similar to: histogam plots

[This feature addresses https://bugs.llvm.org/show_bug.cgi?id=37880 and https://github.com/KSPP/linux/issues/84.] Clang has been ramping up its support of the Linux kernel. We recently added "asm goto with outputs", a long requested feature. We want to continue building our relationship with the Linux community. KSPP is a project to improve security in the Linux kernel, through both

Hi all. I was having some trouble with a for loop and I found the problem is the following. Does anyone have some idea why I got the following R result? Since mone is equal to 3, why mu1 only have 2 components? library(MASS) > p0 <- seq(0.1, 0.9,by=0.1) > m <- 10 > > > p0 <- p0[7] > > ## data generation > > mzero <- p0*m > mone <- m-mzero >

Dear all, I need to do some calculation where the code used are below. I get error message when I choose k to be large, say greater than 25. The error message is "Error in integrate(temp, lower = 0, upper = 1, k, x, rho, m) : the integral is probably divergent". Can anyone give some help on resolving this. Thanks. Hannah m <- 100 alpha <- 0.05 rho <- 0.1 F0

I couldnt find a previous posting on this in the archives, maybe it has already been mentioned. If you use a calculation to generate n observations in random number generators and you don't round to the nearest integer you may be generating n-1 numbers not n numbers as you thought depending on the storage precision of the calculation. e.g. > m <- 1000 > pi0 <- 0.9 >

I have a Master's student working on a project which involves estimating parameters of a certain model via maximum likelihood, with the maximization being done via optim(). A phenomenon has occurred which I am at a loss to explain. If we use certain pairs of starting values for optim(), it simply returns those values as the ``optimal'' values, although they are definitely not

Dear all, I have a function to optimize for a set of parameters and want to set a constraint on only one parameter. Here is my function. What I want to do is estimate the parameters of a bivariate normal distribution where the correlation has to be between -1 and 1. Would you please advise how to revise it? ex=function(s,prob,theta1,theta,xa,xb,xc,xd,t,delta) { expo1=

Dear list I am calling the functiton bugs() provided by R2WinBugs to performs an IRT analysis. The function returns a set of estimated parameters over n replications/iterations. For each replication, two sets of person measures (theta1 and theta2) and two sets of item difficulty parameters (diff1 and diff2) are returned. The code used to obtain these estimates is as follows: sim <-

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

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 =

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

Hi all, I have a function rho.f which gives a list of estimators. I have the following problems. rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give me a different answer, even though corr[4]==0.3. This prevents me from using a for loop. Can someone give me some help? Thank you very much in advance. Hannah >

Hi everybody: I?m trying to rewrite some routines originally written for SAS?s PROC NLMIXED into LME4's glmer. These examples came from a paper by Nelson et al. (Use of the Probability Integral Transformation to Fit Nonlinear Mixed-Models with Nonnormal Random Effects - 2006). Firstly the authors fit a Poisson model with canonical link and a single normal random effect bi ~ N(0;Sigma^2).The

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 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 colleagues, This is not strictly a R question, but I hope it is ok to ask on the list. I fed a vector of p-values from about 20 million anova tests to the package q-value and obtained this output: > qsummary(asso_p.qvalue) Call: qvalue(p = asso_p.vec) pi0: 1 Cumulative number of significant calls: <1e-04 <0.001 <0.01 <0.025 <0.05 <0.1 <1

I am using storey's qvalue package but I keep on getting errors. Why is this? > qvalue(p, lambda=0.5)$pi0 [1] "ERROR: p-values not in valid range." Error in qvalue(p, lambda = 0.5)$pi0 : $ operator is invalid for atomic vectors -- Thanks, Jim. [[alternative HTML version deleted]]

I am trying to fit a nonlinear regression model to data. There are several predictor variables and 8 parameters. I will write the model as Y ~ Yhat(theta1,...,theta8) OK, I can do this using nls() - but "only just" as there are not as many observations as might be desired. Now the problem is that we have a factor "Site" and I want to include a corresponding error

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

Hi all, I'm trying to write a MH algorithm in R for a standard normal distribution, I've been trying for a good week or so now with multiple attempts and have finally given up trying to do it on my own as I'm beginning to run out of time for this, would somebody please tell me what is wrong with my latest attempt: n=100 mu=0 sigma=1 lik<-function(theta) exp(((theta-mu)^2)/2*sigma)