similar to: how to pass extra parameters using call() or similar mechanism ?

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 there a way to call a function, and explicitly set an argument to 'not specified'? My situation is the following. I have a function which passes on most of its arguments to another function. The second function, myfun2, serializes all arguments and is out of my control. myfun <- function(...){ return(myfun2(...)); } now, the value for arguments of myfun are stored in variables.

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 >

Hello all, I'm trying to use the apply function on a data frame, by applying a function that takes a one row data.frame as argument . Here's the example : myfun = function(x) paste(x$f1 , x$f2) df = data.frame(f1 = c(1,4,10),f2 = "hello") apply(df,1,myfun) ==> Does not work (I get "character(0)" ) Though : myfun(df[1,]) works, and myfun(df) works as well. So if

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",

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 <-

Hi All I have a dataframe: myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4)) And I have a function myfun: myfun<-function(x,y) x+y I would like to write a function myfun2 that takes myframe and myfun as parameters and returns a list as below: mylist $first [1] 4 $second [2] 6 Could you please help me with this? Doesn't seem like the

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

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: > > >

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 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 am using nlsLM {minpack.lm} to find the values of parameters a and b of function myfun which give the best fit for the data set, mydata. mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

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

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