*) Do RSiteSearch("keyword") with different keywords (at the R prompt)
to search R functions, contributed packages and R-Help postings.
and
RSiteSearch("measurement error")
gives a number of hits that might be of interest.
If these hits do not pan out, you should let us know why none of the
available packages that have 'measurement error' in their descriptions
satisfy your needs.
HTH,
Chuck
The problem is about linear regression with both y and x have measurement,
and the variance of errors are heterogeneous. The estimated regression
coefficient and its variance are of interest. Is any R package doing this
task ?
Thank you
Carrie--
[[alternative HTML version deleted]]
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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Re: [R] eval-parse and lme in a loop
Connolly, Colm wrote: > > Hi everybody, > > I'm having trouble getting an eval-parse combination to work with lme in a > for loop. > > I have not checked in detail, but it looks like another case for what we call "Ripley's game": http://finzi.psych.upenn.edu/R/Rhelp02a/archive/16599.html Dieter -- View this message in context: http://r.789695.n4.nabble.com/eval-parse-and-lme-in-a-loop-tp2315622p2317292.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply family functions
Hi Steven,
You code has two problems. One is loop structure which should be for(i in
1:length). The second is that you loop the process twice (for and sapply).
Hope followed code will work.
for (i in 1:length(ind)) {
x <- dat$Close_date[i]
dat[["Flag"]][i] <- ifelse((x >= oc[ind[[i]], 1] & x < oc[ind[[i]], 2])
| (x >= oc[ind[[i]], 3] & x < oc[ind[[i]], 4])
| (x >= oc[ind[[i]], 5] & x < oc[ind[[i]], 6]),
"Valid", "Invalid")
}
Yours,
Wu
-
A R learner.
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Re: [R] How to start R
Hi Erik,
Thanks for your advice and URL
q() is on the last line of the printout on starting R
Type 'q()' to quit R.
Sorry I overlook it
B.R.
Stephen
- Original Message
From: Erik Iverson
To: Stephen Liu
Cc: Steve Lianoglou ; R-help@r-project.org
Sent: Sat, August 7, 2010 11:45:44 PM
Subject: Re: [R] How to start R
On 08/07/2010 10:29 AM, Stephen Liu wrote:
> Hi steve,
>
> After starting R which command shall I use to exit it rather than close the
> console. exit/quit did not work. Thanks
What does "did not work" mean?
help.search("quit")
leads you to:
?quit
> quit()
or
> q()
should terminate the R session.
I suggest you read "An Introduction to R" at
http://cran.r-project.org/doc/manuals/R-intro.html, where q() is first
mentioned
in section 1.5.
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Re: [R] How to start R
On 08/07/2010 10:29 AM, Stephen Liu wrote:
Hi steve,
After starting R which command shall I use to exit it rather than close the
console. exit/quit did not work. Thanks
What does "did not work" mean?
help.search("quit")
leads you to:
?quit
> quit()
or
> q()
should terminate the R session.
I suggest you read "An Introduction to R" at
http://cran.r-project.org/doc/manuals/R-intro.html, where q() is first mentioned
in section 1.5.
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Re: [R] How to start R
Hi steve, After starting R which command shall I use to exit it rather than close the console. exit/quit did not work. Thanks B.R. Stephen - Original Message From: Steve Lianoglou To: Stephen Liu Cc: Tim Gruene ; R-help@r-project.org Sent: Sat, August 7, 2010 9:18:42 PM Subject: Re: [R] How to start R Hi, On Sat, Aug 7, 2010 at 8:56 AM, Stephen Liu wrote: > Hi Tim, > > I got it. Thanks for reminding me. It should be R (capital letter) not r. Indeed, you were typing in a "little r", as opposed to a capital R. Now, note the package list you installed: r-base littler r-cran-plotrix r-cran-qtl r-cran-rggobi One of which is "littler" -- aka "little R", which is a scripting front end for R :-) http://dirk.eddelbuettel.com/code/littler.html Enjoy, -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] package for measurement error models
Hi,all, I posted this question couple of days again, but haven't gotten any answers back. I would like to post it again, and if you have any ideas, please let me know. Any helps and suggestions are very much appreciated. The problem is about linear regression with both y and x have measurement, and the variance of errors are heterogeneous. The estimated regression coefficient and its variance are of interest. Is any R package doing this task ? Thank you Carrie-- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
manchester.ac.uk> writes: > > On 07-Aug-10 09:29:41, Michael Bedward wrote: > > Thanks for that clarification Peter - much appreciated. > > > > Is there an R function that you'd recommend for calculating > > more valid CIs ? > > Michael > > It depends on what you want to mean by "more valid"! If you have > a 95% CI for the linear predictor (say L(x) at X=x), then the > probability that the CI will include the true value of L(x) > is 95% (more or less accurately, depending on what approximation, > if any, was used to obtain the CI). Thus, if A(Y) and B(Y) are the > lower and upper limits of a 95% CI for L(x) as functions of the data Y, [snip] > So you can't have everything at once, and it depends on what you > want to mean by "valid"! [snip] Yow. Nothing like r-help on a Saturday morning! Practically speaking, I think the previous recommendations (confint() and boot.ci()) are probably best. I prefer equal probabilities in tails to a symmetric confidence interval. (You could also try for Bayesian credible intervals, which are symmetric in the probability cutoff for each side ...) The other thing to keep in mind is that once you get down to this level of rigor, it's extremely likely that the major source of error/uncertainty in your results will be some other systematic error or violation of the assumptions. Are your data *really* distributed the way you think? Linear on the scale of the linear predictor, independent, homogeneous probabilities/exchangeability within groups, ... ? Or maybe that's just my excuse for not worrying too much. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] image plot but data not on grid.
