similar to: lapply without return values?

Hi I have two processes which take with a certain probability (p1 and p2) x number of years to complete (age1 and age2). As soon as thge first process is completed, the second one begins. I want to calculate the time it takes for the both processes to be completed. I have the following script which gives me the answer, butI think there must be a more elegant way of doing the calculations

We're pleased to announce a one day course covering static and dynamic graphics using R, ggplot and GGobi. The course will be held just before the JSM, on Saturday, 28 July 2007, in Salt Lake City. The course will be presented by Dianne Cook and Hadley Wickham. In the course you will learn: * How to build presentation quality static graphics using the R package, ggplot. We will cover plot

X is a matrix and F is a vector. F2 <- data.frame(cbind(X,F)) F2 V1 V2 V3 F 1 -0.250536332 -1.4755883 1.9580974 -2.136487 2 -0.009856084 0.4953269 0.5486092 -2.744482 3 -0.406962682 0.7729631 0.1861905 -2.891821 4 1.938780097 0.7469251 1.2537781 -1.212992 5 -0.332370358 1.1943637 0.7114278 -1.830441 modF<-formula(F ~ V1 + V2 + V3) #no error message

Hi, I'm having trouble with the "which" or the "seq" function, I'm not sure. Here's an example : > lat=seq(1,2,by=0.1) > lat [1] 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 > which(lat==1) [1] 1 > which(lat==1.1) [1] 2 > which(lat==1.2) [1] 3 > which(lat==1.3) [1] 4 > which(lat==1.4) [1] 5 > which(lat==1.5) [1] 6 >

Hi R, Getting a strange result using ?apply. Please look into the below codes: > d=data.frame(a=c(1,2,3),b=c("A","B","C"),c=c(TRUE,FALSE,FALSE),d=c(T,F,F )) > class(d[,1]) [1] "numeric" > class(d[,2]) [1] "factor" > class(d[,3]) [1] "logical" > class(d[,4]) [1] "logical" >

Hello there, I got a small problem about logical calculation: we can get a sequene from a+b as below: > a<-c(1,2) > b<-c(3,4) > a+b [1] 4 6 but when the sequences are logical. (I want to get (True,False) && (True, True) ==> (True, False), but when I do as below. > e<-c(T,T) > f<-c(F,T) > e [1] TRUE TRUE > f [1] FALSE TRUE >

Hi, I tried several settings by using the "family=gaussian" in "gl1ce", but none of them works. For the case "glm" can work. Here is the error message I got: > glm(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length ,data=iris,family=gaussian()) > gl1ce(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length ,data=iris,family=gaussian()) Error in eval(expr, envir,

Hello, I have a loop with probability computed from a logistic model like this: for (i in 1:300){ p[i]<-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023 +z[i])) x and z generated from normal distribution. I get 300 different probabilities And I want to generate variables from bernulli distribution with P for every observation: T[i]<-rbinom(1,1,p[i]) But i get missing

I want to extra the part of the formula not including the response variable from an lm object. For example if the lm object ABx.lm was created by the call ABx.lm <- lm( y ~ A + B + x, ...) Then ACx.lm is saved as part of a workspace. I wish to extract "~ A + B + x". Later in my code I will fit another linear model of the form z ~ A + B + x for some other response variable z. I

Dear R Users, I implemented a try() statement that looks like: <- function(index) { reduced_model <- try(glm.fit(X4,n,family=poisson(link="log"))) full_model <- try(glm.fit(X5,n,family=poisson(link="log"))) if (inherits(reduced_model,"try-error") || inherits(full_model,"try-error")) return(NA) else { p <-

Dear all: Assume I have 3 distributions, x1, x2, and x3. x1 ~ normal(mu1, sd1) x2 ~ normal(mu2, sd2) x3 ~ normal(mu3, sd3) y1 = x1 + x2 y2 = x1 + x3 Now that the data I can observed is only y1 and y2. It is easy to estimate (mu1+m2), (mu1+mu3), (sd1^2+sd2^2) and (sd1^2+sd3^2) by EM algorithm since y1 ~ normal(mu1+mu2, sqrt(sd1^2+sd2^2)) and y2 ~ normal(mu1+mu3, sqrt(sd1^2+sd3^2)) However, I want

Hey, all. I had a quick question about fitting new glm values and then looking at the error around them. I'm working with a glm using a Gamma distribution and a log link with two types of treatments. However, when I then look at the predicted values for each category, I find for the one that is close to 0, the error (using se.fit=T with predicted) actually makes it overlap 0.

dear R graphics experts---if anyone is running the combination of R 2.7.0 and ghostscript (2.62), could you please run the following and let me know if you get the same strange symbol size that I do, or if there is something weird on my system? regards, /ivo pdf(file = "testhere.PDF", version="1.4", pointsize=14); plot(0, xlim=c(0,26), ylim=c(-1,4.5), type="n"

rlnorm takes two 'shaping' parameters: meanlog and sdlog. meanlog would appear from the documentation to be the log of the mean. eg if the desired mean is 1 then meanlog=0. So to generate random values that fit a lognormal distribution I would do this: rlnorm(N , meanlog = log(mean) , sdlog = log(sd)) But when I check the mean I don't get it when sdlog>0. Interestingly I

Hi. Here are three successive examples of simple quadratic programming problems with the same structure. Each problem has 2*N variables, and should have a solution of the form (1/N,0,1/N,0,...,1/N,0). In these cases, N=4,5,6. As you will see, the N=4 and 6 cases give the expected solution, but the N=5 case breaks down. >cm8 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1 0

Hi R, I have a list of matrices. I need to get the sum of all the matrices in the list. Example: a=b=c=d=matrix(1:4,2,2) l=list(a,b,c,d) I need: > a+b+c+d [,1] [,2] [1,] 4 12 [2,] 8 16 Something like do.call("+",l) is not working...why is this? I may not be knowing the number of matrices in the list... Thanks, Shubha This e-mail

Hi all If I have a string, say "ABCDA", and I want to convert this to the sum of the letter values, e.g. A -> 1 B -> 2 etc, so "ABCDA" = 1+2+3+4+1 = 11 Is there an elegant way to do this? Trying something like which(LETTERS %in% unlist(strsplit("ABCDA", ""))) is not quite correct, as it does not count repeated characters. I guess what I need is

Hi the list, When we use package.skeleton, it create some file in the man directorie. - If we use package.skeleton with namespace=FALSE, it create a file toto-internal.Rd - If we use package.skeleton with namespace=TRUE, it does not create the file toto-internal.Rd Why is that ? Christophe

Hello, R subscribers. I would like to define a function like this one: Fun <- function(x) sin(x)/x with value 1 at x=0. How can I do that? Is there a way to plot it "symbolically", without using x as a vector? e.g. x <- seq(from=-10, to=10, length=100) plot(x, Fun(x)) Thank you in advance.

I tried cast and melt in reshape package, but still can't convert this data frame m m [,1] [,2] [1,] "A" "1" [2,] "A" "2" [3,] "B" "3" to this form. m1 [,1] [,2] [1,] "A" "B" [2,] "1" "3" [3,] "2" NA Please help. [[alternative HTML version deleted]]