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ABCD are four matrix. A * Inverse((Transpose(A)*Tranpose(B)*B*A+C)) * Transpose(A) * Transpose(B) * D how to write in R in an efficient way? --------------------------------- [[alternative HTML version deleted]]

Hello, A collegue of mine has compared the runtime of a linear model + anova in SAS and S+. He got the same results, but SAS took a bit more than a minute whereas S+ took 17 minutes. I've tried it in R (1.9.0) and it took 15 min. Neither machine run out of memory, and I assume that all machines have similar hardware, but the S+ and SAS machines are on windows whereas the R machine is Redhat

Hi, everyone, I was trying to sort a data frame by two columns, one increasing, the other decreasing and got an error. "Error in FUN(left) : invalid argument to unary operator", The command is "BA[order(BA[1],-BA[2]),]". BA is the data frame. It was working if I used increasing on both columns. Why the decreasing symbol "-" is not working here? Thanks. -- Jun

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We had to move our dovecot from one server to another and used rsync to move the mail but it took us some time.. In our log at the new server, we are getting a lot of these..: 2010/10/08 23:28:23 imap(emailadr): Error: Index /opt/vmail/ba-avis.no/bpe/mail/.Nettforum/dovecot.index: Lost log for seq=22 offset=25248 2010/10/08 23:28:23 imap(emailadr): Warning: fscking index file

Hello everybody, In am trying to analyse a BACI experiment and I really want to do it with R (which I find really exciting). So, before moving on I though it would be a good idea to repeat some known experiments which are quite similar to my own. I tried to reproduce 2 published examples but without much success. The first one in particular is a published dataset analysed with SAS by

Hello! Below is my exmample. "myref" is my reference data frame with columns a and b. "temp" is my data with column c analogous to column a in "myref". I am trying to create a new variable b - in "temp" - that matches values from b in "myref" to values in c. If you look at the resulting data frame (temp - at the bottom), you'll notice that

Friends I have an xts that I wish to access. Browse[2]> DATA.ba[[p]]["2012-03-20 00:59:57","bid"] bid 2012-03-20 00:59:57 1.4993 So far so good. Now putting the index into a variable: Browse[2]> Time [1] "2012-03-20 00:59:57 NZDT" Browse[2]> DATA.ba[[p]][Time, "bid"] bid Where has it gone? Looking closer....

OK. About the error it’s a long story, so it’s probably better to pin some source code here. Below is a piece of code related to my problem, clipped from lib/Transforms/Utils/ValueMapper.cpp. I was wondering what “materialized” means here. Value *Mapper::mapBlockAddress(const BlockAddress &BA) { Function *F = cast<Function>(mapValue(BA.getFunction())); // F may not have

For example, strings <- c("aaaa", "bbbb","ccba"). How to get "aaaa", "bbbb" that do not contain "ba" ? _________________________________________________________________ [[alternative HTML version deleted]]

Hello, I am performing a BACI analysis with ANOVA using the following glm: fit1<-glm(log(Cucs_m+1)~(BA*Otter)+BA+Otter+ID+Primary, data=b1) The summary(aov(fit1)) shows significance in the interaction; however, now I would like to determine what combinations of BA and Otter are significantly different (each factor has two levels). ID and PRIMARY substrates are categorical and included in

Hi Aki & Dovecot team, Just continuing with the previous discussion "Migration from one server to 2 new servers". Hi, I have tried the same way as per your suggestion. But it's throwing an error. ------------- [root at production2 log]# doveadm backup -a All -R -u kishore at test.ba.com imapc: dsync(kishore at test.ba.com): Info: imapc(production1.baplc.com:143): Connected

Hi to all the people, I'm working with abundance data of some species, but containing too zero values, and the factors are the ones typical in a BACI experiment (Before-and-After-Control-Impact). Thus, these are two fixed factors. As the data does not holds the normality and homogeneity of variances assumptions of clasiccal ANOVA, I'm trying to fit a zero-altered model using the MCMC glmm

Greetings, I am recoding a dummy variable (coded 1,0) so that 0 = 2. I am using the line sciach$dummyba[sciach$ba==0] <- 2 I notice that it creates a new column dummyba, with 0 coded as 2 but with 1's now coded as NA. Is there a simple way around this in the line I am using, or do I need to have an additional line sciach$dummyba[sciach$ba==1] <- 1 Thanks in advance. David

Not sure if this is the issue but the error says no password database is cfg?d?is that truly the case? Maybe post a redacted config for the community to review? > On Jul 28, 2020, at 7:20 AM, Kishore Potnuru <kishore.reachme at gmail.com> wrote: > > Hi Aki & Dovecot team, > > Just continuing with the previous discussion "Migration from one server to 2 new

I noticed I got a strange connection from what seems to be a user in italy?!? and he connected to my SMB client maybe?? I'm assuming the errors in his logfile ( http://68.48.247.187/log.gustavo.txt ) not finding the service.c file are because he is being denied access.. but how is he connecting in the first place.. And why isnt he being refused by my servers hosts.deny file...? I have

Dear R-ers, I feel I am close, but can't get it quite right. Thanks a lot for your help! Dimitri # I have 2 data frames: x<-data.frame(a=c("aa","aa","ab","ab","ba","ba","bb","bb"),b=c(1:2,1:2,1:2,1:2),d=c(10,20,30,40,50,60,70,80))

I have a data.frame object like: > data.frame(x=c('A','A','B','B'), y=c('Ab','Ac','Ba','Bd')) x y 1 A Ab 2 A Ac 3 B Ba 4 B Bd how could I create a tree structure object like this: |---Ab A---| _| |---Ac | | |---Ba B---| |---Bb Thanks, Zech [[alternative HTML version deleted]]

Dear R-helpers: I am trying to write a script that iterates through a dataframe that looks like this: Example dataset called "sample": names <- c("S1", "S2", "S3", "S4") X <- c("BB", "AB", "AB", "AA") Y <- c("BB", "BB", "AB", "AA") Z <- c("BB",

I wish to display a single-panel Lattice figure with grouped data and fitted regression lines. I don't seem to be able to get the individual regression lines to display, e.g.; d <- data.frame(q = rep(1:6, each=10), cc = rep(seq(10,100, 10),6)) # create data frame with group identifier 'q', independent variable cc d$ba = d$q*0.1*d$cc + 5*rnorm(nrow(d))