similar to: how to initialize random seed properly ?

I don't know the details of pls (in the pls.pcr package, I assume), but if you use validation="CV", that says you want to use CV to select the best number of components. Then why would you specify ncomp as well? Andy > From: ryszard.czerminski at pharma.novartis.com > > When I try to use ncomp parameter in pls procedure I get > following error: > > >

I have a data frame (df) with colums x, y and z. e.g. df <- data.frame(x = sample(4), y = sample(4), z = sample(4)) I can extract column z by: df$z or df[3] I can also extract columns x,y by: df[1:2] or by df[-3]. Is it possible to extract x,y columns in a "symbolic" fashion i.e. by equivalent of df[-z] (which is illegal) ??? Or alternativeley, is there an equivalent of

How can I remove columns with NaN entries ? Here is my simple example: > data <- read.csv("test.csv") > xdata <- data[3:length(data)] > xs <- lapply(xdata, function(x){(x - mean(x))/sqrt(var(x))}) > x <- data.frame(xs) > x C D E F 1 -0.7071068 NaN -0.7071068 -0.7071068 2 0.7071068 NaN 0.7071068 0.7071068

And on that note, here is a function that I use to get around it: -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Jeffrey Horner Sent: Friday, November 03, 2006 10:01 AM To: ryszard.czerminski at novartis.com Cc: r-help at stat.math.ethz.ch Subject: Re: [R] R CMD BATCH: unable to start device PNG ryszard.czerminski

I am trying to produce rather mundane output of the form e.g. pi, e = 3.14 2.718 The closest result I achieved so far with print() is: > print (c(pi, exp(1)), digits = 3) [1] 3.14 2.72 > print(c("pi, e =", pi, exp(1)), digits = 3) [1] "pi, e =" "3.14159265358979" "2.71828182845905" I understand that c() promotes floats to strings and

Is there a function in R, which would return index of maximum value in a vector ? e.g. > v <- round(10*rnorm(8)) > v [1] 6 -3 -6 15 7 9 0 -19 > max(v) [1] 15 ??? index.max(v) ??? 4

In e1071 package/svm default epsilon value is set to 0.1 and not 0.5 as documentation says. R

Consider this: > y <- matrix(1:8, ncol=2) > is.matrix(y[-c(1,2),]) [1] TRUE > is.matrix(y[-c(1,2,3),]) [1] FALSE > is.matrix(y[-c(1,2,3,4),]) [1] TRUE It seems like an inconsistent behaviour: - with 2 or more rows we have a matrix - with 1 row we do not have a matrix and - with 0 rows we have a matrix again I just stumbled on this behaviour, because I had a problem with my

Full_Name: Ryszard Czerminski Version: 1.8.1 OS: GNU/Linux Submission from: (NULL) (205.181.102.120) prcomp(..., scale = TRUE) does not work correctly: $ uname -a Linux 2.4.20-28.9bigmem #1 SMP Thu Dec 18 13:27:33 EST 2003 i686 i686 i386 GNU/Linux $ gcc --version gcc (GCC) 3.2.2 20030222 (Red Hat Linux 3.2.2-5) > a <- matrix(rnorm(6), nrow = 3) > sum((scale(a %*% svd(cov(a))$u, scale

Apparently row names are dropped when I extract single column from a data frame. Why this behaviour ? > y <- as.matrix(df[,1:2]); length(row.names(y)) [1] 324 > y <- as.matrix(df[,1:1]); length(row.names(y)) [1] 0 Best regards, Ryszard

When printing a table it is broken at some point (depending how long are the associated names) >>> see example below. Is there a way to control number of columns being printed for a given chunk of the table? Best regards, Ryszard > z5 AAAAAAA BBBBBBB CCCCCCC DDDDDDD EEEEEEE FFFFFFF GGGGGGG HHHHHHH IIIIIII AAAAAAA 1.00 -0.69 -0.54 -0.88 NA NA NA

How can I select random subsets (rows!) from a data set ? If I generate simple data set > a <- data.frame(x=1:2, y = NaN, z = 2:1) > a x y z 1 1 NaN 2 2 2 NaN 1 I can select random subsets (colums) very easily using sample function: > sample(a, 2) z y 1 2 NaN 2 1 NaN I expected that using transpose of a would do the same for rows, but I am getting rather unexpected

Hi Ryszard! There is a dist function in R. It's in the mva package. You can set the kind of distance that you want. Thanks, Erin mailto:hodgess at gator.uhd.edu

Any ideas what is the problem with this code? > N <- 2; c(Sys.Date(), sprintf('N = %d', N)) [1] "2012-01-10" NA Warning message: In as.POSIXlt.Date(x) : NAs introduced by coercion Best regards, Ryszard Ryszard Czerminski AstraZeneca Pharmaceuticals LP 35 Gatehouse Drive Waltham, MA 02451 USA 781-839-4304 ryszard.czerminski@astrazeneca.com

Any ideas what is wrong? > strsplit("a.b", ".") # generates empty strings with split="." [[1]] [1] "" "" "" > strsplit("a b", " ") # seems to work fine with split=" ", and other characters... [[1]] [1] "a" "b" > > R.Version() $platform [1]

Is there any rational explanation for the bizarre seq() behavior below? > seq(2,8.1, lenght.out=3) [1] 2 3 4 5 6 7 8 > help(seq) > seq(2,8,length.out=3) [1] 2 5 8 > seq(2,8.1,length.out=3) [1] 2.00 5.05 8.10 Except maybe that it is early in the morning :) Best regards, Ryszard Ryszard Czerminski AstraZeneca Pharmaceuticals LP 35 Gatehouse Drive Waltham, MA 02451 USA 781-839-4304

Is it just my installation or bug in 1.9.0 ? The same thing works fine in 1.8.1 Best regards, Ryszard # R-1.9.0 library(pls.pcr) nr <- 8; ndim <- 2 x <- matrix(rnorm(nr*ndim), nrow=nr) y <- as.matrix(x[,1]) for (i in 2:ndim) y <- y + x[,i] y <- y + rnorm(length(y)) m <- pls(x,y,validation='CV') # Error in paste(ncomp, "LV's") : Argument

I am trying to write a function, which would allow to call various methods and would pass to them extra arbitrary parameters. My first attempt was to use call() as illustrated below, but apparently '...' cannot be used in such context. How can this be achieved ? Best regards, Ryszard > myfun <- function(method, x, ...) { + v <- eval(call(method, x, ...)) + } > method =

Any ideas why read.table complains about not correct number of elements in line while readLine/strsplit indicate that all lines have the same number of elements ? R > tbl <- read.table('tmp', header = T, sep = '\t') Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : line 32 did not have 27 elements > lines <-

R-helpers! na.omit() can be used to remove rows with NA's but how can I remove columns ? and remember, which columns have been removed ? I guess I can do t(na.omit(t(o))) as shown below, but this probably creates a lot of overhead and I do not know which columns have been removed. Yours, R > o [,1] [,2] [,3] [1,] 1 NA 7 [2,] 2 NA 8 [3,] 3 NA 9 >