similar to: what does order() stand for in an lme formula?

Dear All, I'm trying to teach myself latex along with the latex function in Hmisc and have hit a roadblock that I can't seem to get around. I'd greatly appreciate any pointers. I'm running R 2.6.0 on Windows XP and have Miktex 2.7 installed. I've reproduced the code below, taken from Frank Harrell's latexsummary introduction. My question relates to getting a pdf

From: Ben Skeggs <bskeggs at redhat.com> We've previously had identical naming between vertex and texture formats, so it mostly made sense to define these together. However, upcoming patches are going to transition the driver over to using updated texture header definitions using NVIDIA's naming, and this will no longer be the case. Signed-off-by: Ben Skeggs <bskeggs at

Why not fix the new names instead to be like the old names? Seems like that would be way simpler... On Feb 15, 2016 12:38 AM, "Ben Skeggs" <skeggsb at gmail.com> wrote: > From: Ben Skeggs <bskeggs at redhat.com> > > We've previously had identical naming between vertex and texture > formats, so it mostly made sense to define these together. > > However,

Hi, I would like to eliminate all the load instructions and replace their uses with the stored values in the following program. The stores and loads are in the same basic block. Is there an optimization pass in LLVM 3.0 that can do this? define void @testFunc() { entry: %sVal = alloca %sTy %f1 = getelementptr %sTy* %sVal, i32 0, i32 0 store i32 789, i32* %f1 %f2 = getelementptr %sTy*

The green book tells:"The basic technique is classic :keep it simple ."A long ,complicated expression or function is less fravorable than" a relatively small computations that combines calls to a few other functions to perform its tasks." But I don't get the point totally.Can anyone give me an example to make me understand this rules totally? ps: Is it mean that f1 is

Hi I am new to R so I apologize if this is trivial. I am trying to predict the resistance or susceptibility of my sequences to a certain drug with a randomForest function from a file with amino acids on each of the positions in the protein. I ran the following: > library(randomForest) > > path <- "C:\\..." > path2 <- "..." > name <-

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On 13.2.2015 16:59, Kevin Laurie wrote: > Dear Jiri, > > I tried the following to try to get the inbox deduplicated. My inbox is > quite large and urgently need to remove the duplicated messages. > Is there an easy way to do this? > Sorry for being so presistant but I need help. > > The command I tried:- > > doveadm deduplicate -u user at domain.net mailbox inbox

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

Buenas tardes a todos, tengo un problema con R: ejecuto el mismo script en el ordenador del trabajo y en mi portátil con los mismos datos y obtengo resultados diferentes (siendo los correctos, los obtenidos en el trabajo): rm(list=ls()) directorio<-"C:\\Documents and Settings\\BDs\\" library(RODBC) library(car) library(gdata) ### DATOS A ###

Hi, May be you can try this: dat1New<-? dat1[!(duplicated(dat1$gene)|duplicated(dat1$gene,fromLast=TRUE)),] dat2<-dat1[duplicated(dat1$gene)|duplicated(dat1$gene,fromLast=TRUE),] ?lst1<-split(dat2,dat2$gene) dat3<-unsplit(lapply(lst1,function(x) {x1<- sum(apply(x[,6:32],2,function(y) y[1]>=y[2]));x2<- sum(apply(x[,6:32],2, function(y) y[1]<=y[2])); if(x1>x2) x[1,] else

On Sat, Dec 19, 2015 at 1:53 PM, Ilia Mirkin <imirkin at alum.mit.edu> wrote: > These are supported on GK20A and GM107. > > Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> > --- > > Was a bit torn on where to place the enums... we're about to gut all > the xml definitions so this seemed appropriate for now. > > Tested on GK20A only. > >

I need to split a given matrix in a sequential order. Let my matrix is : > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10, > T), sample(seq(20, 200, by=20), 10, T)); dat [,1] [,2] [,3] [1,] 200 100 80 [2,] 100 180 80 [3,] 200 150 180 [4,] 200 50 140 [5,] 100 150 60 [6,] 100 50 60 [7,] 100 100 100 [8,] 200 150 100 [9,]

Brilliant, thanks very much!! Works fine. Dan On 6 Dec 2012, at 18:25, arun kirshna [via R] wrote: > Hi, > Your question is not very clear. I hope you are not looking for the subset option in summary(lm()). > If you want just the summary(), then use '&' or '|' > For e.g. > dat1<-read.table(text=" > opcorn Oilamt Batch Yield >

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Um, maybe just dat1$C <- match(dat1$B, unique(dat1$B)) Indexing 1:k with numbers between 1 and k is a bit of a no-op... AFAICT, this even works without stringsAsFactors=FALSE -pd > On 11 May 2018, at 21:30 , MacQueen, Don <macqueen1 at llnl.gov> wrote: > > dat1$C <- seq(length(unique(dat1$B)))[ match( dat1$B, unique(dat1$B) )] -- Peter Dalgaard, Professor, Center for

I have the following code to write the output from auto.arima function. The issue is not in finding the model but to divert its out put fit to a file order_fit.txt. code runs but nothing is written to order_fit.txt where am I going wrong library(forecast) for (i in 1:2) { filen = paste("file",i,".txt",sep="") data <- read.table(filen) dat1 <- data[,1] xt <-

Dear R users, I came up to a problem dealing with percentiles in R. >From my previous questions: I do have a big data.frame, with lots of columns and rows. The following command enables me to calculate means for all data frame. dat1$newID<-rep(1:(nrow(dat1)/12),each=12) #if nrow(dat1)/12 is integer dat2<-with(dat1,aggregate(cbind(dat1[,1:71]),by=list(newID),mean)) What I need is to

Dear all, let suppose I have following vector:   > dat1 <- c(rep("asd", 5), rep("xyz", 12), rep("erd", 17)) > dat1 <- dat1[sample(1:length(dat1), length(dat1), replace=F)] > dat1  [1] "erd" "xyz" "erd" "asd" "asd" "erd" "xyz" "asd" "erd" "erd"

Dear all, I have following two list object, both are basically collection of matrices : dat1 <- matrix(rnorm(25*6), ncol=6) dat1 <- split(dat1, seq(5,25,by=5)) dat1 <- lapply(dat1, matrix, ncol=6) dat2 <- matrix(rnorm(25*4), ncol=4) dat2 <- split(dat2, seq(5,25,by=5)) dat2 <- lapply(dat2, matrix, ncol=4) Now I want to replace last 4 columns of each matrix at "dat1"