search for: temp2

i'm sorry. i had an error in my previous code because i left out a letter in the rownames. while fixing that, i also found a solution. so i'm sorry for the confusion. below is my fix. temp2 <- matrix(rnorm(10),nc=1,nrow=10) rownames(temp2) <- c("a","b","c","d","e","f","g","h","i","j") print(temp2) temp2 <- as.matrix(temp2[order(temp2[,1,drop=FALSE]),]) print(temp2) On Wed,...

this is a bad question but I can't figure it out and i've tried. if i sort the 2 column matrix , temp1, by the first column, then things work as expected. But, if I sort the 1 column matrix, temp2, then it gets turned coerced to a vector. I realize that I need to use drop=FALSE but i've put it in a few different places with no success. Thanks. temp1 <- matrix(rnorm(10),nc=2,nrow=5) rownames(temp1) <- c("a","b","c","d","e") print(...

R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the "temp" (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency ("Freq" | counts) of each time each "delayValue" occurs in the cells. I've tried this command without luck, it seems to be droppi...

...ently this same thing happens in 1.8.0, though I never experienced it. I had a hard time reproducing it, and I don't know what's setting it off, but the code below seems to do it for me. (I'm using R on Windows XP, either 1.8.0 or 1.9.1) Thanks for any help, Elizabeth Purdom > temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4) > colnames(temp2)<-paste("A",as.character(1:4),sep="") > temp2<-as.data.frame(temp2) > newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,drop=F];return(xmat[1,])}) > print(newtemp2)...

...y dbh age 801 0 2977.196 3090.225 6 36.0 802 0 2951.892 3083.769 8 40.6 803 0 2919.111 3075.557 8 40.6 804 0 2914.123 3072.700 9 42.9 805 0 2925.353 3074.675 8 40.6 How many rows (nft) shall be removed? > nft<- rpois(1, 2) > nft [1] 2 Ok remove 2 rows: > temp2<- temp[-sample(nrow(temp), nft), ] > temp2 occ x y dbh age 801 0 2977.196 3090.225 6 36.0 803 0 2919.111 3075.557 8 40.6 805 0 2925.353 3074.675 8 40.6 No problem. However, sometimes rpois(1, 2) lead to nft=0, and in that case I do not want > temp2<-...

I am using R 2.1.1 in an windows XP environment. I have 2 dataframes, temp1 and temp2. Each dataframe has 20 variables (“cocolumns") and 525 observations (“rows”). All variables are numeric. I want to create a new dataframe that also has 20 columns and 525 rows. The values in this dataframe should be the sum of the 2 other dataframe. (i.e. temp1$column 1+tem...

hi, i have a list of data.frame that has same structure. i would like to know a efficient way of rbind-ing it. right now, i write: n = length(temp) # 'temp' is a list of data.frames temp2 = data.frame() for (i in 1:n) temp2 = rbind( temp2, temp[[i]]) return(temp2) but this is not an efficient way since we keeping overwriting temp2. i wonder if there's faster way. thanks -- View this message in context: http://www.nabble.com/rbind-ing-list-tf3140137.html#a8703373 Sent from t...

I think need to do something like this: dat<-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000, replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000)) rle.dat<-rle(dat$state) temp<-1 out<-data.frame(id=1:length(rle.dat$length)) for(i in 1:length(rle.dat$length)){ temp2<-temp+rle.dat$length[[i]] out$V1[i]<-mean(dat$V1[temp:temp2]) out$V2[i]<-sum(dat$V2[temp:temp2]) out$state[i]<-rle.dat$value[[i]] temp<-temp2 } to a very large dataset. I want to apply a few summary functions to some variables within a data.frame for given states. to complicate...

Hi, I am a new novice of LLVM, and I want know how to get the left-hand operand of an instruction? For example: how to get the %temp2 operand in the next instruction: %temp2 = malloc i8, i32 %n Thanks a lot! Best Regards! -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.llvm.org/pipermail/llvm-dev/attachments/20100406/be299269/attachment.html>

Okay, someone explain this behaviour to me: Browse[1]> replace(rep(0, 4000), temp1[12] , temp2[12])[3925] [1] 0.4462404 Browse[1]> temp1[12] [1] 3926 Browse[1]> temp2[12] [1] 0.4462404 Browse[1]> replace(rep(0, 4000), 3926 , temp2[12])[3925] [1] 0 For some reason, R seems to shift indices along when doing this replacement. Has anyone encountered this bug before? It seems to crop u...

Hi, Here is the summary-output of the Coxph-model I used (the output is based on the best final model i.e. all significant explanatory variables and their interactions are included): coxph(formula = Y ~ LT + Food + Temp2 + LT:Food + LT:Temp2 + Food:Temp2 + LT:Food:Temp2) n= 555 coef exp(coef) se(coef) z Pr(>|z|) LT 9.302e+02 Inf 2.822e+02 3.297 0.000979 *** Food...

