similar to: help with contrasts.

Hi, all. Thanks Timo for answering my previous post. It's starting fine now :) I must be doing something wrong again but I can't connect from any imap client. If I telnet to the server I get: $ telnet carpa 11143 Trying 143.107.209.25... Connected to carpa.ciagri.usp.br. Escape character is '^]'. * OK dovecot ready. a001 login marcelo ##### a001 OK Logged in. a002 select INBOX

Hi. Sorry for the basic question but... how do I start the server? From inetd or in standalone mode (which binary)? Thanks

hello, I am working with some data which i would like to find the best fit model for them. First, this data are in a matrix. Each of the columns are treatments and the raws are repetions for the each tratment. So, i've already maden one chart with all the boxplot graphic, for each treatment. Now, i need to find the best fit model for them, i know that exponetial or log model would be perfect

Hi, I'm an Apache James committer and we are curious to use imaptest in order to validate our IMAP protocol implementation. I'm using the nightlybuild : imaptest-20170506 I follow the examples given in the /T/ /est IMAP server compliancy/ examples https://imapwiki.org/ImapTest/Examples. And I wanted to analyse the errors reported, here are some: Error: user1 at james.org[11]: seq too

Hi, I use samba for 5 years in a small enterprise network. We have recently changed our windows client and migrate from windows 95 to windows XP :-). Samba is running on a debian woody. Since we are using windows XP home edition I can see the following log messages : [2004/04/02 15:27:50, 0] smbd/oplock.c:oplock_break(758) oplock_break: receive_smb error (Success) oplock_break failed for

I have 3 basic questions (sorry, but R is a bit new to me) First) The following code will work on C, but something happens with R: > x<-c(0,1,1,0,0,1,0,0,1) > dim(x)<-c(3,3) > types<-numeric(3) > for (i in 1:3) { + j<-1 + while (x[i,j]!=1) j<-j+1 + if (j==4) types[i]<-0 + else types[i]<-1 + } Error: subscript out of bounds Any ideas about what is wrong?

Hi R-users! I'd like to know how to avoid automatic coercion of numeric variables passed to array() to character, when there is one categorical variable between others. I have tried I() function, but it is just functional in data.frame. Best Regards! Eduardo Dutra de Armas

Suppose I have a simple function that returns a matrix, such as: test <- function(x){ return(matrix(c(x,x^2,x^3,x^4),2,2)) } so that test returns: [ x x^3 ] [ x^2 x^4 ] Is it possible for me to get the derivative of an expression such as: c(1,0) %*% test() %*% c(0,1) The vectors are used just to "index" the matrix. I don't want a value, but the expression to work with

dears R-users, I'm interested in model selection problem, and i have faced some problems that i would like to ask for help. well, this is a very small example with 4 variable (just one var. is the response - z) with 100 individuals i would like to do a stepwise search, for the "best" model, and a use BIC criteria. I know when I have a lot of variables, let's say 120, I know,

Hello everyone, I'm using the very good imaptest [0] tool to test my little imap server implementation. I've tried to use the dovecot-crlf [1] file, but it looks like there are some major issues : $ grep -n "In-Reply-To.*;" tests/data/dovecot-crlf 479:In-Reply-To: <20020806175441.GA7148 at linux.taugt.net>; from rueckert at informatik.uni-rostock.de on Tue, Aug 06, 2002

Hi, I have a problem concerning discrepances between R (which I use) and Statistica (which uses my supervisor). I can't say what is the origin of these differences but unfortunately my supervisor doesn't know that either. Our response variable is number (or presence/absence) of parasites in rodents and explanatory variables are presence/absence of several alleles. The rodents were

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

On 6/23/05, RenE J.V. Bertin <rjvbertin at gmail.com> wrote: > Hello, > > I was just having a look at the aov function source code, and see that when the model used does not have an Error term, Helmert contrasts are imposed: > > if (is.null(indError)) { > ... > } > else { > opcons <- options("contrasts") >

Am 23.03.2015 15:08 schrieb Andreas Kasenides: > I am usually emotionally (at least) against of open-source projects > loosing their independence to large corporations. Possibly due to bad > experiences in the past when OSS were driven from Open to Obscure in > the process of trying to make money out of them. 2ct from me-- I put it this way: if dovecot would have been a

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees of freedom,

Dear group, Here is my function : position<-function(x) { pose<-read.csv2((paste(c("LSCPos",x,".csv"),collapse="")),dec=".",sep=",",as .is=T,h=T,skip=1)[,c(4,8,14,15)] pose$CREATED.DATE<-as.Date(pose$CREATED.DATE,"%d/%m/%y") futures<-pose[-grep("USD",pose[,1]),]

Hi all, After a few days of looking into why rsync-2.5.6 would not work when using the "::" syntax on AIX 4.3.3, I discovered it is only the case when the system is not patched. I patched my system to maintenance level 10 and it works fine. AIX 4.3.3.0 - not working # uname -a AIX host1 3 4 00070F3F4C00 # instfix -i | grep ML <stuff deleted> All filesets for 4.3.3.0_AIX_ML

With R and the hist function, is there a way to make a histogram in which the y axis denotes propotion with respect to a separate sample dataset of the same range instead of frequency? [[alternative HTML version deleted]]

Hello I have the following problem : group the lines of a dataframe when no information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays) and when the value of Debut of lines i = value Fin of lines i-1 I can obtain it with a do loop. Is it possible to avoid the loop ? The dataframe initial is df1 dput(df1) structure(list(Matricule = c(1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L, 6L, 8L, 8L, 8L,

On 23 May 2017, at 12.10, aduprat <aduprat at linagora.com> wrote: > > Hi, > > I'm an Apache James committer and we are curious to use imaptest in order to validate our IMAP protocol implementation. > I'm using the nightlybuild : imaptest-20170506 > > > I follow the examples given in the /T/ /est IMAP server compliancy/ examples