similar to: Help with the code

I am trying to plot fitted lme values as a smooth line of a graph showing the exponential relationship between temperature and soil respiration. In the plot, the x-axis has temperature, and the y-axis has soil respiration. When I try to add a line showing temperature versus the fitted values, it is jagged and not smooth. Here is the code I used: lme.1<-lme(fixed=LnFlux~Temp,

I need to order a long vector of integers with rather few unique values. This is very slow: > x <- sample(rep(c(1:10), 50000)) > system.time(ord <- order(x)) [1] 189.18 0.09 190.48 0.00 0.00 But with no ties > y <- sample(500000) > system.time(ord1 <- order(y)) [1] 1.18 0.00 1.18 0.00 0.00 it is very fast! This gave me the following idea: Since I don't care

Say, I just picked this up on my messages! There are a whole host of these requests! Anyone know whow there people are? Is there a way to report them? Any suggestions as to how to block them? [Aug 23 10:34:16] NOTICE[1010] chan_sip.c: Registration from '"912" <sip:1 at 41.1.1.1>' failed for '184.106.217.112' - Wrong password [Aug 23 10:34:16] NOTICE[1010]

Dear All, Here is a minimal working example : library(AMModels) mymodels <- amModelLib(description = "Simple AM Model library") dataset1 <- 1:100 dataset2 <- seq(2,200,2) model1 <- lm(dataset2~1) model2 <- lm(dataset2~ dataset1) m1 <- amModel(model = model1,comment = "Initial model.") m2 <- amModel(model = model2,comment = "Final model.")

Dear list, I have two functions created in the same environment, fun1 and fun2. fun2 is called by fun1, but fun2 should use an object which is created in fun1 fun1 <- function(x) { ifelse(somecondition, bb <- "o", bb <- "*") ## mymatrix is created, then myresult <- apply(mymatrix, 1, fun2) } fun2 <- function(idx) { if (bb == "o) { #

Hello I need some help in figuring Bravington’s debugger out. Ok I have 2 functions, fun1 and fun2 saved in a ASCII file say filename is funs. Fun1 has a loop which calls fun2, fun2 has a loop which fails and I need to find out the value of the variables of the fun2 and fun1 loops at the specific iteration that fails. Both fun1 and fun2 loops will iterate thousands of times so line by line debug

Hello, I'm building a package. My code is stored in foo.R. This code has two functions FUN1 and FUN2. FUN1 calls FUN2. FUN1 is listed in export() under the package NAMESPACE but NOT FUN2. After building the package when I call FUN1 is giving me an error that cannot find FUN2. I solved this by adding FUN2 in the export() NAMESPACE. However, what is puzzling me is that I have other examples

I'm not sure if this is what you want, but simply add ... to the list of arguments for fun1 and fun2 would eliminate the error. Andy From: Hans-Joerg Bibiko > > Dear all, > > I wrote some functions using the special argument '...'. OK, it works. > > But if I call such a function which also called such a > function, then I get an error message about unused

I have a list object, in which I have stored n lme4-models. For example: library(lme4); myModels <- list(); myModels[1] <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy) myModels[2] <- lmer(Reaction ~ Days + (1|Subject) + (0+Days|Subject), sleepstudy) Now I would like to perform an anova over all models in the list. However, the anova function requires that every model is inserted

Dear R-users, i have to integrate the following function `fun1` <- function (a, l1, l2) { exp(log(l1) * (a - 1) - l2 * lgamma(a)) } but if l1 is large, i get the "non-finite function value" error, so my idea is to rescale with exp(-l1) `fun2` <- function (a, l1, l2) { exp(log(l1) * (a - 1) - l2 * lgamma(a) - l1) } but it seems this doesn't solve the problem, when

Dear list, I know I have seen this discussed before but I haven't been successful in searching for "ellipsis", "dots", "..." in the archives. I would like to filter "..." arguments according to their name, and dispatch them to two sub-functions, say fun1 and fun2. I looked at lm() but it seemed more complicated than I need as it modifies the calling

Hi, I have a vector a=c(NA, 3, 4, 4, NA, NA, 3) and I would like to use is.na(a) function to get a vector like this: wy=(1,2,2,2,1,1,2) - you know, this vector create 1 or 2 depends on value in vector "a" This is my short code but something is wrong and I don't know what... for (i in 1:7){ a=c( NA, 3, 4, 4, NA, NA, 3) fun1=function(x){ x=1 print(x) } fun2=function(x){

Hi R users: I'm trying to make a function where two of the parameters are functions, but I don't know how to put each set of parameters for each function. What am I missing? I try this code: f2<-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){ force(fun1) force(fun2) force(n) p1<-unlist(par1) p2<-unlist(par2) force(p1) force(p2)

Dear Friends, Let's assume there are three parameters that were passed into fun1. In fun1, we need to modify one para but the remains need to be untouched. And then all parameters were passed into fun2. However, I have failed to achieve it. Please see the following code. ########################################## fun1 <-function(x, y, z=10) {x+y+z;} fun2 <-function(aa, ...) {

> On Feb 25, 2015, at 5:35 PM, Benjamin Tyner <btyner at gmail.com> wrote: > > Actually, it depends on the number of cores: Under current semantics, yes. Each 'stream' of function calls is lazily capturing the last value of `i` on that core. Under Luke's proposed semantics (IIUC), the result would be the same (2,4,6,8) for both parallel and serial execution. This is

Actually, it depends on the number of cores: > fun1 <- function(c){function(i){c*i}} > fun2 <- function(f) f(2) > sapply(mclapply(1:4, fun1, mc.cores=1L), fun2) [1] 8 8 8 8 > sapply(mclapply(1:4, fun1, mc.cores=2L), fun2) [1] 6 8 6 8 > sapply(mclapply(1:4, fun1, mc.cores=4L), fun2) [1] 2 4 6 8 > >/ On Feb 24, 2015, at 10:50 AM,

Dear R users, I'm pretty new on using lmer package. My response is binary and I have fixed treatment effect (2 treatments) and random center effect (7 centers). I want to test the effect of treatment by fitting 2 models: Model 1: center effect (random) only Model 2: trt (fixed) + center (random) + trt*center interaction. Then, I want to compare these 2 models with Likelihood Ratio Test.

Dear all, I run the example of "censboot" contained in "boot" package. But, I can't find the confidence interval of the resulted "censboot" object. Any idea ? > aml.fun <- function(data) { + surv <- survfit(Surv(time, cens)~group, data=data) + out <- NULL + st <- 1 + for (s in 1:length(surv$strata)) { +

Dear list! ? I have question of?'correct function formation'. Which function (fun1 or fun2; see below) is written more correctly? Using ''structure'' as output or creating empty ''data.frame'' and then transform it as output? (fun1 and fun1 is just for illustration). ? Thanks a lot, OV ? code: input <- data.frame(x1 = rnorm(20), x2 = rnorm(20), x3 =

If a function needs to be passed as an argument to another function, default arguments to the function being passed are lost. Consider this example: fun1 <- function(x, A=c("power","constant")) { arg <- match.arg(A) cat(paste("A is:",paste(A,collapse=", "),"\narg is:",arg,"\n")) cat("formals:\n") print(formals())