similar to: intigrate function and absolute 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]

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

Hi, I posted this message earlier in "Rmetrics" and I don't know whether I posted in the wrong place, so I'm posting it again in Rhelp. I have a function in x and y and let's call it f(x,y). I need to get the Hessian matrix. i.e I need (d^2f/dx^2), (d^2f/dxdy), (d^2f/dydx), (d^2f/dy^2).I can get these using the D function. now I need to evaluste the hessian matrix for

I have a file: x y z 0.025 0.025 1.65775 0.025 0.050 1.62602 0.025 0.075 1.63683 0.025 0.100 1.91847 0.025 0.125 2.00913 0.025 0.150 1.82222 0.025 0.175 1.70901 0.025 0.200 1.39759 0.025 0.225 1.39089 0.025 0.250 1.04762 If I read the file like this: data<-read.table("file.dat") How do I access the vectors x,y,z that are inside the dataframe data? I studied Venables and

Are there any plans to move the schema definition from migrations to the model, where it should be? Like og does, which I think is used by the ruby Nitro framework... ObjectGraph: http://www.rubygarden.org/index.cgi/Libraries/og_tutorial.rdoc Example code: [code] class Post property :title, String, :sql => ''VARCHAR2(32) NOT NULL'' property :body, String property

Why fails nls with "singular gradient" here? I post a minimal example on the bottom and would be very happy if someone could help me. Kind regards, ########### # define some constants smallc <- 0.0001 t <- seq(0,1,0.001) t0 <- 0.5 tau1 <- 0.02 # generate yy(t) yy <- 1/2 * ( 1- tanh((t - t0)/smallc) * exp(-t / tau1) ) + rnorm(length(t))*0.01 # show the curve

Dear R listers, I am learning the MLE utility optim() in R to program ordered logit models just as an exercise. See below I have three independent variables, x1, x2, and x3. Y is coded as ordinal from 1 to 4. Y is not yet a factor variable here. The ordered logit model satisfies the parallel regression assumption. The following codes can run through, but results were totally different from what I

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 =

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

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 :

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

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

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-help, I am trying to estimate a Cox model with nested effects basing on the minimization of the overall AIC; I have two frailties terms, both gamma distributed. There is a error message (theta2 argument misses) and I don?t understand why. I would like to know what I have wrong. Thank you very much for your time. fitM7 <- coxph(Surv(lifespan,censured) ~ south + frailty(id,

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