...r(mfrow=c(4,3)) Temp1 <- IBM.Close[which(format(time(IBM.Close),"%Y")==yr),] Temp3 <- tapply(Temp1[,1],as.yearmon(time(Temp1)),FUN=mean) for(i in Smth:length(Temp3)){ i <- ifelse(i < 10, paste(0,i,sep=""),i) Date <- paste(i,yr,sep="-") Temp2 <- IBM.Close[which(format(time(IBM.Close),"%m-%Y")==Date),] plot(time(Temp2),Temp2,type="l",main=paste(factor(as.numeric(i), labels = month.name[as.numeric(i)]),yr,sep="-")) } } # my sweave code (pass in IBM.Close) \pagebreak \subsection{Graph} \begin{fig...

Hi Qiuping yi, > I am a new novice of LLVM, and I want know how to get the left-hand > operand of an instruction? > > For example: > how to get the %temp2 operand in the next instruction: > > %temp2 = malloc i8, i32 %n there is no left-hand side, temp2 is just a name for the instruction. Since the LLVM IR is in SSA form, registers have exactly one definition, and thus there is no point in distinguishing between a register and the instruction t...

...s in it. Here is code, but of course it requires my original, 601x601 symmetric matrix called mat > any(is.na(mat)) [1] FALSE > any(is.na(d)) [1] FALSE > dim(mat) [1] 601 601 > length(which(d==0)) [1] 5 > d<-rowSums(mat) > temp1<-eigen(diag(d)-mat,symmetric=T) > temp2<-eigen(diag(d)-mat,symmetric=T,EISPACK=T) > any(is.na(temp1$vec)) [1] TRUE > any(is.na(temp1$vec[,-52])) [1] FALSE > any(is.na(temp2$vec)) [1] FALSE > all.equal(abs(temp1$vec[,-52]),abs(temp2$vec[,-52])) [1] "Mean relative difference: 0.3278133" > all.equal(temp1$v...

temp = "~aparch(" temp1 = paste(temp,1, sep = "") temp2 = paste(temp1,1, sep = ",") temp3 = paste(temp2, ")",sep = "") temp 3 is a character but I want to convert to formula object. How do I do this? -- View this message in context: http://r.789695.n4.nabble.com/how-to-convert-string-to-object-tp2999281p2999281.html Sent...

...mp, + stop = 10*temp + test1$time, + status = rep(test1$status,4), + x = c(test1$x+ 1:7, rep(test1$x,3)), + epoch = rep(1:4, rep(7,4))) > > fit1 <- coxph(Surv(start, stop, status) ~ x * factor(epoch), stest) ## New lines > temp1 <- fit1$linear.predictor > temp2 <- as.matrix(temp1) > match(temp1, unique(temp1)) [1] 1 2 3 4 4 5 6 7 7 7 6 6 6 8 8 8 6 6 6 9 9 9 6 6 > match(temp2, unique(temp2)) [1] 1 2 3 4 4 5 6 7 7 7 6 6 6 NA NA NA 6 6 6 8 8 8 6 6 ----------------------- I've solved it for my code by not calling match on...

I have a data frame with both positive and negative values, and I want to make all the negative values equal zero, so i can eventually take an average. I've tried temp2 <- ifelse(tempr<0, 0, tempr) but it doesn't seem to work. Any suggestions? Thanks!

...a bigger change the potential for breakage is higer, however. Terry T. tmt45% R --vanilla R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) > load('test.rda') > ls() [1] "temp2" > temp2 <- round(temp2, 15) > length(unique(temp2)) [1] 954 > length(table(temp2)) [1] 942 > .Machine$double.eps [1] 2.220446e-16 > range(temp2) [1] 0.0000 26.0397 Terry T.

...en I backup /home). All directories that are not writable--even though unchanged--will be sent again. [Note that unwritable files are not transferred again--this only occurs with directories]. -- Set of commands to reproduce problem -- mkdir -p temp1/some/set/of/subdirectories/ mkdir temp2 chmod -R a-w temp1/some/set/of rsync -av temp1/ temp2/ rsync -av temp1/ temp2/ -- Output from those commands -- # Note the 2nd rsync run and how unwritable directories are listed # again 1190 newren@amr:~$ mkdir -p temp1/some/set/of/subdirectories/ 1191 newren@amr:~$ mkdir temp2...

Hi Irucka, I tried it and was able to plot it without any errors.? Here, your code indicates you need two lines. temper[[i]][1] ?temper[[1]][1] # which is the column 1. ? Month 1???? 1 2???? 2 3???? 3 ?temper[[1]][2] #? Data1 #1?? 1.5 #2? 12.3 #3? 11.4 Suppose I use names(temper) instead of seq_along(temper) pdf("irucka.pdf") ?lapply(names(temper),function(i)