> If it's regular, but incomplete, that method will work. If it's > irregular, then no image method will work without first interpolating > a regular grid. Thanks Hadley. As I suspected, but ggplot2 is so very clever that I thought it was worth asking :) Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] image plot but data not on grid.
On Sat, Aug 7, 2010 at 2:54 AM, Michael Bedward wrote: > On 7 August 2010 06:26, Hadley Wickham wrote: > >> library(ggplot2) >> qplot(x, y, fill = z, data = df, geom = "tile") > > Hi Hadley, > > I read the original question as being about irregularly spaced data. > The above method doesn't seem to for me in such a case but perhaps I'm > constructing my example incorrectly. If it's regular, but incomplete, that method will work. If it's irregular, then no image method will work without first interpolating a regular grid. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to start R
Hi, On Sat, Aug 7, 2010 at 8:56 AM, Stephen Liu wrote: > Hi Tim, > > I got it. Thanks for reminding me. It should be R (capital letter) not r. Indeed, you were typing in a "little r", as opposed to a capital R. Now, note the package list you installed: r-base littler r-cran-plotrix r-cran-qtl r-cran-rggobi One of which is "littler" -- aka "little R", which is a scripting front end for R :-) http://dirk.eddelbuettel.com/code/littler.html Enjoy, -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to start R
Hi Tim, I got it. Thanks for reminding me. It should be R (capital letter) not r. On console:- ~$ R R version 2.10.1 (2009-12-14) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. B.R. Stephen - Original Message From: Tim Gruene To: Stephen Liu Cc: R-help@r-project.org Sent: Sat, August 7, 2010 8:45:04 PM Subject: Re: [R] How to start R What exactly do you mean with 'just hanging there'? Do you expect anything else? When I start R on Debian squeeze (with the command 'R'), I see R version 2.11.1 (2010-05-31) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > and I can start working from there. Tim On Sat, Aug 07, 2010 at 05:30:29AM -0700, Stephen Liu wrote: > Hi folks, > > Just finished installing R on Ubuntu 10.04, running on a VM, download > following > > packages on repo; > > r-base > littler > r-cran-plotrix > r-cran-qtl > r-cran-rggobi > > But I could not get R started. r is on /usr/bin/r > > On console evoking it just hanging there. Any additional packages I need to > install? Thanks > > B.R. > > > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen GPG Key ID = A46BEE1A __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to start R
What exactly do you mean with 'just hanging there'? Do you expect anything else? When I start R on Debian squeeze (with the command 'R'), I see R version 2.11.1 (2010-05-31) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > and I can start working from there. Tim On Sat, Aug 07, 2010 at 05:30:29AM -0700, Stephen Liu wrote: > Hi folks, > > Just finished installing R on Ubuntu 10.04, running on a VM, download > following > packages on repo; > > r-base > littler > r-cran-plotrix > r-cran-qtl > r-cran-rggobi > > But I could not get R started. r is on /usr/bin/r > > On console evoking it just hanging there. Any additional packages I need to > install? Thanks > > B.R. > > > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen GPG Key ID = A46BEE1A signature.asc Description: Digital signature __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to start R
Hi folks, Just finished installing R on Ubuntu 10.04, running on a VM, download following packages on repo; r-base littler r-cran-plotrix r-cran-qtl r-cran-rggobi But I could not get R started. r is on /usr/bin/r On console evoking it just hanging there. Any additional packages I need to install? Thanks B.R. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
