similar to: environments

I would like to assign an value to an element of a list contained in an environment. The list will contain vectors and matrices. Here's a simple example: # create toy environment testEnv = new.env(parent = emptyenv()) # create list that will be in the environment, then assign() it x = list(a=1,b=2) assign("xList",x,testEnv) # create new element, to be inserted into xList c = 5:7

Hi Arnaud, Hi Scott, Hi list, Here is a patch to support the Zinto D from ONLINE USV-Systeme AG. I already sent a version to Russell Kroll (2006-04-09), without no response and I cannot find support for Zinto in svn until now. I found a discussion on this list about the Xanto from ONLINE, but the Zinto seems to use different commands. The commands are quite similar to those for Opti-UPS, so I

Hi all, I''m getting an error and I really don''t know why, because I seem to be doing everything well..I pass "testvar" as a variable to render partial: <%= render(:partial => "main/categorysmallitem", :collection => group_of_four, :locals => {:testvar=>"AA"})%> But then, when I try to use testvar inside the partial, it

Kay Sievers wrote: > Hi, > this simple program compiled with klibc: > > #include <stdio.h> > int main(int argc, char* argv[]) > { > setenv("TESTENV", "happy", 1); > printf("TESTENV='%s'\n", getenv("TESTENV")); > } > > prints: > > ./env > TESTENV='(null)' > >

Here's what I'm doing: 1. I set __FOO=0 in the current context. 2. I use Dial() and the M(testvar) macro call. 3. The macro-testvar can see ${FOO} just fine. 4. the macro-testvar modifies FOO. I tried: Set(__FOO=1) Set(_FOO=1) Set(FOO=1) 5. When the Dial()ed call hangs up, and the original context continues, FOO is still 0.

Hi Can anyone tell me what''s wrong with this one. Sorry I''m new and I just want to print some message to the the puppet log on the client side. --------------------------------------------------- $testvar = "Hello world!" class mytest { exec { "/bin/echo $testvar" } } I keep getting: err: Could not parse; using old configuration: Syntax error at

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

I am lost: > plot(testError, col="red") > lines(testVar, col="black") Only one plot (the red one) appear on the Window, the black line did not appear...what's wrong? Thanks a lot! [[alternative HTML version deleted]]

Dear All, In R 3.6.3 (and earlier), method dispatch used to work for methods stored in local environments that are attached to the search path. For example: myfun <- function(y) { out <- list(y=y) class(out) <- "myclass" return(out) } print.myclass <- function(x, ...) print(formatC(x$y, format="f", digits=5)) myfun(1:4) # prints: [1]

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hi there fellow hackers, geeks and others who want the UNIX desktop to succeed. I bring forward the question whether or not a (freedesktop.org) specification is necessary for application scheduling software. The current UNIX Desktop user does not have an easy way to configure the tasks which he wants to schedule. Sure we have such a utility. And sure we can shout to our Desktop users

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hello, I got the hint to use nesting environments (for a GUI problem), but I'm not sure whether I understand what he means (and I did not really succeed with reading documentation - probably I looked for the wrong keywords?). That's what he wrote: > Another good way to control the scoping is nesting environments (e.g. through closures). That is, your object does not need to be

Hello, If the signature of a method defines which generic it implements then I'm confused about why this minimal example I invented won't work : setGeneric("myFun", function(rs, ...){standardGeneric("myFun")}) setGeneric("myFun", function(cs, ...){standardGeneric("myFun")}) setMethod("myFun", "numeric", function(rs, colour =

I declared the environment of the function myfun to be NULL as follows: environment(myfun) <- NULL Later on I called that myfun and got an error message because the function index() in the zoo package was called inside myfun and was not visible: Error in myfun(args) : couldn't find function "index" I tried to use zoo::index() instead of index(), but that did not work. In fact,

Dear Rers, I have a function to barplot() a matrix, eg myfun <- function(x, ...) { barplot(x , ... )} (The real function is more complicated, it does things to the matrix first.) So I can do: m1 <- matrix(1:20,4) myfun(m1) myfun(m1, main="My title") I'd like to be able to add the number of rows of the matrix passed to the function to the "..." argument, eg

Hello R users, i'm wondering how to trigger variable creation. Whenever a variable is created i want my own function myFun(...) to be started. if (exists("x")) {rm(x)} # after removal of x # any of these calls x<-10 # should call myFun x=10 # should call myFun assign(x,10) # should call myFun etc.

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje