similar to: passing a list where names arguments are expected

hi, get.vars.name.prefix<-function(prefix){ result<-list() len<-nchar(prefix) var.names<-ls(name=1,pattern=prefix) #name=1 probably wrong option print(var.names) for(i in 1:length(var.names)){ name<-var.names[i] field<-substr(name,len+1,nchar(name)) result[[field]]<-get(name) } result } for example x.1<-1 x.2<-2 get.vars.name.prefix("x.") should

hello, the example below does not work. (i know it's not supposed, but it makes it clear what i'm trying to achieve) par(mfrow=c(2,1)) xyplot(y~x2|x1,data=dataframe1,pch=20) xyplot(y~x2|x1,data=dataframe2,pch=20) i know i could probably merge the two datasets and do something like xyplot(y~x2|x1+dataset,data=merged) any other suggestion? thanks. [[alternative HTML version deleted]]

hello, i need help with: data$f1<-as.factor(data$f1) data$f2<-as.factor(data$f2) s3<-equal.count(data$s2,number=3) densityplot(~y| f1 + f2 + s3, data=mydata ) this produces 3 plots, *successively*, one for each value of s3. i was hoping it would produce one plot. is trellis limited to 2 conditional variables? if not, what are the appropriate commands? thanks! [[alternative HTML

hi, a) i have something like: ecdfgrp1<-ecdf(subset(mydata,TMT_GRP==1)$Y); ecdfgrp2<-ecdf(subset(mydata,TMT_GRP==2)$Y); how can i plot the difference between these 2 step functions? i could begin with ecdfrefl<-function(x){ecdfgrp2(x)-ecdfgrp1(x);} ... what next? b) if i have a vector with repeated numeric values how can i get the subset without repeated values .e.g (0,4,0,2,2)

hello, i have something like: out<-list() for(i in 1:n){ data<-gen(...) #fixed size data out[[i]]<- fun(data) } > object.size(out[[1]]) 6824 In principle 1 GB should allow n = 1024^3/6824 = 157347? i have about 2GB are not taken by other processes. however, I can see the memory shrinking quite rapidly on my system monitor and have to stop the simulation after only n=300. why

hi, here's what i have: \lstset{ basicstyle=\ttfamily, keywordstyle=\bfseries, showstringspaces=false, columns = fullflexible, mathescape = true, language=R } \begin{lstlisting} lst$val<-val \end{lstlisting} ./software.tex:16:Extra }, or forgotten \endgroup. lst$ the culprit here is the $ sign. thanks. ps: i'm posting here rather than Latex is bec i guess an R user is

hi, library(lattice) x<-c(-1,-1,-1,0,0,0,0,1,1,2) rng<-range(x) histogram(x,endpoints=c(rng[1]-0.5,rng[2]+0.5),nint=length(unique(x))) instead of contiguous bins, i'd like spaces between them to indicate that the data is discrete, or even better, line segments positioned at integer values. i know how to do this with plot, but i need it with histogram, so that in more complicated

I'd like to avoid looping through an array in order to change values in the array as it takes too long. I red from an earlier post it can be done by "do.call" but never got it to work. The Idea is to change the value of "y" according to values in "x". Wherever "x" holds the value 3, the corresponding value in "y" should be set to 1. So I

Hello, My app needs to know where to store files but the path is different for dev and production. What is the best way to do it? - something like $path = ''/home/myrailsapp/public/files/'' in development.rb and $path = ''/home/my_remote_prod_sapp/public/files/'' in production.rb? - do a test to check the env and add the path in application.rb? - do a test to

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hello, I'm a novice user trying to figure out how to retain NA aggregate values. For example, given a data frame with data for 3 of the 4 possible factor colors("orange" is omitted from the data frame), I want to calculate the average height by color, but I'd like to retain the knowledge that "orange" is a possible factor,

I've written a series of functions that evaluates an integral from -inf to a or b to +inf using equally spaced quadrature points along a normal distribution from -10 to +10 moving in increments of .01. These functions are working and give very good approximations, but I think they are computationally wasteful as I am evaluating the function at *many* points. Instead, I would prefer to use

Hi, I wonder if anyone could advise me with this: I've been trying to make a standard curve in R with lm() of some standards from a spectrophotometer, so as I can express the curve as a formula, and so obtain values from my treated samples by plugging in readings into the formula, instead of trying to judge things by eye, with a curve drawn by hand. It is a curve and so I used the

Dear R-helpers, I've got two functions; callTimes() calls times(), passing it an optional argument (bar) by name (bar=harry). times() then believes it has been passed a name, rather than a value ? but I want the value, not the name. Worse, if I evaluate the name, it is evaluated in the environment times() was defined, not where it is called. How can I call times(), defining its optional

Hi all, I have roughly fifty dataframes and a dataframe with the names of the fifty dataframes. I want to perform the same set of manipulations on all fifty dataframes, but can't find a way to batch process from a list with the dataframe names using a loop. Is there a way to read the file names from the dataframe with the names and then call the referenced dataframe? This would save me a

Hello all Currently I want to use the accurate liveness information when writing a *target independent* FunctionPass based on LLVM. The one I can find is LiveValues, a FunctionPass. But it doesn't use classic dataflow equation and can only provide approximate and conservative result. The another one is LiveVariables which use classic data flow equation, but it comes from Clang's analysis

I have a function in R that takes another function as argument: f <- function(g, ...) { #g is expected to be a function } I want to see if there is a way to implement "f" in C and calling it from R using ".C" interface. I know that I can use function pointers for my C implementation, but I imagine it's going to be nearly impossible to pass a function from R to C. Are

I am trying to develop a prognostic model using logistic regression. I built a full , approximate models with the use of penalization - design package. Also, I tried Chi-square criteria, step-down techniques. Used BS for model validation. The main purpose is to develop a predictive model for future patient population. One of the strong predictor pertains to the study design and would not

I am 100% certain that there is an easy way to do this, but after experimenting off and on for a couple of days, and searching everywhere I could think of, I haven't been able to find the trick. I have this piece of code: ... attach(d) if (ORDINATE == 'ds') { lo <- loess(percent ~ ncms * ds, d, control=loess.control(trace.hat = 'approximate')) grid <-

Hello, I am stuck in a relatively simple procedure and was wondering if anybody knows the answer. I am a relatively new R user. How do I use an argument of a custom function in the name of a dataset in R? For example, I have the function: MyF <- function(Tic, price){ xxxxx xxxxx xxxxx Ratio.Tic<- SharpeRatio.annualized(roc) } I would

Hello, Today I was investigating ks.test() with two numerical arguments (x and y) and was left a bit confused about the policy behind handling ties. I might be missing something, so sorry in advance, but here is what confuses me: The documentation states: "The presence of ties always generates a warning, since continuous distributions do not generate them" But when I run a test with