similar to: Problems with foreign

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I suggest to add a panel function to levelplot (or perhaps to an other 3d lattice function) which is able to translate the z values into the size of the rectangles. It could be used to display categorical data. I append the proposed code and two examples: - panel.catlevelplot() - example1.catlevelplot.esoph() - example2.catlevelplot.esoph() Wolfram Fischer #------ CODE

Full_Name: Axel Benz Version: 1.7.1 OS: Windows Submission from: (NULL) (137.251.33.43) Hello, I guess many people will answer me again that this is a S language feature, but I am only a stupid computer scientist and I simply do not understand this logic, despite of reading a lot about S: > test field tuckey 4 Kreis2 -1 5 Kreis5 -2 9 Metall -3 17

I have a dataframe with the following variables: idnum area gender race etc. I would like to make a table that looks like area gender race M F B W A 1 4 5 3 5 1 2 6 7 4 6 3 etc. Basically, I want to make a single broad table with a number of sub-set tables. I have tried: cbind(table(area, gender), table(area, race)) But, when I do this, I lose the labels gender / race. This makes it

Full_Name: Ed Borasky Version: R-beta 2.1.0 2005-04-08 OS: Linux 2.6.11 GCC 3.3.5 Submission from: (NULL) (24.21.57.139) I downloaded the latest R-beta tarball and did a build with the default options. OS is Linux 2.6.11 and compiler is GCC 3.3.5. "make check-all" failed with the following message: make[3]: Entering directory `/home/znmeb/R-beta/tests' running code in

Here is an example: The legends (mainly the factor level names) are cut off on the right. Somehow the internal calculation which computes horizontal space for the legend is not flexible enough. ## Call a new graphics window {with default par()s !}: get(getOption("device"))() par(mfrow = c(2,2)) ## part of example(interaction.plot) _improved_ using with() : data(OrchardSprays)

This arose in testing [.terms and has me confused. data(esoph)?? # use a standard data set t0x <- terms(model.frame( ~ tobgp, data=esoph)) t1 <-? terms(model.frame(ncases ~ agegp + tobgp, data=esoph)) t1x <- (delete.response(t1))[-1] > all.equal(t0x, t1x) [1] TRUE # the above is wrong, because they actually are not the same > all.equal(attr(t0x, 'dataClasses'),

Dear List, I try to make a xyplot with different colors and symbols, I came this far: library(DAAG) xyplot(csoa~it|sex*agegp,data=tinting,groups=target,pch=list(1,2),auto.key=list(space = "right")) this produces a plot with different colors and symbols but unfortunately the legend does only follow the color scheme and not the different symbols. Any suggestions what to change? And

R-1.7.0 built on NetBSD 1.6, but the validation test suite failed: Machinetype: Intel Pentium III (600 MHz); NetBSD 1.6 (GENERIC) Remote gcc version: gcc (GCC) 3.2.2 Remote g++ version: g++ (GCC) 3.2.2 Configure environment: CC=gcc CXX=g++ LDFLAGS=-Wl,-rpath,/usr/local/lib make[5]: Entering directory `/local/build/R-1.7.0/src/library' >>> Building/Updating

This will probably seem very simple to experienced R programmers: I am doing a snp association analysis and am at the model-fitting stage. I am using the Stats package's "drop1" with the following code: ##geno is the dataset ## the dependent variable (casectrln) is dichotomous and coded 0,1 ## rs743572_2 is one of the snps (which is coded 0,1,2 for the 3 genotypes)

On 05/04/2019 9:03 a.m., Therneau, Terry M., Ph.D. via R-devel wrote: > This arose in testing [.terms and has me confused. > > data(esoph)?? # use a standard data set > > t0x <- terms(model.frame( ~ tobgp, data=esoph)) > t1 <-? terms(model.frame(ncases ~ agegp + tobgp, data=esoph)) > t1x <- (delete.response(t1))[-1] > > > all.equal(t0x, t1x) > [1]

On 05/04/2019 10:19 a.m., Therneau, Terry M., Ph.D. wrote: > Duncan, > ? I should have included it in my original note, but > > ?? ? all.equal(unclass(t0x), unclass(t1x)) > > returns TRUE as well.? I had tried that as well. ? But a further look at > all.equal.default shows the following line right near the top: > ??? if (is.language(target) || is.function(target))

On 05/04/2019 10:46 a.m., Therneau, Terry M., Ph.D. wrote: > > > On 4/5/19 9:39 AM, Duncan Murdoch wrote: >> On 05/04/2019 10:19 a.m., Therneau, Terry M., Ph.D. wrote: >>> Duncan, >>> ?? I should have included it in my original note, but >>> >>> ??? ? all.equal(unclass(t0x), unclass(t1x)) >>> >>> returns TRUE as well.? I had

I've been asked by someone else whom I originally taught `` to just work with substitute() and then all will be fine'' ... But it looks to me that I've been caught here. Is it possible to make this work along the way we thought it should? 1) Inside a function, say tst() with the 'formula' and a 'data' argument, 2) call another modeling function using

Hi, > -----Original Message----- > From: Shoultz, Gerald [SMTP:bjz9 at cdc.gov] > Sent: Monday, February 09, 2004 3:38 PM > To: r-help at stat.math.ethz.ch > Subject: [R] Another question, unfortunately. . . .(Installing > "foreign"/trying to import/export SAS files) > > > I am trying to use read.ssd and then read.xport (to get a SAS data set). > I have

Is there a standardized way in R to perform post-hoc comparisons for main adn interaction effects of between- and within-subject factors? For instance, I have a data set of performance of adults of different age groups (20-30, 60-70,70-80) performing a WM task (n-back, with n=1,2,3,4) in two different conditions (while sitting or walking). The corresponding ANOVA produces the following output

Hi. I am new here, and I am writing this Winbugs code with BRugs. n=length(bi.bmi) Lagegp=13 Lgen=2 Lrace=5 Lstra=15 Lpsu=2 #model gen x race bi.bmi.model=function(){ # likelihood for (i in 1:n){ bi.bmi[i]~ dbern(p[i]) logit(p[i])<- a0 + a1[agegp[i]]+a2[gen[i]]+a3[race[i]] + a12[agegp[i], gen[i]] + gam[stra[i]]+ u[psu[i],stra[i]] } # constraints for a1, a2, a3, a12 a1[1]<-0.0

Dear all, I have a data frame containing the results of an experiment. Like this: a<-seq(1,4,by=1) b<-seq(1,2,by=1) test<-expand.grid(b,a,a) colnames(test)<-c("replicates","bins", "groups") test$abc <- rnorm(32) test$def <- rnorm(32) test$ghi <- rnorm(32) test The following code snippet aggregates the data for one variable and then draws a

Hi everyone, I'm trying to import .sps (SPSS portable file) file. the read.spss function (library foreign) doesn't allow to import such files. should I import in spss and then save as sav file? there is not other solutions available? what I mostly like from spss file is that they have variable labels. want is really wish to keep are the variable.labels from the spss file; so, if there is a

Dear All, I wonderif there is a simple way to draw a regression line in the xyplot: more specifically, let: age <- c(20:30, 31:40 ) age.cut <- cut(age, breaks = 2 ) y<- rnorm(20) x <- rnorm(20,4,1) xyplot(y ~ x| age.cut, xlab="x", ylab="y") How to draw (in the plot given by xyplot) the two regression lines (y ~ x) corresponding to the two category