similar to: having problems re-ordering a dataframe

Dear all: I have this: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3 Example: m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<- rep(c(5,10,15),c(4,4,4)) data.frame(v,u) This is the result I want but I would like to learn a simpler way to do it. Any clue?

We have a wireless network that uses 802.1x authentication, in which domain joined computers use their machine credentials to connect. Windows machines do this automatically, and until recently Linux computers could join using wicd, wpa-supplicant, and a simple script that would retrieve the machine password with tdbdump. ( specifically tdbdump -k SECRETS/MACHINE_PASSWORD/DOMAIN

#I have a data set that looks like this. A bit more complicated actually with # three factor levels but these calculations need to be done on one factor at a #I then have a set of different rates that are applied #to it. #dataset cata <- c( 1,1,6,1,1,2) catb <- c( 1,2,3,4,5,6) doga <- c(3,5,3,6,4, 0) data1 <- data.frame(cata, catb, doga) rm(cata,catb,doga) data1 # start rates #

Dear all:   I have this:  A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3   And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3   Example:   m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<-

Hi there, I run two regressions: y = a1 + b1 * x + e1 y = a2 + b2 * z + e2 I want to test against the null hypothesis: b1 = b2. How do I design the test? I think I can add two equations together and divide both sides by 2: y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2). or just y = a3 + 0.5*b1 * x + 0.5*b2 * z + e3 If I run this new regression, I can test against

See if the following helps: > m <- outer(letters[1:5], 1:4, paste, sep="") > m [,1] [,2] [,3] [,4] [1,] "a1" "a2" "a3" "a4" [2,] "b1" "b2" "b3" "b4" [3,] "c1" "c2" "c3" "c4" [4,] "d1" "d2" "d3" "d4" [5,]

I downloaded R v2.1.1 earlier this morning to compile under Fedora Core 4. It compiled without incident, but 'make check' failed. Below is the relevant part of its report. Is this a known problem? I used a locally compiled version of GCC v4.0.0 that reports [kent at d89h102 R-2.1.1]$ gcc -v Using built-in specs. Target: i686-pc-linux-gnu Configured with: ../gcc-4.0.0/configure

Dear developeRs, Compilation of the latest version (0.9-5) of my actuar package fails with r-release MacOS_X ix86 on CRAN; see http://www.R-project.org/nosvn/R.check/r-release-macosx-ix86/actuar-00check.html All errors come from accented letters in comments in latin-1 encoded files (except hierarc.R which is in UTF-8, my bad). Encoding is declared as latin-1 in DESCRIPTION. The package

Hello R guru's I am a newbie to R, In my research work I usually generate a lot of ELISA data in form of absorbance values. I ususally use Excel to calculate the concentrations of unknown, but it is too tedious and manual especially when I have 100's of files to process. I would appreciate some help in creating a R script to do this with minimal manual input. s A1-G1 and A2-G2 are

Hi. I have tried to use Package "SEM". As a learning, I try to convert a program running well of EQS which is as follows to SEM: ### EQS ### /SPECIFICATION CAS=100; VAR=5 MAT=COR; ANA=COR; /EQUATIONS V1=*F1+E1; V2=*F1+E2; V3=*F1+*F2+E3; V4=**F1+*F2*E4; V5=*F2+E5; /VAR E1 TO E5=*; F1*1.0; F2=1.0; /COV E1,E2=*; F1,F2=*: /PRINT FIT ALL; /MATRIX ...... /END This is the converted SEM

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

Hi, I'm back on the list ;-) I seem to be having some trouble getting W2K machines to join the domain in 3.0.1rc2. I haven't looked at this in detail since 2.2.8a but it looks like the account gets created in LDAP and then it has trouble setting the password appropriately. I believe this is the relevant part of the log: api_rpcTNP: samr op 0x3a - api_rpcTNP: rpc command:

Hi, I have a column (V4) in a midi event list which includes tone pitch names, i.e. "A4, E4, C#4, A3...": > compo[1:10,] V1 V2 V3 V4 V5 V6 V7 1 1 1 0 A4 96 2 0 2 1 1 0 E4 96 2 0 3 1 1 0 C#4 96 2 0 4 1 1 0 A3 96 2 0 5 1 3 0 B4 96 1 0 6 1 3 0 E4 96 1 0 7 1 3 0 B3 96 1 0 8 1 3 0 G#3 96 1 0 9 1 4 0 C#5 96 1 0 10

Hello subscribers, I have a very strange question regarding SSL setup on gluster storage. I have create a common CA and sign certificate for my gluster nodes, placed host certificate, key and common CA certificate into /etc/ssl/, create a file called secure-access into /var/lib/glusterd/ Then, I start glusterd on all nodes, system work fine, I see with peer status all of my nodes. No problem.

Hi, people. I met a little problem for which someone might have a solution. Let's say I have an executable file (named "pp.R") with this contents: #!/usr/bin/Rscript options(echo=TRUE) a <- 1 Sys.sleep(3) a <- 2 If I execute "./pp.R" at the shell prompt, the output shows the timely progress of the script as expected. If I use "./pp.R | tee

Hi, The lab in which I send my samples return the results in a format that is difficult for me to run my analysis. The lab outputs the results where each parameter is its own row and it’s not consistently in the same order (and not each sample is tested for the same suite of variables). e.g. >dataset site parameter value a e2 3 a e1 1 a e3 5 b e3

Hi, people. The output produced by "?options" contains: 'expressions': sets a limit on the number of nested expressions that will be evaluated. Valid values are 25...500000 with default 1000. [...] and a bit further down: The 'factory-fresh' default settings of some of these options are [...] 'expressions'

Hi, people. Executing the following command: hist(rpois(100,5), labels=TRUE) yields a graphic in which some labels are truncated (on an X11 device). The truncated labels are those over the highest bars. The hist function should ideally manage enough room for the labels, automatically. (Specifying ylim solves my problem, but yet, hist could be frienlier.) --please do not edit the

Hello, I use: R version 2.9.2 (2009-08-24) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 on Ubuntu 9.10, I usually run R from ESS (5.4 on current Unbuntu) from Emacs-22.2.1. But I also tried the following from the console and it gave the same results. I have a data file containing lots of European characters, French, German, Italian and so on. I can read it

+ > p2<-function(r){ + gama=0 + for(i in 1:1000){ + c=caminho[[4]] + for(i in 1:caminho[[3]]+1) { + c=c+caminho[[i+3]]*((r[i])^(i-1)) + d=(abs(c))*exp(-(x^2/2))} + gama=gama + ( d/(h(r[i])) ) } + return(gama)} > e3<-p2(r) OBS: r is a rnorm(1000,0,1) > caminho theta_chapeu f_estrela k a0 a1 a2 a3 1 3.2 1.2 3 2 1 4 5 > question i wanted gama to be