similar to: deriv3 example on Venables/Ripley page 263

Dear list! I am interested in learning about MLE and I wonder whether it is possible to use the examples for maximum likelihood estimation given in 8.7 in MASSv3 with R? AFAIU R does not have a direct replacement for S-PLUS's ms() which the examples use for the fitting, but optim() may be of help for me. However, I am not sure how I can convert the use of ms() using optim()--is there an

Hi everyone, I am using R 2.4.1 on a MacOS X ("Tiger") operating system. In the last few day I was trying to estimate the parameters of a mixture of two normal distributions using Maximum Likelihood. The code is from Modern Applied Statistics with S (4th edition), chapter 16 ("Optimization"), the dataset is available under MASS in R. Unfortunately, when I tried out the

Using S+ 6 for Linux and R 1.4.1 Taking verbatim some lines from S-Plus that work perfectly, and running them on R, I get an unexpected error message: R : Copyright 2002, The R Development Core Team Version 1.4.1 (2002-01-30) > library(MASS) > data <- read.table("/usr/local/data/mcmanus.dat",header=T) > mcmanus.frm <-

Hi, in the code below, I am drawing 1000 samples from two beta distributions, each time using the same random number seed. Using set.seed(80) produces results I expect, in that the differences between the distributions are very small. Using set.seed(20) produces results I can't make sense of. Around half of the time, it behaves as with set.seed(80), but around half of the time, it behaves

Dear all, I would like to solve the following problem, which can be done with optimal control theory or dynamic programming: max(x,y) a*u1+b*u2+c*f1(u2) s.t. 0<u1<x, 0<u2<f2(x,u2), x'=f3(u1,u2,x) which can be rewritten if optimal control theory should be applied as H=a*u1+b*u2+c*f1(u2)+lambda*(x') s.t. 0<u1<x, 0<u2<f2(x,u2) The maximum principle

OS: CentosOS 7 I have installed samba + openldap + smbldap-tools + pam by: yum --enablerepo=extras install -y epel-release yum install -y smbldap-tools yum install -y samba openldap openldap-clients openldap-servers migrationtools yum install -y nss-pam* I know that smbldap-tools is a dead project, but I'm interested in it and would like research on it. I create users and

Hi, In grid 4.0, adding two complex units by `sum()` seems to give wrong results. In the following example, `u1 + u2` gives the correct result, but `sum(u1, u2)` also `sum(unit.c(u1, u2))` give the wrong results. ``` library(grid) u1 = 0.4*sum(unit(1, "inch"), unit(1, "mm")) u2 = 0.1*sum(unit(1, "inch"), unit(1, "mm")) u1 # [1] 0.4*sum(1inches, 1mm)

Dear all, I am using the function 'D' in the 'stats' package to perform symbolic derivation. This works very well and it is much faster than e.g. Mathematica (at least for my purposes). First, I would like to thank the development team for this excellent function. However, I run into trouble in some cases, particularly when I am to do some operations on long expressions

Hello, could You please help: I am looking for a way to formulate test accuracy measures such as test sensitivity, specificity, predictive values, and correct classification rate using p/d/qnorm. The tests' primary values follow a bimodal distribution, which is modelled by a mixture of two normal distributions: p * dnorm ((x - u1) / s1) / s1 + (1 - p) * dnorm ((x - u2) / s2) / s2)

If this is posted elsewhere I cannot find it. I need to perform multiple integration where some of the variables are in the bounds of the other variables. I was trying to use R2Cuba function but cannot set the upper and lower bounds. My code so far is : int <- function(y){ u2 = y[1] z2 = y[2] u1 =y[3] z1 = y[4] ff <- u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)) return(ff) }

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

> On Mar 11, 2018, at 3:32 PM, Neha Aggarwal <aggarwalneha2000 at gmail.com> wrote: > > Hello All, > I am facing a unique problem and am unable to find any help in R help pages > or online. I will appreciate your help for the following problem: > I have 2 data-frames, samples below and there is an expected output > > R Dataframe1: > C1 C2

Hi, If I have an R object UUU, where the second element is U2, based on "g" column of my.table my.table of UUU is: mmm ggg gindex map Info aaa123 U1 1 1 1 aaa124 U1 1 2 1 bbb1378 U2 2 1 1 bbb8888 U2 2 2 0 bbb1389 U2 2 3

Hello everybody! I’m trying to define the optimal number of surveys to detect the highest number of species within a monitoring season/session. To do this I want to run all the possible combinations between a set of samples and to calculate the total number of species for each combination of 2, 3, 4 …n samples events, so that at the end I will be able to define which is the lowest number of

Hello: Why is deriv3() functioning differently in R from that in Splus using library(MASS) ? For example deriv3(~(t1*log(t2)+lgamma(t1)+(1-t1)*log(y)+y/t2),c("t1","t2"),function(y,t1,t2)NULL) complains of lgamma. Mervyn

Hello All, I am facing a unique problem and am unable to find any help in R help pages or online. I will appreciate your help for the following problem: I have 2 data-frames, samples below and there is an expected output R Dataframe1: C1 C2 C3 C4...... CN R1 0 1 0 1 R2 1 0 1 1 R3

How can I divide a unit by an number or average a vector of units, e.g.: u1 <- unit( 3, 'npc' ) u2 <- unit( 6, 'npc' ) u1 / 2 ( u1 + u2 ) / 2 mean( unit.c(u1,u2) ) I would use that e.g. to to calculate the coordinates of the midpoint of a line. Wolfram

Dear all, Suppose I have a very long expression e. Lets assume, for simplicity, that it is e = expression(u1+u2+u3) Now I wish to replace u2 with x and u3 with 1. I.e. the 'new' expression, after replacement, should be: > e expression(u1+x+1) My question is how to do the replacement? I have tried using: > e = parse(text=gsub("u2","x",e)) > e =

Dear all, I need to do a loop in R, but I am not sure the software is generating "n" times the variables I request differently. When I ask to print the last matrix created, I just can see the loop for n=1. To be more precise, supose I need to simulate 10 times one variable and I want to fit the 10 variables simulated in a matrix. I dont really know what I am doing wrong, but I just

There are two users u1 & u2 in the cluster. Some files are created by u1, and they are read only for u2. Of course u2 can read these files. Later these files are renamed by u1. Then I switch to the user u2. I find that u2 can't list or access the renamed files. I see these errors in log: [2018-01-15 17:35:05.133711] I [MSGID: 109045] [dht-common.c:2393:dht_lookup_cbk] 25-data-dht: