Re: [R] retrieve p-value from a cox.obj
On Wed, 2007-08-29 at 08:40 -0700, clearsky wrote: > I have a cox.obj named obj, > obj <- coxph( Surv(time, status) ~ group, surv.data) > now I want to retrieve the p-value from obj, so that I can run this hundreds > of times and plot out the distribution of the p-value. could anyone tell me > how to get p-value from obj? > > thanks, Hi Clearsky try: summary(obj)$waldtest[3] -- Bernardo Rangel Tura, M.D,Ph.D National Institute of Cardiology Brazil __ R-help@stat.math.ethz.ch 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] Sequential Rank Test
Hi R-Masters
I need use a sequential approach in serie of cases, but may data is not
normal.
If data is normal distribution is very easy create analysis using
likelihood ratio like of Wald test.
But in my case I need use a non-parametric test (Mann-Whitney).
I was use: RSiteSearch("sequential rank test") but not solve my
problem.
Do you know routine or package implement sequential rank test in R?
Thanks in advance
--
Bernardo Rangel Tura, M.D,Ph.D
National Institute of Cardiology
Brazil
__
R-help@stat.math.ethz.ch 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] Flow Cytometry Standard, fcs format in R.
On Fri, 2007-07-13 at 16:06 -0300, Horacio Castellini wrote: > Hi all. > How do I extract date from fcs format file with R. I.e I'd like > make statistical analysis using R-program, but I don't know if there > are R-packets for fcs format file, and using examples. > > Thanks. Hi Horacio! Is possible using rflowcyt or prada available in http://www.bioconductor.org In Rnews have article about this: http://cran.r-project.org/doc/Rnews/Rnews_2006-5.pdf Bernardo Rangel Tura, MD, Ph.D National Institute of Cardiology Rio de Janeiro - Brazil __ R-help@stat.math.ethz.ch 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] enable-R-shlib ?
Hi R Masters! Recently I migrated to Ubuntu Linux [I a former windows user]. Now I think compile a 64-bit R version for my computer [Turion AMD], but I not sure if using the configure option --enable-R-shlib. I note this option is usefull for some GUI like gnomeGui and JGR, but this will go penalty performace so I ask: 1- I must using this option? 2- If I using this option, have an another option wich i gain performace [like --enable-Blas]? Thanks for all -- Bernardo Rangel Tura, M.D, PhD National Institute of Cardiolgy Rio de Janeiro - Brasil __ R-help@stat.math.ethz.ch 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] compile errors with rgl-0.70.570 on FreeBSD
On Wed, 2007-04-04 at 21:12 +0200, Rainer Hurling wrote: > Thank you for your work on rgl. > > Reading in another thread about your new (inofficial) version of rgl > (see below) I tried it with R version 2.5.0 alpha (2007-03-31 r40986) > under FreeBSD 7.0-CURRENT. Unfortunately I got the following error: > > --- > #R CMD INSTALL rgl_0.70.570.tar.gz > * Installing to library '/usr/local/lib/R/library' > * Installing *source* package 'rgl' ... > checking for libpng-config... yes > configure: using libpng-config > configure: using libpng dynamic linkage > checking for X... libraries /usr/X11R6/lib, headers /usr/X11R6/include > checking GL/gl.h usability... no > checking GL/gl.h presence... no > checking for GL/gl.h... no > checking GL/glu.h usability... no > checking GL/glu.h presence... no > checking for GL/glu.h... no > configure: error: missing required header GL/gl.h > ERROR: configuration failed for package 'rgl' > ** Removing '/usr/local/lib/R/library/rgl' > ** Restoring previous '/usr/local/lib/R/library/rgl' > --- Rainer, I have same problem in my Ubuntu. I solve instaling OpenGl in system, in my case I install this debian poackages: freeglut3, freeglut3-dev, libgl1-mesa-dev, libglu1-mesa-dev, libice-dev, libsm-dev, libx11-dev, libxau-dev, libxdmcp-dev, libxext-dev, libxt-dev, mesa-common-dev, x11proto-core-dev, x11proto-input-dev, x11proto-kb-dev, x11proto-xext-dev, xlibmesa-gl-dev and xtrans-dev If you install similar packages in your system I think your problem will be solve. -- Bernardo Rangel Tura, M.D, PhD National Institute of Cardiolgy Rio de Janeiro - Brasil __ R-help@stat.math.ethz.ch 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] error installing packages
On Wed, 2007-03-07 at 09:33 -0800, Bricklemyer, Ross S wrote:
> I was finally able to get R to 'configure', 'make', and 'install' on Mandriva
> 2007. Itried to install gnomeGUI and I received an error. See below. At
> what step do I make R a shared library? Where did I go wrong?
>
> Ross
>
> ==
> downloaded 74Kb
>
> * Installing *Frontend* package 'gnomeGUI' ...
> Using R Installation in R_HOME=/usr/local/lib64/R
> R was not built as a shared library
> Need a shared R library
> ERROR: configuration failed for package 'gnomeGUI'
> * Removing '/usr/local/lib64/R/library/gnomeGUI'
>
> The downloaded packages are in
> /root/tmp/RtmpkHUeyA/downloaded_packages
> Warning message:
> installation of package 'gnomeGUI' had non-zero exit status in:
> install.packages(c("gnomeGUI"))
> =
Hi Ross!
I use Ubuntu and instaling R using Synaptic. In this software have this
observation about GnomeGUI:
As of R 2.1.0, this interface is no longer provided with the upstream
sources. As such, this package is now an empty stub that will be removed
in a subsequent revision of the Debian package.
So I think gnomeGUI not instalable in R now...
--
Bernardo Rangel Tura,M.D.,Ph.D
National Institute of Cardiology
Rio de Janeiro - Brazil
__
R-help@stat.math.ethz.ch 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] Sigmoidal fitting
On Sat, 2007-03-03 at 16:30 -0500, Cressoni, Massimo (NIH/NHLBI) [F] wrote: > I am trying to write a function that fits a sigmoid given a X and Y vector > guessing the start parameters. > I use nls. What I did (enclosed) seems to work well with many data points but > if I want to fit small > vectors like : > > pressure <- c(5,15,9,35,45) > gas <- c(1000,2000,3000,4000,5000) > > it do not work. The help page says that it do no not work on zero residual > data. > > Massimo Cressoni Hi Massimo I think which this script solve your question pressure <- c(5,9,15,35,45) gas <- c(1000,2000,3000,4000,5000) plot(gas,pressure) model.1 <- nls(pressure ~ SSlogis(gas, ASym, xmid, scal))) coef.sig<-coef(summary(model.1))[,1] est.p<-coef.sig[1]/(1+exp((coef.sig[2]-gas)/coef.sig[3])) points(gas,est.p,col=2) ---- If you need more help just need a mail -- Bernardo Rangel Tura,M.D.,Ph.D National Institute of Cardiology Rio de Janeiro - Brazil __ R-help@stat.math.ethz.ch 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] prop.trend.test issue
At 03:55 AM 12/3/2006, Ethan Johnsons wrote: >I have the clinical study data. > >Year 0 Year 3 >Retinol (nmol/L)N Mean +-sd Mean +-sd >Vitamin A group 73 1.89+-0.36 2.06+-0.53 >Trace group57 1.83+-0.31 1.78+-0.30 > >where N is the number of male for the clinical study. > >I want to test if the mean serum retinol has increased over 3 years >among subjects in the vitamin A group. If You desire check mean serum retinol has increased over 3 years in vitamin A group. You may use t.test Look this example: #Generate random Data set.seed(123) VitA1<-rnorm(73,1.89,.36) Trace1<-rnorm(57,1.83,0.31) VitA2<-rnorm(73,2.06,.53) Trace2<-rnorm(57,1.78,0.30) # Calculate diference Year 3 - Year 0 dVitA<-VitA2-VitA1 dTrace<-Trace2-Trace1 # Testing diference t.test(dVitA,dTrace) Welch Two Sample t-test data: dVitA and dTrace t = 2.2762, df = 117.746, p-value = 0.02464 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.02756874 0.39659494 sample estimates: mean of x mean of y 0.15905162 -0.05303022 Bernardo Rangel Tura, MD, Phd National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch 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] Colour-coded Editor for R Code
At 07:11 AM 11/6/2006, Jon Minton wrote: >Do any of you know any simple programming editors for R scripts which offer >basic colour-coding and bracket-matching facilities? > >Dregging through scripts to find a missing comma or parentheses is something >I'd rather do less of... > > > >Jon Minton Well If you use windows I recomend Tinn-R https://sourceforge.net/projects/tinn-r Bernardo Rangel Tura, MD, PHD National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch 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] plotCI
Hi R masters! I need a Help with plot confidence intervals for one equation. I use library gplots and plotCI command in this script: require(gplots) ano <-1980:2002 rf<-exp(91.37162-0.04720281*ano) ciw.f<-sqrt(1.766073e-08) plotCI(ano,rf,uiw=ciw.f) But in the graph not shown the errors bar and I have this error msg zero-length arrow is of indeterminate angle and so skipped Well, where is my eror? Thanks in advance Bernardo Rangel Tura, MD, PhD National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch 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] Call for Beta Testers: R+ (read R plus) for Solaris and Linux:
At 12:21 PM 8/8/2006, [EMAIL PROTECTED] wrote: >We look forward to hearing your comments and inputs on R+ ... please >feel free to suggest a final name for our commercially supported R. I don´t understanding this mail. Is possible exist a commercial version of R? If R source is licensed for GPL as free software other people can make a commercial version? []s Tura __ R-help@stat.math.ethz.ch 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] Excel to R
>Hi peolple!
I have a many excel tables with mode than 100 variables. And I want
use R to analize that.
But I have a problem, a group of this variables (more than 50) in any
table is a factor and other part is a number.
Tha factors variables have tha values enconde this form (1=Yes,2=No and 9 = NA)
Well I use this scripts to import the database
require(RODBC)
channel <- odbcConnectExcel("f:/teste.xls")
data <- sqlFetch(channel, "Sheet1")
summary(data)
qw ee
Min. :1.000 Min. :1.000
1st Qu.:1.000 1st Qu.:1.500
Median :1.000 Median :2.000
Mean :1.333 Mean :2.429
3rd Qu.:1.750 3rd Qu.:3.500
Max. :2.000 Max. :4.000
NA's :1.000
But qw is a factor (and is colnum type isvtext)
Is possible modify my script for this utcome
> summary(data)
qw ee
1 :4 Min. :1.000
2 :2 1st Qu.:1.500
NA's:1 Median :2.000
Mean :2.429
3rd Qu.:3.500
Max. :4.000
Thanks in advance
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] generalized hypergeometric function
Hi R-masters I need compute generalized hypergeometric function. I look in R-project and R-help list and not find nothing about generalized hypergeometric function Is possible calculate the generalized hypergeometric function? Somebody have a script for this? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to compare areas under ROC curves calculated with ROCR package
At 11:54 AM 3/23/2006, Frank Samuelson wrote: >The seROC routine you included is an very good approximation to the >standard error of the Mann-Whitney-Wilcoxon/Area under the ROC curve >statistic. It is derived from negative exponential models, but works >very well in general (e.g. Hanley and McNeil, Diagnostic Radiology, >1982, v. 143, p. 29). >A more general estimator of the variance is given by Campbell, >Douglas and Bailey, Proc. Computers in Cardiology, 1988, p.267) >I've implemented that in R code included below. It is not an unbiased >estimator, but it is very close. > >The cROC function is probably not what you want, however. >It assumes that the data from the two different area measures >are independent. You said your measures are "from the same dataset." >Your different AUC measures will be highly correlated. >There are a number of methods to deal with correlated ROC curves >in existence. > >If you are interested in performing hypothesis testing on the difference >in AUC of two parameters, I would suggest a permutation test. >Permuting the ranks of the data between parameters is >simple and works well. Laurent, Jarek, Frank My routines seROC and cROC was developed based in the paper A method of comparing the areas under Receiver Operating Characteristic curves derived from the same cases. Hanley JA & McNeil BJ, Radiology 1983 148 839-43 The seROC is calculation of standart error of ROC and cROC is calculation of statistical significance for two AUC in same sample. I don´t documented this rotine but it assume that the data from the two different area measures are DEPENDENT. A dependency of two measure is fix with incorporation of Pearson correlation among this measures. I wait that this information helps yours to fix this problem. Well... If you s think I can help yours just talk. Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Hierarchical log-linear models
Hi R-masters!
I try using package gRbase for fit a Hierarchical log-linear models
and show as graphical model.
I using data about patients of my Hospital (file sm.csv) and my script is :
setwd("F:/INCL/TURA")
dados<-read.csv("sm.csv")
attach(dados)
tab1<-table(obesidade,HAS,TG,DM,HDL)
require(gRbase)
dados.g <- as.gmData(tab1)
m2 <-hllm(~obesidade*HAS*TG*DM*HDL,dados.g)
m2.f <- fit(m2,engine="loglm")
Well my doubt is why
dynamic.Graph(m2) is the same model of dynamic.Graph(m2.f)
Where I wrong?
Thanks in advance
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil obesidade,HAS,TG,DM,HDL,SM
N,S,S,S,S,N
N,S,N,S,S,N
N,S,N,S,N,N
S,S,S,S,S,S
S,S,S,S,S,S
N,S,N,S,S,N
S,S,S,N,S,S
N,S,N,S,S,N
N,N,N,N,S,N
N,N,N,S,S,N
N,S,N,S,N,N
N,N,S,S,S,N
N,S,S,S,S,N
N,S,N,S,S,N
N,S,N,N,S,N
S,S,N,S,S,S
N,S,N,N,S,N
N,S,S,S,S,N
S,S,N,S,N,S
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,N,S,N
N,S,N,S,S,N
N,S,N,S,N,N
N,S,N,N,N,N
N,S,N,N,S,N
S,S,N,S,N,S
S,S,N,S,N,S
N,N,S,S,N,N
N,S,S,N,S,N
N,N,N,S,S,N
N,S,N,S,S,N
S,S,N,S,S,S
N,S,N,S,N,N
N,S,N,S,S,N
N,S,N,S,N,N
S,S,S,S,N,S
N,S,N,S,S,N
N,S,N,S,N,N
N,S,N,S,S,N
N,S,S,S,S,N
N,S,N,S,N,N
S,S,N,S,N,S
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N,N,N,N,S,N
N,S,N,N,S,N
N,S,N,S,N,N
N,N,S,S,S,N
S,S,N,S,S,S
N,S,S,S,N,N
N,N,S,S,N,N
N,N,S,N,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,N,S,N
S,S,N,N,S,S
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N,N,N,S,S,N
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N,S,N,S,S,N
N,S,N,N,S,N
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N,S,S,S,S,N
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N,S,N,S,S,N
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N,S,N,S,S,N
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N,N,N,S,S,N
N,N,N,N,S,N
N,S,N,S,N,N
N,S,N,N,N,N
N,N,S,S,N,N
N,S,S,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,S,S,S,N
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N,S,N,N,S,N
N,S,N,N,S,N
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N,S,N,S,S,N
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N,S,N,S,N,N
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N,S,N,S,S,N
N,N,N,N,S,N
N,N,N,N,S,N
N,S,N,N,S,N
S,S,N,S,S,S
N,N,N,N,S,N
N,S,N,N,S,N
N,S,N,N,S,N
N,S,N,S,S,N
N,N,N,S,S,N
N,S,S,S,S,N
N,S,N,N,S,N
N,S,S,N,S,N
N,S,S,N,S,N
N,N,S,S,S,N
N,S,S,S,S,N
N,S,N,S,N,N
N,S,N,S,S,N
N,N,S,S,S,N
N,S,N,S,S,N
N,S,N,N,N,N
N,N,S,N,S,N
N,S,N,S,S,N
N,S,S,S,S,N
S,S,N,S,S,S
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N,S,S,N,N,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
S,S,N,N,S,S
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,S,S,S,N
N,S,N,N,S,N
N,N,S,S,S,N
N,S,N,S,N,N
N,S,N,N,S,N
N,S,N,S,N,N
N,S,S,S,S,N
S,S,S,S,S,S
N,S,S,S,S,N
N,S,S,S,S,N
N,S,S,S,S,N
S,S,N,N,S,S
N,S,N,S,S,N
N,S,S,S,N,N
N,S,N,N,S,N
N,S,N,S,S,N
S,S,S,S,S,S
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,S,N,S,N
N,S,N,S,S,N
N,S,N,N,S,N
N,S,S,S,S,N
N,S,N,S,N,N
N,N,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
S,S,N,N,S,S
N,S,N,S,S,N
N,N,N,S,S,N
S,S,S,S,S,S
N,S,N,N,S,N
N,S,N,S,N,N
N,N,N,S,N,N
N,N,N,S,S,N
N,S,S,N,S,N
N,S,S,N,S,N
N,S,N,N,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,N,S,N
S,S,S,S,S,S
N,S,S,N,S,N
N,S,S,S,N,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,N,S,N
N,S,S,N,S,N
N,S,S,N,N,N
N,S,S,N,S,N
N,S,N,N,S,N
N,S,N,N,S,N
N,S,N,N,S,N
N,S,N,S,S,N
N,S,N,N,S,N
N,S,N,N,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,S,N,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
N,S,N,S,S,N
S,S,N,S,S,S
S,S,N,S,N,S
N,S,N,N,N,N
N,N,S,S,N,N
N,N,N,S,N,N
N,S,N,S,S,N
N,N,N,N,S,N
Re: [R] How can I use r-cran-lmtest?
