similar to: enquiry

Hi, here comes my problem, say I have the following functions (example case) #------------------------------------------------------------ function1 <- function (x, theta) {a <- theta[1] ( 1 - exp(-theta[2]) ) * theta[3] ) b <- x * theta[1] / theta[3]^2 return( list( a = a, b = b )) } #----------------------------------------------------------- function2<-function (x, theta) {P

R-help - I'm trying to create axis breaks similar to this : http://www.r-bloggers.com/wp-content/uploads/2010/08/bar-chart-natural-axis-split1.png . Is there a way to do this in R? Here's my code thus far: structure(list(condition = structure(c(2L, 1L, 3L), .Label = c("con", "exp", "unedit"), class = "factor"), trial.avg = c(4.04583333333333,

Hi, I am trying to do parameter estimation with optim, but I can't get it to work quite right-- I have an equation X = Y where X is a gaussian, Y is a multinomial distribution, and I am trying to estimate the probabilities of Y( the mean and sd of X are known ), Theta1, Theta2, Theta3, and Theta4; I do not know how I can specify the constraint that Theta1 + Theta2 + Theta3 + Theta4 = 1 in

I have to study censured datas concernig the occurence of infection at the point of insertion of catheter in patients with renal disease. Catheter may be removed for ather reasons than infection, in this case, the observation is censored. Each has exactly 2 observations. The question are the pronostic factors of infection (the other variables are; age, sex, type of renal disease...). Could you

m0<-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x) params<-all.vars(m0) this reads all the params from m0 so theta1,2 and 3 correct? params<-params[-which(params=="x")] checks which params are multiplied by x? np<-length(params) for(i in 1:6){ esp<-get(sprintf("m%d",i-1))

I have a problem with the use of locfit with censured data, when I carry out locfit by: fitbmt<-locfit(~recur,data=BMTAGE11,cens=df.status,family="hazard",alpha=0.5) it does not give me any message, but if I want to obtain the graph or even if I ask for (fitbmt) made it gives me the following message: > fitbmt31 Problem: Object "fitbmt31" not found, while calling

Hi there, this is my first posting to this list, so let me quickly introduce myself. I'm Alex and currently working on a new version of the rsync package for the eisfair Linux distribution?. I have some problems understanding the behaviour of the 'auth users' option in the rsyncd.conf file when running rsync in daemon mode. I set up a module and a secrets file. This is the behaviour

Hi I am trying to find a parametric bootstrap confidence interval and when I used the boot function I get zero bias and zero st.error? What could be my mistake? Thank you and take care. Laila [[alternative HTML version deleted]]

Hi, Could you please let me know to use a list in a for loop here geneset is a loop.I am trying to match the names of the list with 1st row of the output. result<- list() for(i in 1:length(output) { result[[i]] <- geneset(which(geneset %n% output[,1])) } Kindly help me out -- View this message in context: http://www.nabble.com/For-loop-tp18163665p18163665.html Sent from the R

Dear Bert, Thank you so much for your kind and valuable feedback. I tried finding the starting values using the approach you mentioned, then did the following to fit the nonlinear regression model: nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x), start = list(theta1 = 0.37, theta2 = exp(-1.8), theta3 =

Dear friends, This is the dataset I am currently working with: >dput(mod14data2_random) structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L, 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4, 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4 ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34, 28)), row.names = c(NA, -15L), class =

Can any one help me to solve problem in my code? I am actually trying to find the stopping index N. So first I generate random numbers from normals. There is no problem in finding the first stopping index. Now I want to find the second stopping index using obeservation starting from the one after the first stopping index. E.g. If my first stopping index was 5. I want to set 6th observation from

Oh, sorry; I changed signs in the model, fitting theta0 + theta1*exp(theta2*x) So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 = +.055 as starting values. -- Bert On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you so much for your kind and valuable feedback. I tried finding the > starting

I am having the following error in my function function(theta,reqdIRR) { theta1<-theta[1] theta2<-theta[2] n<-length(reqdIRR) constant<- n*(theta1+theta2) sum1<-lapply(reqdIRR*exp(theta1),FUN = sum) sum2<-lapply(exp(theta2 - reqdIRR*exp(theta1)),FUN = sum) sum = sum1 + sum2 log.fcn = constant - as.numeric(sum) result = - log.fcn return(result) } *error :

Hi I need to find the Hessian matrix for a complicated function from a certain kind of data but i keep getting this error Error in f1 - f2 : non-numeric argument to binary operator the data is given by U<-runif(n) Us<-sort(U) tau1<- 2 F1tau<- pgamma((tau1/theta1),shape,1) N1<-sum(Us<F1tau) X1<- Us[1:N1]

Dear Bert, Thank you for your extremely valuable feedback. Now, I just want to understand why the signs for those starting values, given the following: > #Fiting intermediate model to get starting values > intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random) > summary(intermediatemod) Call: lm(formula = log(y - 0.37) ~ x, data = mod14data2_random) Residuals: Min

Basic algebra and exponentials/logs. I leave those details to you or another HelpeR. -- Bert On Sun, Aug 20, 2023 at 12:17?PM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you for your extremely valuable feedback. Now, I just want to > understand why the signs for those starting values, given the following: > > #Fiting intermediate model to get

I got starting values as follows: Noting that the minimum data value is .38, I fit the linear model log(y - .37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37, exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2 in the nonlinear model. This converged without problems. Cheers, Bert On Sun, Aug 20, 2023 at 10:15?AM Paul Bernal <paulbernal07 at

Dear R users, I have the following two problems, related to the function sub, grep, regexpr and similia. The header of the file(s) I have to import is like this. c("y (m)", "BD (g/cm3)", "PR (Mpa)", "Ks (m/s)", "SP g./g.", "P (m3/m3)", "theta1 (g/g)", "theta2 (g/g)", "AWC (g/g)") To get rid of spaces and

Hi there, I ran the following code: vols=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Swaption vols.csv" , header=TRUE, sep=",") X<-ts(vols[,2]) #X dcOU<-function(x,t,x0,theta,log=FALSE){ Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t) Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2]) dnorm(x,mean=Ex,sd=sqrt(Vx),log=log) }