Michael Bedward wrote: > On 7 August 2010 19:56, Martin Maechler wrote: > >> I'm coming late to the thread, >> but it seems that nobody has yet given the advice which I would >> very *strongly* suggest to anyone asking for confidence >> intervals in GLMs: >> >> Use confint() > > confint was actually mentioned in the second post on the thread :) > But it's not what the OP is after. Yep. The problem is that we cannot (yet? easily?) do confint on a transformation of the parameters, hence not on the predicted values (on whatever scale). Another thing is that confint() isn't quite a cure-all because it relies on the distribution of the (signed square root of) the LRT, and if that is substantially different from chi-square(1) (or N(0,1)), then you might still not obtain improved coverage, etc. That being said, one should probably say that it is quite common to base CI's on the linear predictor scale, and there's nothing REALLY wrong with that procedure. As will have transpired by now, once you leave this rather well-trodden road, you can easily find yourself in a rather impenetrable thicket... As far as I recall the asymptotics, all of the "+/- 2*se" procedures have approximation errors of the same order as that caused by the discrete distribution of the response, so it is not like one method is going to "win" by orders of magnitude. > Michael -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] several figures from one Sweave chunk? [solved]
(on-list)
Hello Cameron
On Fri, 6 Aug 2010 19:02:22 +0100
Liviu Andronic wrote:
> On Fri, 6 Aug 2010 11:30:59 -0600
> Cameron Bracken wrote:
> > There is no real good way to deal with this in Sweave. Sweave does
> > not actually know anything about the graphics it is creating, all
> > it knows it that a graphic is or is not being created. The problem
> > lies with the graphic devices, each has its own way of creating a
> > new "page." That is when you create a new plot without closing the
> > graphics device. For example, pdf() createS a new page in the pdf
> > document, png() just overwrites the old image and tikz() creates a
> > whole new tikzpicture. These are all done without Sweave knowing a
> > bit about it.
> >
> > So the answer is no, pgfSweave nor cacheSweave have any way of
> > dealing with this. In my experience, you are much better off
> > either creating separate chunks for each image, or using something
> > like layout:
> >
> > <>=
> > layout(matrix(c(1,3,2,4),nrow=2))
> > for (i in 1:4) plot(rnorm(100)+i)
> > @
> >
Unfortunately neither creating separate chunks for each image nor using
layout are options in my case: I need to automate the repetitive task
of describing the variables in my data set (15 for one df, about 45
for the second), using graphs and summary stats. This would require to
loop over mixed R/LaTeX code, something not allowed by Sweave.
Fortunately, I found a way to work around, based on the Sweave FAQ [1].
If you need not only graphs, but some LaTeX markup as well, Dieter
Menne proposed a very elegant solution that involved defining a
\newcommand containing most LaTeX code and calling it in the R loop
via a cat() statement [2]. Another alternative is to put all LaTeX code
in cat() statements, but this tends to get messy.
The final hurdle was to output some 'verbatim' text from within a
<>=
@
code chunk, which contained the entire loop. This can
be done by printing the object within
cat("\\begin{verbatim}\n")
[..]
cat("\\end{verbatim}\n")
I would have wanted to include this in the \newcommand call, but since
my object was a list, it couldn't be passed via cat(). I would love to
hear a workaround to this.
Here are the links to the LyX [3] and Sweave [4] versions and the PDF
[5] result of my file, extending Dieter's example.
Another potential solution, that I haven't yet fully investigated, is
brew. Contrary to my first impressions, its syntax is simpler than that
of Sweave. Moreover, it "natively" handles looping over mixed R/LaTeX
code and is thus perfect for generating repetitive reports. Combined
with a new 'weave' method [6], it seems that brew could easily be used
instead of Sweave.
To echo Dieter's comments, I had some headaches when thinking
about the hen and the egg. Hope this helps
Liviu
[1] http://www.stat.uni-muenchen.de/~leisch/Sweave/FAQ.html#x1-11000A.9
[2] https://stat.ethz.ch/pipermail/r-help/2008-June/164783.html
[3] http://s000.tinyupload.com/index.php?file_id=22646520644600938098
[4] http://s000.tinyupload.com/index.php?file_id=73594116524789598605
[5] http://s000.tinyupload.com/index.php?file_id=00010585455841885948
[6] http://biostatmatt.com/archives/573/comment-page-1#comment-770
> > Hope that helps. If you have a specific application where you need
> > this type of functionality, I may be able so suggest a workaround.
