similar to: strsplit() does not split on "."?

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

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

Is there an easy way to control smoothness of the contour lines? In the plot I am working on due to the undersampling the contour lines I am getting are jugged, but it is clear "by eye" these should be basically straight lines. In maps package I found smooth.map function, but maybe there is a more generic way of accomplishing the same thing. Ideally there would be an option to control

RF trains fine with X, but fails on prediction > library(randomForest) > chirps <- c(20,16.0,19.8,18.4,17.1,15.5,14.7,17.1,15.4,16.2,15,17.2,16,17,14.1) > temp <- c(88.6,71.6,93.3,84.3,80.6,75.2,69.7,82,69.4,83.3,78.6,82.6,80.6,83.5,76 .3) > X <- cbind(1,chirps) > rf <- randomForest(X, temp) > yp <- predict(rf, X) Error in predict.randomForest(rf, X) : subscript

I getting "Error in matrix(0, n, n) : too many elements specified" while building randomForest model, which looks like memory allocation error. Software versions are: randomForest 4.5-25, R version 2.7.1 Dataset is big (~90K rows, ~200 columns), but this is on a big machine ( ~120G RAM) and I call randomForest like this: randomForest(x,y) i.e. in supervised mode and not requesting

I just started using nice package "sets" and I wonder if there are utilities to convert (some) sets to data frame (as in the example below) > library(sets) > a <- gset(elements = list(e('A', 0.1), e('B', 0.8))) > lst <- as.list(a) > nr <- length(lst) > rnames <- character() > for (i in 1:nr) rnames[i] <- lst[[i]] > df <-

I want to fit a linear model with fixed slope e.g. y = x + b (instead of general: y = a*x + b) Is it possible to do with lm()? Regards, Ryszard -------------------------------------------------------------------------- Confidentiality Notice: This message is private and may ...{{dropped:11}}

I would like to use greek "tau" as a symbol of variable to integrate over in plotmath expression(integral(f(tau)*dtau, 0,t)) but nothing seems to work. I tried d{\tau}, d\tau, etc., without any success Is it possible? How can I accomplish this? Best regards, Ryszard -------------------------------------------------------------------------- Confidentiality Notice: This message is

I am looking for an efficient way to find near neighbors... More specifically... I have two sets of points: A & B and I want to find points in set B which are closer to set A than some cutoff (or n-closest) I will appreciate very much any pointers... Ryszard -------------------------------------------------------------------------- Confidentiality Notice: This message is private and may

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

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

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: > > >

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 >

Hi, It is most probably just my R-ignorance, but I have following problem with using predict(). I train the model using 164 cases and then I try to use it on the data set with 35 cases, but I am getting 164 predictions ? R-code below illustrates in more detail what I am doing. Truly yours, R train = read.csv("train.csv", header = TRUE, row.names = "mol",

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

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

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

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

I want to start R processes on multiple processors from single shell script and I want all of them to have different random seeds. One way of doing this is sleep 2 # (with 'sleep 1' I am often getting the same number) ... set.seed(unclass(Sys.time())) Is there a simpler way without a need to sleep between invoking different R processes ? Ryszard