similar to: Re: [S] [R/S] question re solution

Hi, Erin: A cleaner way is to pass "n2" to "outer" as a "..." argument, as in the following modification of your code: boot1 <- function(y,method="f",p=1) { n1 <- length(y) n2 <- n1*p n3 <- n2 - 1 a <- 0.5*(outer(1:n3,1:n3,function(x,y, n2.){n2. - pmax(x,y)}, n2.=n2)) return(a) } y1 <- c( 9, 8, 7, 3, 6) boot1(y=y1,p=4)

Dear R and S People: First, thank you to so many people for your help to my problem. Here is the solution: a <- 0.5*(outer(1:n3,1:n3,function(x,y,n2.){n2. - pmax(x,y)},n2.=n2)) I have one final pesky question, please: During my experiments, I tried the following: a <- 0.5*(outer(1:n3,1:n3,function(x,y,n2.=n2){n2. - pmax(x,y)})) Why doesn't this work please? thank you! Sincerely,

Dear R and S People: I have run across something very strange. Here is a function that I wrote for R: boot1 <- function(y,method="f",p=1) { n1 <- length(y) n2 <- n1*p n3 <- n2 - 1 a <- 0.5*(outer(1:n3,1:n3,function(x,y){n2 - pmax(x,y)})) return(a) } and here is the R output: > y1 [1] 9 8 7 3 6 > source("boot1.R") > boot1(y=y1,p=4) [,1] [,2]

Dear R and S People: I ended up using the "assign" command, and things work in S+. boot1 <- function(y,method="f",p=1) { n1 <- length(y) #n2 <- n1*p assign("n2",n1*p) n3 <- n2 - 1 a <- 0.5*(outer(1:n3,1:n3,function(x,y){n2 - pmax(x,y)})) return(a) } thanks for listening! Sincerely, Erin H mailto: hodgess at gator.uhd.edu

Hello...i want to find the empirical rate for type 1 error using fixed trimmed mean. To make it easy, i'm referring to journal given by this website http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf. I already run the programme and there is no error in it but i got zero for the empirical rate of type 1 error. The empirical rate for the type 1 error given in the journal

Hello, I haven't found errors in your code. I implemented the test in the paper (the first, fixed symetric mean) and it also gives me zero Type I errors, when alpha = 0.05. Try to see the value of min(pv) or to plot the histogram of 'pv', hist(pv) and you'll see that there are no significant p-values, at that level. Anyway I'll continue to look at it, but my first

Thank you David. Using xtabs operation simplifies the code very much, many thanks ;) On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > > Hi Bogdan, > > Kinda messy, but: > > > > N <-

Simple matrix indexing suffices without any fancier functionality. ## First convert M and N to character vectors -- which they should have been in the first place! M <- sort(as.character(M[,1])) N <- sort(as.character(N[,1])) ## This could be a one-liner, but I'll split it up for clarity. res <-matrix(NA, length(M),length(N),dimnames = list(M,N)) res[as.matrix(C[,2:1])] <-

Hi, I run this code to get the power of the test for modified bartlett's test..but I'm not really sure that my coding is right.. #normal distribution unequal variance asim<-5000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) n1<-20 n2<-20 n3<-20 mu<-0 sd1<-sqrt(25) sd2<-sqrt(50) sd3<-sqrt(100) g1<-rnorm(n1,mu,sd1) g2<-rnorm(n2,mu,sd2)

> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > Hi Bogdan, > Kinda messy, but: > > N <- data.frame(N=c("n1","n2","n3","n4")) > M <- data.frame(M=c("m1","m2","m3","m4","m5")) > C <-

Here's another approach: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400)) # Rebuild the factors using M and N C$m <-

I am trying to solve one of my graduate level physics problems with ruby... Here is what I have so far... a6=0.0 for n1 in -10..10 for n2 in -10..10 for n3 in -10..10 if n1!=0 and n2!=0 and n3!=0 p=Math.sqrt(n1**2+n2**2+n3**2+n1*n2/1.414+n1*n3/1.41+n2*n3/1.414) a6+=(1/p)**6 end end end end puts a6 What I''ve got here is a 10x10x10 face-centered

Dear R help, I am trying solve an MLE convergence problem: I would like to estimate four parameters, p1, p2, mu1, mu2, which relate to the probabilities, P1, P2, P3, of a multinomial (trinomial) distribution. I am using the mle2() function and feeding it a time series dataset composed of four columns: time point, number of successes in category 1, number of successes in category 2, and

Dear R Help, I have two nested negative log-likelihood functions that I am optimizing with the spg function [BB package]. I would like to perform model selection on these two objective functions using AIC (and possibly anova() too). However, the spg() function returns a list and I need a fitted model object for AIC(), ICtab() [bbmle package], or anova(). How can I perform AIC-based model

Dear Contributors, thanks for collaboration. I am trying to reorganize data frame, that looks like this: n1.Index Date PX_LAST n2.Index Date.1 PX_LAST.1 n3.Index Date.2 PX_LAST.2 1 NA 04/02/07 1.34 NA 04/02/07 1.36 NA 04/02/07 1.33 2 NA 04/09/07 1.34 NA 04/09/07

Dear R-users, Is there an R function to compute the multinomial beta function? That is, the normalizing constant that arises in a Dirichlet distribution. For example, with three parameters the beta function is Beta(n1,n2,n2) = Gamma(n1)*Gamma(n2)*Gamma(n3)/Gamma(n1+n2+n3) Thanks in advance for any assisstance. Regards, Greg [[alternative HTML version deleted]]

Hi, how can I create an array of lists of three components? This approach does not work: n1 <- 2 n2 <- 4 n3 <- 5 res <- array(rep(vector("list",3), n1*n2*n3), dim = c(n1,n2,n3)) res[1,1,1] # is not a list with three components... The goal is that res[1,1,1] is a list with three components. Also, appending the components didn't work. For example, I tried: component

Dear R Help - I am analyzing data from an ecological experiment and am having problems with the ANOVA functions I've tried thus far. The experiment consists of a blocked/split-plot design, with plant biomass as the response. The following is an overview of the treatments applied (nitrogen addition, phosphorus addition, and seeded/not seeded) and at what level (block, main-plot, and sub-plot):

Hi Bogdan, Kinda messy, but: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))

Hi: I am looking for some help in making two boxplots next to each other. I have a data like this: N1 T1 N2 T2 N3 T3 N4 T4 ... Nn Tn 7 8.2 4 5 8 10 4 5 ..... 10 11 I want to have box plot for all Normal samples (N1,N2,N3,N4,,,,Nn) and another box plot for all tumors (T1,T2,T3,T4,...Tn). I have data in a numeric class. If data is represented as N1