At 12:28 AM 3/2/2006, Savio Ramos wrote: >Hi, > >I installed the package r-cran-lmtest in a Debian Sid but I can't >use it. I typed "lmtest" but nothing occur. > >Any help? > >Thanks. Hi Savio! The Lmtest is a package for Testing Linear Regression Models. This is a organized group of rotines, data files, help files etc abou this theme. for use this package You have type: require(lmtest) help(package="lmtest") Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Fitdistr and MLE for parameter lambda of Poisson distribution
At 09:35 AM 2/10/2006, Gregor Gorjanc wrote:
>Hello!
>
>I would like to get MLE for parameter lambda of Poisson distribution. I
>can use fitdistr() for this. After looking a bit into the code of this
>function I can see that value for lambda and its standard error is
>estimated via
>
>estimate <- mean(x)
>sds <- sqrt(estimate/n)
>
>Is this MLE? With my poor math/stat knowledge I thought that MLE for
>Poisson parameter is (in mixture of LaTeX code)
>
>l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).
>
>Is this really equal to (\sum^n_{i=1} x_i) / n
>
>--
>Lep pozdrav / With regards,
> Gregor Gorjanc
Gregor,
If I understood your LaTeX You is rigth.
If you don´t know have a command wich make this for you: fitdistr()
Look:
> d<- rpois(50,5)
> d
[1] 6 4 6 4 5 5 4 11 7 5 7 3 5
10 4 9 4 2 4 5 4 4 9 3 10
[26] 4 3 9 6 7 5 4 2 7 3 6 7 8 6 6 3 3 3 2 5 4 3 8 5 7
> library(MASS)
> fitdistr(d,"Poisson")
lambda
5.320
(0.3261901)
[]s
Tura
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Re: [R] Coefficient of association for 2x2 contingency tables
At 04:48 PM 12/6/2005, Alexandre Santos Aguiar wrote:
>Hi,
>
>Found no measure of association or correlation for 2x2 contingency tables in
>fullrefman.pdf or google. Can someone point to a package that implements such
>calculations?
>
>Thanx.
>
>--
>
> Alexandre Santos Aguiar
IN vcd poackage have many measure.
If You Know formulae do do you create a function, example
epitable<-function(exposure, outcome,OR=F,RISK=F)
{
tab <- table(exposure, outcome,deparse.level = 2)
a <- tab[1,1]; b <- tab[1,2]; c <- tab[2,1]; d <- tab[2,2]
m1<-a+b; m2<-c+d; p1<-a/m1; p2<-c/m2
if(RISK){
rr <- (a / (a + b)) / (c / (c + d))
se.log.rr <- sqrt((b / a) / (a + b) + (d / c) / (c + d))
lci.rr <- exp(log(rr) - 1.96 * se.log.rr)
uci.rr <- exp(log(rr) + 1.96 * se.log.rr)
rd<-p1-p2
se.rd.mle<-((p1*(1-p1))/m1)-((p2*(1-p2))/m2)
se.rd.ub<-((p1*(1-p1))/(m1-1))-((p2*(1-p2))/(m2-1))
lci.rd.mle<- rd - 1.96*se.rd.mle
uci.rd.mle<- rd + 1.96*se.rd.mle
lci.rd.ub<- rd - 1.96*se.rd.ub
uci.rd.ub<- rd + 1.96*se.rd.ub
}
if(OR){
or <- (a / b) / (c / d)
se.log.or <- sqrt(1 / a + 1 / b + 1 / c + 1 / d)
lci.or <- exp(log(or) - 1.96 * se.log.or)
uci.or <- exp(log(or) + 1.96 * se.log.or)
}
print(tab)
chi<-chisq.test(tab,correct=F)
if (OR | RISK){
cat("_")
cat("\n Estimate 95% CI ")
cat("\n_\n")
}
if (RISK){
cat("RR",round(rr,3),"",
round(lci.rr,3), round(uci.rr,3), "\n")
cat("RD",round(rd,3),"",
round(lci.rd.mle,3), round(uci.rd.mle,3), " (MLE)\n")
cat(" ", round(lci.rd.ub,3),
round(uci.rd.ub,3), " (UB)\n")}
if (OR){cat("OR", round(or,3), " ", round(lci.or,3),
round(uci.or,3), "\n")}
if (OR | RISK){cat("_\n")}
print(chi)
}
FA<-rbinom(250,1,.3)
obito<-rbinom(250,1,.1)
epitable(FA,obito,RISK=T)
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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Re: [R] Distribution fitting problem
At 10:32 2/11/2005, you wrote: >I am using the MASS library function > >fitdistr(x, dpois, list(lambda=2)) > >but I get > >Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : > Function cannot be evaluated at initial parameters >In addition: There were 50 or more warnings (use warnings() to see the first >50) > >and all the first 50 warnings say > >1: non-integer x = 1.45 >etc > >Can anyone tell me what I am doing wrong. p.s. the data was read in from >a .csv file that I wrote using octave Mark, Try fitdistr(x, "Poisson") I think this is enough for fit Poisson distribuition for your data Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Converting PROC NLMIXED code to NLME
At 20:40 7/10/2005, Singh, Jatinder wrote: >NLME > >kidney.nlme<-nlme(model=rtime~ >(event* >((b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u)+log(delta)+log(gamma)+(gamma-1)*lo >g(rtime)) >+((-exp(b1*age+b2*sex+b3*gn+b4*an+b5*pkn+u))*delta*(rtime**gamma)) >), >fixed=list(delta~1,gamma~1,b1~1,b2~1,b3~1,b4~1,b5~1), >random=u~1|patient, >start=c(delta=0.03,gamma=1.1,b1=-0.003,b2=-1.2,b3=0.09,b4=0.35,b5=-1.43) >, >data=(kidney), >method="ML", >na.action=na.include >) Hi! Try change "na.action=na.include" for "na.action=na.omit" The singularity occurs because the fixed effects matrix is not of full rank due to the unused factor levels. Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] missing values in step procedure
At 11:11 7/10/2005, you wrote: >Hi, >I have the problem that for the step procedure stops due to missing >values. There are no options in Step or stepAIC to handle missing >values. Is there any way to run stepwise modelselection in R in an >automated way in this case? > >Here is the last step before it stops. Hope someone knows. Best regards, >Andreas > >Step: AIC= 1999.16 > EF ~ SF120_KS + SF120_PS + HADA0 + SOZU0 + LVEDD + logPROBNP + > ALTER + SD0_01 + ASE_UK + DS140POS + RSQSICH0 + SD0_01:ASE_UK + > SD0_01:DS140POS + SD0_01:RSQSICH0 + ASE_UK:DS140POS + >ASE_UK:RSQSICH0 + > DS140POS:RSQSICH0 + SD0_01:ASE_UK:RSQSICH0 + >SD0_01:DS140POS:RSQSICH0 + > ASE_UK:DS140POS:RSQSICH0 > >Df Sum of Sq RSS AIC >- SOZU0 1 3.0 25356.0 1997.2 >- HADA0 1 7.6 25360.6 1997.3 >- ALTER 1 13.0 25365.9 1997.4 >- SF120_PS 1 14.7 25367.6 1997.5 >- ASE_UK:DS140POS:RSQSICH0 1 20.1 25373.1 1997.6 >- SD0_01:DS140POS:RSQSICH0 1 44.8 25397.7 1998.0 >- SD0_01:ASE_UK:RSQSICH01 54.4 25407.4 1998.2 > 25352.9 1999.2 >- LVEDD 1 382.2 25735.1 2004.6 >- SF120_KS 1 476.4 25829.3 2006.4 >- logPROBNP 1 891.9 26244.9 2014.4 >Error in step(mod2, direction = "back") : > number of rows in use has changed: remove missing values? Andreas Try data<-na.omit(original database) before you run step() or stepAIC() Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Fisher's method in discriminant analysis
At 13:15 5/9/2005, you wrote:
>Hi,
>
> I'm using mda library to solve a discriminant
>analysis. I get results, but the thing is that I want
>to use Fisher's method to obtain the classification
>functions and I'm lost in what I should do: libraries
>to use, ... Can anybody give me a clue??
Hi Carlos,
I think you need something this
require(mda)
data(iris)
attach(iris)
irisfit <- mda(Species ~ ., data = iris)
irisfit$fit$coef[,1:2]
r1<-
-4.23481161+0.37972423*Sepal.Length+0.59682846*Sepal.Width+0.01575609*Petal.Length+0.11009570*Petal.Width
r2<-0.31169082-0.05826293*Sepal.Length-0.23855482*Sepal.Width+0.18922693*Petal.Length+0.03917652*Petal.Width
plot(r1,r2,pch = 21, bg = c("red", "green3", "blue")[unclass(iris$Species)])
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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[R] loglinear model selection
Hi R-masters!
I have a problem and need your help.
I have 9 discrete variables with 2 levels each.
In exploratory analisys I generate one matrix with chi-square for tables
with 2 ariables each with this script
setwd("F:/")
dados<-read.csv("log.csv")[,2:10]
dados.x<-matrix(NA,ncol=9,nrow=9)
for(i in 1:8){
for(j in (i+1):9){
tab<-table(dados[,i],dados[,j])
dados.x[i,j]<-round(as.numeric(chisq.test(tab)$statistic),3)
dados.x[j,i]<-round(as.numeric(chisq.test(tab)$statistic),3)
}
}
dados.x
The result is:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,]NA 3.589 6.351 3.957 4.269 0.851 2.955 1.188 1.975
[2,] 3.589 NA 9.664 24.596 12.510 26.284 8.580 3.608 6.574
[3,] 6.351 9.664 NA 25.054 10.300 12.189 19.811 0.192 11.744
[4,] 3.957 24.596 25.054 NA 50.032 35.587 22.401 1.950 26.631
[5,] 4.269 12.510 10.300 50.032 NA 80.876 54.954 0.127 61.573
[6,] 0.851 26.284 12.189 35.587 80.876 NA 57.346 0.741 49.738
[7,] 2.955 8.580 19.811 22.401 54.954 57.346 NA 1.520 80.615
[8,] 1.188 3.608 0.192 1.950 0.127 0.741 1.520NA 0.311
[9,] 1.975 6.574 11.744 26.631 61.573 49.738 80.615 0.311 NA
Now I need fit a loglinear model with this variables, but I need know have
a command with generate ALL models with the set this 8 vairables (ALL minus
[,8]) incluindind the interactions.
Can Anyone Help me?
Thanks in advance
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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Re: [R] R: to the power
At 10:11 12/7/2005, [EMAIL PROTECTED] wrote: >hi all > >why does R do this: > >(-8)^(1/3)=NaN > >the answer should be : -2 Allan In my computer: > (-8)^(1/3) [1] NaN > -8^(1/3) [1] -2 > -(8^(1/3)) [1] -2 The problem is -8 or the problem is (-8) ? []s Tura -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Tree Structured Survival Analysis
Hi People! Have someone package with: Tree Structured Survival Analysis in R? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problem in lme longitudinal model
Hi R-masters! I trying model Heart disease mortality in my country with a lme model like this: m1.lme<-lme(log(rdeath)~age*year,random=~age|year,data=dados) where: rdeath is rate of mortality per 10 person per age and year age: age of death (22 27 32 37 42 47 52 57 62 67 72 77 82) year: year of death (1980:2002) I don´t have problem to fit the model, but in residual analysis I have one problem. If i type acf(m1.lme$residuals) the graph show 4 plot with a sin curve (a example in attach) and I get this variogram: > Variogram(m1.lme) variog dist n.pairs 1 0.39390651 276 2 0.64864522 253 3 0.96798703 230 4 1.27650944 207 5 1.41581475 184 6 1.42646856 161 7 1.40486087 138 8 1.19026138 115 9 0.96193309 92 10 0.7037427 10 69 11 0.8591257 11 46 12 1.2215657 12 23 Well I need help for solution this problem Somebody can help me? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] help im lme()
Hi R people! I have a doubt in lme(). I use this model: m2.lme<-lme(log(cmort)~idade+ano,random=~idade|ano,data=dados) If i use summary I recive this output: > summary(m2.lme) Linear mixed-effects model fit by REML Data: dados AIC BIClogLik 1139.313 1170.554 -562.6563 Random effects: Formula: ~idade | ano Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 3.131076e-03 (Intr) idade 3.515018e-05 -0.042 Residual5.687940e-01 ...etc... I know the value of Residula random effects is m2.lme$sigma but how do I find the value of idade or intercept random effects Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] problem with legend
I have problem with legend command. Please look this script:
>dcbv.fm
Time Series:
Start = 1980
End = 2002
Frequency = 1
[1] 2994.023 3388.414 3111.762 2990.967 3077.438 3058.274 3049.934 2974.130
[9] 2889.659 2801.790 2631.391 2661.700 2312.526 2518.968 2567.044 2443.952
[17] 2117.638 2042.461 2025.816 1939.560 1640.775 1583.609 1659.912
> dcbv.ms
Time Series:
Start = 1980
End = 2002
Frequency = 1
[1] 3700.239 4076.438 3856.495 3680.345 3871.887 3789.770 3831.173 3768.876
[9] 3585.572 3754.374 3372.859 3419.667 3185.194 3319.215 3445.845 3265.214
[17] 2773.961 2661.904 2669.835 2569.190 2187.719 2217.756 2196.378
>plot(dcbv.ms,ylim=c(min(dcbv.fm),max(dcbv.ms)))
>lines(dcbv.fm,col=2)
>legend(1984,2500,c("DCVB-MS","DCBV-FM"),col=c(1,2),cex=.6,fill=T)
At end of script the legend of plot have only one color: black. I think the
legend will have two colors: black and red.
Where I make mistake?
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor0.1
year 2004
month11
day 15
language R
Thanks in advance
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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[R] Database reorganization
Hi people, I need reorganization several databases. I would like to know if that is possible in R. My problem is like that: original data base: Age x1980 x1981 1 5 8 3 7 9 5 9 15 7 11 20 future data base yearage x 19801 5 19803 7 19805 9 19807 11 19811 8 19813 9 19815 15 19817 20 Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Lecture
Hi R-masters! I you receive an invitation to make a lecture about R in Medical School of Federal University of Rio de Janeiro. The goal of the lecture is to stimulate the use of R instead of others statiscal software in the Medical School. It would like to hear tour suggestions on which topics to present in the lecture. Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] update.packages
Hi R-masters! First of All: Happy new Year ! I have one doubt. I one time for week command in R update.packages() for recive new versions of packages in my R, I also subscribe R-pkgs for recive notices of new packages. But I discover other packages in CRAN without notices in R-pkgs. So How to I mantain my R- system update with all packages in CRAN (New versions of my packages and install New packages) Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] AUC for logistic regression [was: (no subject)]
At 20:21 15/12/2004, Kerry Bush wrote: Your formula is only a trapzoidal approximation to the theoretical AUC. It might be downward biased. I am also wondering if there is a funcion in R that can compute the MLE estimate for AUC. Kerry I agree with you. But I dont Know compute the MLE estimate for AUC and its standatr error. Second I dont know MLE(AUC)/SE is normal so I dont konw wich test use. Third I dont konw compute the correction is necessary because the test in same population. If You explain me this topics I upgrade the funciotn with pleasure. But same with a possible downward bias the function solve Roman problem... Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] AUC for logistic regression [was: (no subject)]
At 17:07 15/12/2004, Spencer Graves wrote:
Dear R-helper,
I would like to compare the AUC of two logistic regression models (same
population). Is it possible with R ?
Thank you
Roman Rouzier
Roman
If I understand your question You have 2 ROC curve from same dataset. In
this case you can use a routine create for me :
seROC<-function(AUC,na,nn){
a<-AUC
q1<-a/(2-a)
q2<-(2*a^2)/(1+a)
se<-sqrt((a*(1-a)+(na-1)*(q1-a^2)+(nn-1)*(q2-a^2))/(nn*na))
se
}
cROC<-function(AUC1,na1,nn1,AUC2,na2,nn2,r){
se1<-seROC(AUC1,na1,nn1)
se2<-seROC(AUC2,na2,nn2)
sed<-sqrt(se1^2+se2^2-2*r*se1*se2)
zad<-(AUC1-AUC2)/sed
p<-dnorm(zad)
a<-list(zad,p)
a
}
The first function (seROC) calculate teh standart error of ROC curve, the
second function (cROC) compare ROC curves .