> >
> > -Cameron
> >
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Re: [R] Confidence Intervals for logistic regression
On 07-Aug-10 09:29:41, Michael Bedward wrote:
> Thanks for that clarification Peter - much appreciated.
>
> Is there an R function that you'd recommend for calculating
> more valid CIs ?
> Michael
It depends on what you want to mean by "more valid"! If you have
a 95% CI for the linear predictor (say L(x) at X=x), then the
probability that the CI will include the true value of L(x)
is 95% (more or less accurately, depending on what approximation,
if any, was used to obtain the CI). Thus, if A(Y) and B(Y) are the
lower and upper limits of a 95% CI for L(x) as functions of the data Y,
P(A(Y) < L(x) < B(Y)) = 0.95 (to within approximation)
and this may be asymmetrical in that we may have
P((A(Y) > L(x)) = 1 - P(B(Y) < L(x)) != 0.025
(e.g. it may come out as a 1%:4% split of the 5%).
The response probability P(Y=1 | X=x) will be a monotonic function
F(L(x)) of x -- e.g. for the logistic exp(L)/(1+exp(L)), increasing
from 0 to 1. Then {F(A(Y)), F(B(Y)} is a 95% CI for P = F(x),
since P[A(Y) < L(x) < B(Y)] = P[F(A(Y)) < F(L(x))=P(x) < F(B(Y))]. Also,
P[A(Y) > L(x)] = P(F(A(Y)) > F(L(x) = P(x)] and
P[B(Y) < L(x)] = P(F(B(Y)) < F(L(x) = P(x)]
for exactly the same reason (monotonicity of F). Hence the split
of the 5% between left tail and right tail ion the response scale
P(x) = F(L(x)) is exactly the same as the split on the linear
predictor scale L(x).
Therefore, on that front (comparison of probabilities of coverage),
the CI transformed to the response scale {F(A(Y)), F(B(Y)} is exactly
as valid as the CI {A(Y),B(Y)} on the original linear predictor scale.
In particular, if the latter is "equi-tailed" (2.5% on either side)
then the former will be too. If that is what you mean by "valid",
then you're finished.
However, possibly you may want "valid" to mean "extending to equal
distances on either side of the point estimate" -- e.g. as you
do with Estimate +/- 1.96*SE. It may be that, on the linear predictor
scale, you achieve this and also equi-tailed (2.5% either way).
But then, when you transform to the response scale, you will lose
that symmetry: F(Est - 1.96*SE) and F(Est + 1.96*SE) will not
be equidistant from F(Est) (though the equi-tailed 2.5%:2.5%
of the tail probability will be preserved).
If you have a reason for wanting to, you can start with a 95% CI
for L(x) which is not equi-tailed, but does have the property of
symmetry in the response scale: F(Est - 1.96*SE) and F(Est + 1.96*SE)
will be equidistant from F(Est). So you could set up the CI for
L(x) as {A(x) = Est - c0(x)*SE, B(x) = Est + c1(x)*SE} where c0(x)
and c1(x) (which in general depend on x) are chosen so that
you get symmetry on the response scale. But then you will lose
the equi-tailed property on the linear predictor scale, hence also
on the response scale.
So you can't have everything at once, and it depends on what you
want to mean by "valid"!
However, in the case of response being the probability of Y=1,
you might want to be careful about symmetry on the response
scale, since that could result in a CI which goes above 1 or
below 0, which would not be "valid" ...
For large samples, asymptotically all these issues tend to dwindle
into near-irrelevance, since locally the reponse is close to linear
and whatever you achieve on one scale will be (close to) achieved
on the other scale.
Hoping this helps,
Ted.
> On 7 August 2010 18:37, Peter Dalgaard wrote:
>>
>> Probably, neither is optimal, although any transformed scale is
>> asymptotically equivalent. E.g., neither the probability scale
>> nor the logit scale stabilizes the variance of a simple proportion
>> (the arcsine transform does), so test-based CIs should really be
>> asymmetric in both cases rather than just +/- 1.96se.
>>
>> However, working on the linear predictor scale has the advantage
>> that CIs by definition will not cross the boundaries of the
>> parameter space. (For the "usual" link functions: logit, probit,
>> cloglog, that is; it's not true for the identity link, obviously.)
