similar to: Error when calling residual.plots() on an ANOVA object

Dear all, here is the result of R.Version(): > R.Version() $platform [1] "powerpc-apple-darwin8.6.0" $arch [1] "powerpc" $os [1] "darwin8.6.0" $system [1] "powerpc, darwin8.6.0" $status [1] "" $major [1] "2" $minor [1] "3.1" $year [1] "2006" $month [1] "06" $day [1] "01" $`svn rev`

Dear *, I would like to obtain for each factor of my anova model the "response variable vs factor" plot with means and 95% Tukey HSD intervals. I would appreciate any information on how to do that. Cheers -------------------------------------------------------------------- Max MANFRIN Tel.: +32 (0)2 650 3168 IRIDIA - CoDE, CP 194/6

Dear Spencer and Prof. Fox, Thank you for your replies. I'll very appreciate, if you have any ideas concerning the problem described below. First, I'd like to describe the model in brief. In general I consider a model with three equations. First one is for annual GRP growth - in general it looks like: 1) GRP growth per capita = G(investment, migration, initial GRP per

Hi R-users, I try to put image legend via image.plot function, but I do not know put this in log scale. The last line of code put legend in linear scale, but I want put this in log scale. This is a image of the ocean color web site 8-day composite L3 4KM. Anybody colud get me advise please. This is de code... uno<-schl[1,,] rn_cl<-exp(range(uno,na.rm=T)) x11()

Hi. I would appreciate your advice about these TWO PROBLEMS: ======================================================= Find the simplest way to A) configure a Dovecot 2.2.10 IMAP server to access (existing) email users' (mbox-format) mail folders ****in read-only mode.**** B) Do the above while allowing a master_user to access mail on behalf of real users. (I've got master_user access

Hi. I would appreciate your advice about this PROBLEM: ======================================================= Find the simplest way to configure a Dovecot 2.2.10 IMAP server to access existing email users' (mbox-format) mail folders via a master_user ****in read-only mode.**** (I've got the master_user access working.) Simplest because while I'm proficient at UNIX systems

I've benefited from this list with input on how to build up a symmetrical matrix. The purpose of that query was to work with the output from the MINE routine posted at www.exploredata.net To the extent it helps others, here is the script that I was working on an which turns a given MINE output column (in the case below, the third column corresponding to "MIC") into a matrix.

Dear Ladies and Gentlemen! On a CD-ROM-server at the University Koblenz-Landau we are using Samba (2.0.5a) with Red Hat Linux (6.0, Kernel 2.2.5). The server is located here in Koblenz, but also used from another department of the university at Landau (about 80 miles from Koblenz; they have their own class B network). A traceroute from both locations gave a response time of about 10-50 msec. On

Terry - your example didn't demonstrate the problem because the variable that interacted with strata (zed) was not a factor variable. But I had stated the problem incorrectly. It's not that there are too many strata terms; there are too many non-strata terms when the variable interacting with the stratification factor is a factor variable. Here is a simple example, where I have

Terry - your example didn't demonstrate the problem because the variable that interacted with strata (zed) was not a factor variable. But I had stated the problem incorrectly. It's not that there are too many strata terms; there are too many non-strata terms when the variable interacting with the stratification factor is a factor variable. Here is a simple example, where I have

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5

> datafilename="E:/my documents/r/sex/bysex1.csv" > data.sex=read.table(datafilename,header=T) > data.sex y.sex.age.region.c.n 1 1980,F,A,N,-18.15,13.61 2 1980,F,A,N,-18.61,13.04 3 1980,F,A,N,-18.81,12.32 4 1990,F,A,N,-21.12,11.7 5 1990,F,A,N,-20.77,11.58 6 1990,F,A,N,-21.6,13.34 7 1990,F,A,N,-21.78,12.6 > model.anova<-aov(c~age*sex,data=data.sex)

Hi Dennis, look at the help page for summary.survfit, the Value n.event. G?ran On 2024-05-15 22:41, Dennis Fisher wrote: > OS X > R 4.3.3 > > Colleagues > > I have created objects using the Surv function in the survival package: >> FIT.1 > Call: survfit(formula = FORMULA1) > > n events median 0.95LCL 0.95UCL >

Hi there, I checked google for aov. usually one uses summary to see whether the p-value is small. but I want to put aov in my script. how can I get the p-value, (0.1115, 0.6665, 0.6665 in the following example)? thanks YU > datafilename="http://personality-project.org/r/datasets/R.appendix2.data" > data.example2=read.table(datafilename,header=T) > aov.ex2 =

Dear list, I am trying to use the SSOAP package to access a very simple SOAP service, included as a demonstrator in the axis2 Apache package: "Version" (takes no input parameters and should return a string with the version number). The web service itself seems to be working as I could access it using an other client (eclipse Web Server tester). You will find the service wsdl here:

Terry Therneau has been very helpful on r-help but we can't figure out what change in R in the past months made extra columns appear in model.matrix when the terms object is subsetted to remove stratification factors in a Cox model. Terry has changed his logic in the survival package to avoid this issue but he requires generating a larger design matrix then dropping columns. A simple

Dear R contributors, I have a problem in selecting in an efficient way, rows of a data frame according to a condition, which is a part of a row name of the table. The data frame is made of 64 rows and 2 columns, but the row names are very long but I need to select them according to a small part of it and perform calculations on the subsets. This is the example: X Y "Unique to

Quoting Francesca PANCOTTO <f.pancotto at unimore.it>: > Dear R contributors, > > I have a problem in selecting in an efficient way, rows of a data > frame according to a condition, > which is a part of a row name of the table. > > The data frame is made of 64 rows and 2 columns, but the row names > are very long but I need to select them according to a small

(Re-)read the discussion of indexing (both `[` and `[[`) and be sure to get clear on the difference between matrices and data frames in the Introduction to R document that comes with R. There are many ways to create numeric vectors, character vectors, and logical vectors that can then be used as indexes, including the straightforward way: df[ c( "Unique to strat ", "Unique

I am currently trying to access a SOAP web service (hosted on a JBoss5 server and defined as a J2EE-ws service) via the SSOAP R-package. But all I can get is a NULL answer. The web service seems to work as it can be accessed using a general web-service tester. The web service wsdl is here: http://smg8.ulb.ac.be:8080/web2?wsdl. This is how I try to call the web service from R: --- >