The parameters:
AUC - area under curve
na - number of positives results
nn - number total tests (positives +negatives)
r - correlation of two numeric variables
Best wishes
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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Re: [R] noncommutative addition: NA+NaN != NaN+NA
At 06:17 07/09/2004, you wrote: Hi guys. Check this out: NaN +NA [1] NaN NA + NaN [1] NA I thought "+" was commutative by definition. What's going on? In my version, both cases is NA: > NaN +NA [1] NA > NA + NaN [1] NA > R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor9.1 year 2004 month06 day 21 language R Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Cox regression for prevalence estimates
At 09:27 07/09/2004, you wrote: Bernardo Rangel Tura wrote: At , The Michaelson Institute wrote: How can R be used to calculate the prevalence ratios using Cox regression + robust variance estimates ? Well, In Design package have a command: cph This command have a option "robsut" with default=FALSE, but in help is write: " ... robust if TRUE a robust variance estimate is returned. ..." I think that is your response... bye Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil No, robust is an option to coxph, not cph. cph uses 'after the fit' correction using the robcov or bootcov functions in Design. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University Sorry Frank! I made a digitation mistake, but I think yours package answers the Tomas Karpati´s need. Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Cox regression for prevalence estimates
At , The Michaelson Institute wrote: How can R be used to calculate the prevalence ratios using Cox regression + robust variance estimates ? Well, In Design package have a command: cph This command have a option "robsut" with default=FALSE, but in help is write: " ... robust if TRUE a robust variance estimate is returned. ..." I think that is your response... bye Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] newsgroup on R
At 08:51 02/09/2004, you wrote: HI, by newsgroup, I mean a kind of user forum like MATLABs http://newsreader.mathworks.com/WebX?14@@/comp.soft-sys.matlab where anyone can post a question, without receiving unnecessary emails daily Hi Bin Jiang! I visit the url: http://gmane.org/info.php?group=gmane.comp.lang.r.general and in this site is possible read and post messages in R-help list without recive e-mails dialy. So is not necessary newsgroups for solve your problem. Cheers. Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Probability(Markov chain transition matrix)
At 17:25 30/04/2004, you wrote:
>So we come up with Transition matrix stating probabilities of each state
>
>FromTo NC 0 1 2 3
>NC 0.79 0.21 0 00
>0 0.09 0.73 0.18 00
>1 0.09 0.51 0 0.40 0
>2 0.09 0.38 0 0 0.55
>3 0.06 0.32 0 0 0.62
>
>
>Thus if one starts with all the accounts having no credit ð0 =(1,0,0,0,0), after one
>period the distribution of account is ð1=(0.79, 0.21, 0, 0, 0)
>after subsequent periods, it becomes
>
>Ð2=(0.64, 0.32, 0.04, 0, 0),
>Ð3= (0.540, 0.387, 0.058, 0.015, 0)
>Ð4=(0.468,0.431, 0.070, 0.023, 0.008)
>Ð5=(0.417, 0.460, 0.077, 0.028, 0.018)
>and
>Ð10=(0.315, 0.512, 0.091, 0.036, 0.046)
Hi
I create a rotine for a problem like this, I hope this is useful for you
#
##MARKOV CHAIN ##
## ##
## ini- initial state ##
## trans- transition matrix##
## n - number of transitions ##
## f - ini%*%trans ##
## fase - f consolidate##
## ##
#
# initial state
ini<-matrix(c(10,0,0,0,0,0,0,0,0),nrow=3,ncol=3)
# transition matrix
#
# [,1] [,2] [,3]
# [1,] 0.85 0.1 0.05
# [2,] 0.00 0.7 0.3
# [3,] 0.00 0.0 1.0
#
# In R the command is
trans<-matrix(c(.85,0,0,.1,.7,0,0.05,0.3,1),ncol=3)
markov<-function(ini,trans,n){
l<-ncol(ini)
fase<-matrix(0,nrow=l,ncol=l)
fases<-array(0,dim=c(nrow(ini),ncol(ini),n))
for (i in 1:n){
f<-ini%*%trans
for (w in 1:l){fase[w,w]<-sum(f[,w])}
ini<-fase
fases[,,i]<-fase
# print(fase)
#
# Fase allow calcule de value for transition:
# If sate 1 value 0,95, state 2 value 0,3 and state 3
# value 0 QALY.
# I´m calculate de value of any transition if
# print(sum(fase%*%c(.95,.3,0)))
#
#
}
return(fases)
}
a<-markov(ini,trans,5)
For your case:
ini<-matrix(c(1,rep(0,24)),ncol=5)
trans<-matrix(c(0.79,0.21,0,0,0,0.09,0.73,0.18,0,0,0.09,0.51,0,0.40,0,0.09,0.38,0,0,0.55,0.06,0.32,0,0,0.62),ncol=5)
solv<-markov(ini,trans,10)
> solv[,,10]
[,1] [,2] [,3] [,4] [,5]
[1,] 0.3173366 0.000 0.000 0.000 0.000
[2,] 0.000 0.2876792 0.000 0.000 0.000
[3,] 0.000 0.000 0.2875772 0.000 0.000
[4,] 0.000 0.000 0.000 0.2886571 0.000
[5,] 0.000 0.000 0.000 0.000 0.2810951
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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Re: [R] Changing Gui preferences
At 06:05 24/04/2004, you wrote: >I'm not sure how to find 'HOME' but I presume that it is the path that appears in the >'Change directory' box when you open this through the gui. In which case for me >'HOME' is > >C:\apps\R\rw1090 > >Now I have 3 versions of Rconsole near here. They are in > >C:\apps\R\rw1090\etc >C:\apps\R\rw1090 and >C:\apps\R > >I still don't know how to change my Gui preferences! Murray, I don´t use win XP, I use win 98 SE. In my system I know de 'HOME' with a simple procedure: 1- Look R shortcut 2- Right Click in R shortcut 3- Choose properties 4- Look satrt in box 5- de directorie in box is the 'HOME' Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fit system of equations
Hi R-masters I have this system of equations R(x,t)= a(t)+b(x,t) a(t) = c + d*t b(t) = e + f*t where x is a vetor of age<-c(37,42,47,52,57,62,67,72,77,83), t is vetor of year (1980:2000) R(x,t) = Rate of mortality in age x on year t a(t) = base mortality on year t b(x,t) = exponential rate of mortality for age x on year t b(t) = exponential rate of mortality on year t I wish is possible fit this system in R? Case positive Which the best method? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] multiple plots in different windows
At 14:29 05-02-2004, you wrote:
>Hi all,
>I'd like to generate a number of plots to compare different
>vectors I have stored in a list. To do this I do something like
>(in a linux system):
>
>for(i in 1:L) {
> X11()
> plot(listOfFunctions[[i]])
>}
>
>First question is: is this the right way to create several plots (in
>different windows) ?
Hi
I think wich the correct code is:
>for(i in 1:L) {
>windows()
> plot(listOfFunctions[[i]])
>}
This work in my computer.
I dont know thje answer for your second question
Best wishes
Bernardo Rangel Tura, MD, MSc
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil
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[R] NLS mensagem error...
Hi R-masters, I have a problem with nls() and my research data. Look this example: X2000<-c(1.205268,2.850695,5.100860,8.571610,15.324513,25.468599,39.623418,61.798856,91.470006,175.152509) age<-c(37,42,47,52,57,62,67,72,77,82) fit <- nls(X2000~R*exp(A*age),start=list(R=.1,A=.1)) Error mensage: Error in nls(X2000 ~ R * exp(A * age), start = list(R = 0.1, A = 0.1)) : singular gradient In addition: Warning message: no finite arguments to min; returning Inf How I fix this problem? Other command? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fit non-linear regressor
Hi R masters, Sorry for first mensage, this is orignal text... y<-c(2.8150,3.5239,4.0980,4.5845,5.0709,5.4824,5.8427,6.3214,6.7349,7.3651) x<-c(37,42,47,52,57,62,67,72,77,82) I need fit R and A in y=f(x)=R*exp(A*x), with minimize sd= sqrt(SRR/(n-2)) where SRR is Sum of the Square of the Residuals and n is number of data points (in this case 10) How do I make this? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fit non-linear regressor
Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fitdistr Error in new VR version
>Hi R-Masters I found a strange error in fitdistr(): In case of VR Version 7.1-11 k21stsList<-c(0.76697,0.57642,0.75938,0.82616,0.93706,0.77377,0.58923,0.37157,0.60796,1.00070,0.97529,0.62858,0.63504,0.68697,0.61714,0.75227,1.16390,0.66702,0.83578) fitdistr(k21stsList, "normal",list(mean = 0.5, sd = 0.1)) mean sd 0.74584591 0.17908744 (0.04108548) (0.02904255) In case of VR Version 7.1-13 k21stsList<-c(0.76697,0.57642,0.75938,0.82616,0.93706,0.77377,0.58923,0.37157,0.60796,1.00070,0.97529,0.62858,0.63504,0.68697,0.61714,0.75227,1.16390,0.66702,0.83578) fitdistr(k21stsList, "normal",list(mean = 0.5, sd = 0.1)) mean sd 0.74584591 0.17908744 (0.04108548) (0.02904255) Error in fitdistr(k21stsList, "normal", list(mean = 0.5, sd = 0.1)) : supplying pars for the Normal is not supported Why this occur? []s Tura --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] MASS fitdistr()
At 15:49 27-11-2003, you wrote: >Dear R experts, > >I am trying to use the R MASS library fitdistr() to fit the following >list: > >k21stsList<-c(0.76697,0.57642,0.75938,0.82616,0.93706,0.77377,0.58923,0.37157,0.60796,1.00070,0.97529,0.62858,0.63504,0.68697,0.61714,0.75227,1.16390,0.66702,0.83578) > >as follows, > >library(MASS) >fitdistr(k21stsList, "normal") > >But, I get > >Error in fitdistr(k21stsList, "normal") : 'start' must be a named list Hi You don´t put the start list, something like this fitdistr(k21stsList, "normal",list(mean = 0.5 sd = 0.1)) mean sd 0.74584591 0.17908744 (0.04108548) (0.02904255) Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] stepAIC problem
At 09:46 05-10-2003, Hiroto Miyoshi wrote: >Dear R-users > >I have a probelm running stepAIC in R1.7.1 >(...) >--small example >> library(MASS) >> x1<-runif(100) >> x2<-runif(100) >> x3<-runif(100) >> x4<-runif(100) >> x5<-runif(100) >> y<-x1+x2+x3+runif(100) >> t<-data.frame(y=y,x1=x1,x2=x2,x3=x3,x4=x4,x5=x5) >> x<-lm(y~x1+x2+x3+x4+x5,data=t) >> stepAIC(x) >Start: AIC= -247.61 > y ~ x1 + x2 + x3 + x4 + x5 > > Df Sum of Sq RSS AIC >- x51 3.747e-067.456 -249.608 >- x41 0.0267.483 -249.254 > 7.456 -247.609 >- x11 4.866 12.322 -199.375 >- x21 8.182 15.639 -175.543 >- x31 8.597 16.054 -172.922 >Error in as.data.frame.default(data) : can't coerce function into a >data.frame Hiroto, In My computer with R 1.7.1 e R 1.8.0 (win 98 SE) don´t have this error... []s Tura __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Help with Temporal series
Good afternoon R-masters, I am with some doubts in the R, see the script below: m<-c(69.6,67.3,75.6,74.3,64.7,60,65.7,62.5,66.5) d<-c(11.6,15,17.8,18.3,11.2,11,4.6,5.8,7) year<-c(1994,1995,1996,1997,1998,1999,2000,2001,2002) male<-ts(m,start=c(1994)) death<-ts(d,start=c(1994)) data<-data.frame(year,death,male) require(ts) d100<-HoltWinters(data$death,gamma=0) m100<-HoltWinters(data$male,gamma=0) par(mfrow=c(3,1)) plot(d100,main="Death") plot(m100,main="Male") ccf(male,death) I have 2 doubts: 1 - How to I should interpret the third graph? 2 - Has a hypothesis test to evaluate the cross-correlation it is significant in R? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