>> --
>> Peter Dalgaard
>> Center for Statistics, Copenhagen Business School
>> Phone: (+45)38153501
>> Email: pd@cbs.dk _Priv: pda...@gmail.com
E-Mail: (Ted Harding)
Fax-to-email: +44 (0)870 094 0861
Date: 07-Aug-10 Time: 11:42:09
-- XFMail --
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Re: [R] Confidence Intervals for logistic regression
On 7 August 2010 19:56, Martin Maechler wrote: > I'm coming late to the thread, > but it seems that nobody has yet given the advice which I would > very *strongly* suggest to anyone asking for confidence > intervals in GLMs: > > Use confint() confint was actually mentioned in the second post on the thread :) But it's not what the OP is after. Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
> "PD" == Peter Dalgaard > on Sat, 07 Aug 2010 10:37:49 +0200 writes: PD> Michael Bedward wrote: >>> I was aware of this option. I was assuming it was not ok to do fit +/- 1.96 >>> se when you requested probabilities. If this is legitimate then all the >>> better. >> >> I don't think it is. I understood that you should do the calculation >> in the scale of the linear predictor and then transform to >> probabilities. >> >> Happy to be corrected if that's wrong. PD> Probably, neither is optimal, although any transformed scale is PD> asymptotically equivalent. E.g., neither the probability scale nor the PD> logit scale stabilizes the variance of a simple proportion (the arcsine PD> transform does), so test-based CIs should really be asymmetric in both PD> cases rather than just +/- 1.96se. PD> However, working on the linear predictor scale has the advantage that PD> CIs by definition will not cross the boundaries of the parameter space. PD> (For the "usual" link functions: logit, probit, cloglog, that is; it's PD> not true for the identity link, obviously.) I'm coming late to the thread, but it seems that nobody has yet given the advice which I would very *strongly* suggest to anyone asking for confidence intervals in GLMs: Use confint() which (for "glm"s) will load the MASS package and use likelihood - profiling, giving (much) more reliable confidence intervals, notably in the case of the infamous Hauck-Donner phenomenon, which has been mentioned many times "here", and notably in MASS (the book!), I think even in its first edition. Even more reliable (probably) would be to use the (recommended) 'boot' package and use bootstrap confidence intervals, i.e., boot.ci() there. Martin Maechler, ETH Zurich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rJava question
In a HTML file, a java applet can be launch by What is the equivalent rJava command(s) to do the same? Thank you Best -- Wincent Rong-gui HUANG Doctoral Candidate Dept of Public and Social Administration City University of Hong Kong http://asrr.r-forge.r-project.org/rghuang.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
Thanks for that clarification Peter - much appreciated. Is there an R function that you'd recommend for calculating more valid CIs ? Michael On 7 August 2010 18:37, Peter Dalgaard wrote: > > Probably, neither is optimal, although any transformed scale is > asymptotically equivalent. E.g., neither the probability scale nor the > logit scale stabilizes the variance of a simple proportion (the arcsine > transform does), so test-based CIs should really be asymmetric in both > cases rather than just +/- 1.96se. > > However, working on the linear predictor scale has the advantage that > CIs by definition will not cross the boundaries of the parameter space. > (For the "usual" link functions: logit, probit, cloglog, that is; it's > not true for the identity link, obviously.) > > -- > Peter Dalgaard > Center for Statistics, Copenhagen Business School > Phone: (+45)38153501 > Email: pd@cbs.dk Priv: pda...@gmail.com > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
Michael Bedward wrote: >> I was aware of this option. I was assuming it was not ok to do fit +/- 1.96 >> se when you requested probabilities. If this is legitimate then all the >> better. > > I don't think it is. I understood that you should do the calculation > in the scale of the linear predictor and then transform to > probabilities. > > Happy to be corrected if that's wrong. Probably, neither is optimal, although any transformed scale is asymptotically equivalent. E.g., neither the probability scale nor the logit scale stabilizes the variance of a simple proportion (the arcsine transform does), so test-based CIs should really be asymmetric in both cases rather than just +/- 1.96se. However, working on the linear predictor scale has the advantage that CIs by definition will not cross the boundaries of the parameter space. (For the "usual" link functions: logit, probit, cloglog, that is; it's not true for the identity link, obviously.) -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] image plot but data not on grid.
