search for: colx

#If I have two lists as follows a1<- array(1:6, dim=c(2,3)) a2<- array(7:12, dim=c(2,3)) l1<- list(a1,a2) a3<- array(1:4, dim=c(2,2)) a4<- array(5:8, dim=c(2,2)) l2<- list(a3,a4) #how can I create a new list with the mean across all arrays within the list, so all components are included? As an example for [[1]]; cbind((l1[[1]][,1]+l2[[1]][,1])/2,

...ll, I would like to ask the following. I have an array of data in an objetct, let's say X. I need to use a for loop on the elements of one or more columns of X and I am having a debate with a colleague about the best memory management. I believe that if I do: col1 = X[,1] col2 = X[,2] ... colx = X[,x] and then for(i in whatever){ do something using col1[i], col2[i] ... colx[i] } my memory management is better that doing: for(i in whatever){ do something using X[i,1], X[i,2] ... X[,x] } BTW, here I *have to* use a for() loop an no nifty tapply, lapply and family. Any comment is wel...

...---------------------------------------------------- ## Author: Martin Maechler, Date: 12 Jan 98, 18:15 lk <- length(k.set); slk <- ceiling(sqrt(lk)) op <- par(mfrow=c(slk,slk), mar = .2+c(3,2,1,1), oma=rep(1,4)) on.exit(par(op)) ## oma does not really work if(is.R()) { colx <- rainbow(lk); coly <- terrain.colors(lk) } else colx <- lk + (coly <- 1:lk) for(i in 1:lk) { k <- k.set[i] if(is.R()) { plot.new() plot.window(xlim= c(0,10^ k), ylim= c(0,10^(k+y.k.add)),"") box(lty='dashed...

Good afternoon Has anyone tried using Dr. Elith's BRT script? I cannot seem to run gbm.step from the installed gbm package. Is it something external to gbm? When I run the script itself <- gbm.step(data=model.data, gbm.x = colx:coly, gbm.y = colz, family = "bernoulli", tree.complexity = 5, learning.rate = 0.01, bag.fraction = 0.5) ... I keep encountering the same error: ERROR: unexpected ')' in "bag.fraction = 0.5)" I've tried all sorts of variations (suc...

Dear All, I am very new to R - trying to teach myself it for some MSc coursework. I am plotting temperature data for two different sites over the same time period which I have downloaded from a university weather station data archive. I am using the following code to create the plot plot ( x = TEMP3[,"TIME"], y = TEMP3[,"TEMP"], type = "p", col =

...t; vtmp <- as.data.frame(lapply(letters,function(x) { rep(x,each=50) })) > > names(vtmp) <- paste("col",LETTERS,sep="") > > head(vtmp) > colA colB colC colD colE colF colG colH colI colJ colK colL colM colN > colO colP colQ colR colS colT colU colV colW colX colY colZ > 1 a b c d e f g h i j k l m n > o p q r s t u v w x y z > 2 a b c d e f g h i j k l m n > o p q r s t u v w x y z > 3 a...

Excelente! Ahora corre muy rápido. No conocía ese método, creo que me va a resultar muy útil. Muchas gracias y saludos. Fernando Macedo El 24/02/15 a las 10:51, Jorge I Velez escribió: Fernando, Podrias intentar R> a <- rep('a', 5) R> b <- rep('b', 5) R> a [1] "a" "a" "a" "a" "a" R> b [1] "b"

Hi, Could you please advice some easy way to do the following for a dataframe (header=F) having unequal column- & row- length. 1. Combine/stack/join contents from - a) multiple rows into one column. b) multiple columns into one row. 2. Stack contents from multiple columns (or, rows) into one column (or, row). Thank you. Cheers, Santana [[alternative HTML

Hi, In a function, I compute 10 (un-named) vectors of reasonable length (4471 in the particular example I have to hand) that I want to combine into a data frame object, that the function will return. This is very slow, so *I'm* doing something wrong if I want it to be quick and efficient, though I'm not sure what the best way to do this would be. I know it is the combining into data