similar to: plot 3 lines with ggplot2

In the example below, from http://www.ling.upenn.edu/~joseff/rstudy/summer2010_ggplot2_intro.html I'd like to make (a) the fitted line thicker and (b) change the background fill color for the confidence envelope around each fitted line to a low-alpha transparent version of the same color used for the separate fitted lines for GENDER, rather than grey for both. How can I do this?

Hi, # For publications, I am not allowed to repeat the axes. I have tried to remove the axes using: # yaxt="n", but it did not work. I have not understood how to do this in ggplot2. Can you help me? # I also do not want loads of space between the graphs (see below script with Dummy Data). # If I could make it look like the examples on the (nice) examples page: #

Hi Simon, Thanks so much for your help. Your advice has been taken to heart. I now pass in blocks of 100,000 records, and it does 100,000 predictions in seconds and returns a logical vector with the predictions to an int array. It works like a charm! I want to reference what we were talking about earlier. Let?s say we evaluate an R expression. Is there a way to just print, verbatim, what R

In splus rexp() and rweibull() are related: > set.seed(153) > rexp(1) [1] 0.0493267 > set.seed(153) > rweibull(1, shape=1) [1] 0.0493267 (you can also try shape =2, then rweibull = sqrt(rexp) ) However, in rw0.64.1 (on Win NT) they are different > .Random.seed <- 1:4 > rexp(1) [1] 1.412030 > .Random.seed <- 1:4 > rweibull(1, shape=1) [1] 2.054032 May be rweibull

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Hello, I call a function via .Call passing to it a raw vector(D) and an integer(I) The vector is a series K1,KData1, V1,VData1, K2, KData2, ... where the integer K1 is the length of Data1 and similarly for Ki (wrt Datai)(similarly for V*) There 2*I such pairs( (Ki,KDatai), (Vi,VDatai)) The numbers Ki(and Vi) are written in network order. I am returning a list of I elements each element a

I am using rexp to generate several exponential distributions. I am passing rexp a vector of rates , r. I am wanting to simulate a sample of size 200 for each rate so the code looks like: rexp(n=200*length(r),rate=r) this gives me a vector of the random exponential variables, but they are all disjointed b/c rexp goes through and simulates an exponential variable for each rate and it does that 200

Dear all, while looking for some inspiration of how to organise some code, I studied the code of random.c and noticed that for distributions with 2 or 3 parameters the user is not warned if NAs are created while such a warning is issued for distributions with 1 parameter. E.g: R version 2.7.0 Under development (unstable) (2008-02-29 r44639) [...] > rexp(2, rate=Inf) [1] NaN NaN Warning

Hi all, I just run into this today. Apparently rexp() sometimes gives different slightly results for the same seed on 32 bit and 64 bit machines. runif() is the same for both, so the problem seems to be in rexp(). 64 bit Linux is the same as 64 bit OSX, and R-devel gives the same results as R-3.0.2. Best, Gabor # --------------------------------------------- > options(digits=22) ;

Trying to get past a frustrating error to do a "simple" Box's M test using Java. The Box's M test says it will work with a data.frame. Here's the setup code: REXP myDf = REXP.createDataFrame(new RList( new REXP[] { new REXPDouble(d1), new REXPDouble(d2), new REXPDouble(d3), new REXPDouble(d4), new REXPInteger(d5) })); Here's the call: REXP boxMResult =

On 27/10/2017 8:10 AM, Morkus via R-devel wrote: > Trying to get past a frustrating error to do a "simple" Box's M test using Java. > > The Box's M test says it will work with a data.frame. > > Here's the setup code: > > REXP myDf = REXP.createDataFrame(new RList( > new REXP[] > { > new REXPDouble(d1), > new REXPDouble(d2), > new

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Hi Sadaf, Out of curiosity, what sorts of things have you tried to fix this? For example, after playing around with this a bit, if I remove your "eps" parameter from your `ranges` list, it works. Perhaps you should try tweaking the values you pick for your parameters. You don't even have to put it in the `tune` function to get an idea of the ranges you should use: R>

Dear R-helpers, Can you help me to see why "code 1" gives error while "code 2" runs fine? The only difference in the data is the distribution of age categories. I am attaching the session after the code. Many thanks. XL library(survival) library(rpart) # code 1 n <- 20 age <- rep(1:3, c(2, 3, 15)) eg<- data.frame(rexp(n), rbinom(n,1,prob=.3), age=age)

This might be a very beginner question, but I'm looking for an example for something that I have never done. Suppose that I wanted to express actions with respect to lifted semantics of CPU instructions to an intermediate representation, BAP IL or LLVM IR. How might I go about providing a Backus Naur Form specification and then dynamically interpreting those lifted instructions by also

Does it work if you supply the closing parenthesis on the call to boxM? The parser says the input is incomplete and a missing closing parenthesis would cause that error.. // create a string command with that variable name.String boxVariable = "boxM(boxMVariable [,-5], boxMVariable[,5]"; // try to execute the command... // FAILS with org.rosuda.REngine.Rserve.RserveException: eval

Thanks! I just figured it out (thanks to "Beyond Compare") and was coming here to post back. The boxM test doesn't work with that (now, finally working) REXP structure, but I probably now need to create a table or something and parse that structure. So much fun! :) Thanks again. - M Sent from [ProtonMail](https://protonmail.com), Swiss-based encrypted email. > --------

Hi R-helpers, I am using rpart package for decision tree using R.We are invoking R environment through JRI from our java application.Hence, the result of R command is returned in REXP and we use geterrMessage() to retrieve the error. When we execute the following command, cnr_model<-rpart(as.factor(Species)~Sepal Length+Sepal Width+Petal Length, method="class",

Dear all, First, let's create some data to play around: set.seed(1) (df <- data.frame(Group=rep(c("Group1","Group2","Group3"), each=10), Value=c(rexp(10, 1), rexp(10, 4), rexp(10, 10)))[sample(1:30,30),]) ## Now we need the empirical distribution function: edf <- function(x) ecdf(x)(x) # empirical distribution function evaluated at x ##

Hi, Wondering if anyone could help me out with this error.Im trying to fill a matrix with random numbers taken from an exponential distribution using a loop: x.3<-matrix(rep(0,3000),nrow=1000,byrow=T)for(i in 1:1000){x[i,]<-rexp(3,rate=2/3)} I get the error message: Error in x[i, ] <- rexp(3, rate = 2/3) : incorrect number of subscripts on matrix Any ideas??? Appreciate any thoughts.