On 7 August 2010 06:26, Hadley Wickham wrote: > library(ggplot2) > qplot(x, y, fill = z, data = df, geom = "tile") Hi Hadley, I read the original question as being about irregularly spaced data. The above method doesn't seem to for me in such a case but perhaps I'm constructing my example incorrectly. Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
Stefano, I was aware of this option. I was assuming it was not ok to do fit +/- 1.96 se when you requested probabilities. If this is legitimate then all the better. Thanks! Troy On 6 August 2010 22:25, Guazzetti Stefano wrote: > a closer look to the help on predict.glm will reveal that the function > accepts a 'type' argument. > In you case 'type = response' will give you the results in probabilities > (that it seems to be what you are looking for). > There also is an example on use of the 'type' argument at the end of the > page. > > Stefano > > -Messaggio originale- > Da: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org]per conto di Troy S > Inviato: Friday, August 06, 2010 6:31 PM > A: Michael Bedward > Cc: r-help@r-project.org > Oggetto: Re: [R] Confidence Intervals for logistic regression > > > Michael, > > Thanks for the reply. I believe Aline was sgiving me CI's on coefficients > as well. > > So c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * > pred$se.fit) gives me the CI on the logits if I understand correctly? > Maybe > the help on predict.glm can be updated. > > Thanks! > > On 6 August 2010 01:46, Michael Bedward wrote: > > > Sorry about earlier reply - didn't read your email properly (obviously :) > > > > You're suggestion was right, so as well as method for Aline below, > > another way of doing the same thing is: > > > > pred <- predict(y.glm, newdata= something, se.fit=TRUE) > > ci <- matrix( c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * > > pred$se.fit), ncol=2 ) > > > > lines( something, plogis( ci[,1] ) ) > > lines( something, plogis( ci[,2] ) ) > > > > > > > > On 6 August 2010 18:39, aline uwimana wrote: > > > Dear Troy, > > > use this commend, your will get IC95% and OR. > > > > > > logistic.model <- glm(formula =y~ x1+x2, family = binomial) > > > summary(logistic.model) > > > > > > sum.coef<-summary(logistic.model)$coef > > > > > > est<-exp(sum.coef[,1]) > > > upper.ci<-exp(sum.coef[,1]+1.96*sum.coef[,2]) > > > lower.ci<-exp(sum.coef[,1]-1.96*sum.coef[,2]) > > > > > > cbind(est,upper.ci,lower.ci) > > > > > > regards. > > > > > > 2010/8/6 Troy S > > > > > >> Dear UseRs, > > >> > > >> I have fitted a logistic regression using glm and want a 95% > confidence > > >> interval on a response probability. Can I use > > >> > > >> predict(model, newdata, se.fit=T) > > >> > > >> Will fit +/- 1.96se give me a 95% of the logit? And then > > >> exp(fit +/- 1.96se) / (exp(fit +/- 1.96se) +1) to get the > probabilities? > > >> > > >> Troy > > >> > > >>[[alternative HTML version deleted]] > > >> > > >> __ > > >> R-help@r-project.org mailing list > > >> https://stat.ethz.ch/mailman/listinfo/r-help > > >> PLEASE do read the posting guide > > >> http://www.R-project.org/posting-guide.html > > >> and provide commented, minimal, self-contained, reproducible code. > > >> > > > > > >[[alternative HTML version deleted]] > > > > > > __ > > > R-help@r-project.org mailing list > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > PLEASE do read the posting guide > > http://www.R-project.org/posting-guide.html > > > and provide commented, minimal, self-contained, reproducible code. > > > > > > >[[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > Rispetta l'ambiente: Se non ti è necessario, non stampare questa mail. > > > "Le informazioni contenute nel presente messaggio di posta elettronica e in > ogni suo allegato sono da considerarsi riservate e il destinatario della > email è l'unico autorizzato > ad usarle, copiarle e, sotto la propria responsabilità, divulgarle. > Chiunque riceva questo messaggio per errore senza esserne il destinatario > deve immediatamente rinviarlo > al mittente cancellando l'originale. Eventuali dati personali e sensibili > contenuti nel presente messaggio e/o suoi allegati vanno trattati nel > rispetto della normativa > in materia di privacy ( DLGS n.196/'03)". > > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence Intervals for logistic regression
> I was aware of this option. I was assuming it was not ok to do fit +/- 1.96 > se when you requested probabilities. If this is legitimate then all the > better. I don't think it is. I understood that you should do the calculation in the scale of the linear predictor and then transform to probabilities. Happy to be corrected if that's wrong. Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
