similar to: R help for invoking nmmin()

Hi, I'm trying to get a toy program making use of nmmin to run successfully. I've gotten to the point of compiling. However, when I attempt to run my executable, I guess a bus error. I see that someone else has asked about using nmmin before <http://tolstoy.newcastle.edu.au/R/help/06/03/23944.html>, but I haven't come across any replies. Is there some documentation on how to

Hello, right now, i have a program to collect data into a table. right now, my table is table1 <- data.frame(trial = NA, x = NA, y = NA) for each time when i want to add data into my data, i have to copy data of table into an array for each column, and then i add new data into my array, then i copy my array into the table one column by one column. For example temptrial <- table1$trial;

Hi, I have read R's Writing Extensions manual and am still confused about how to use some of the routines there when I call from C. Specifically, I am writing a little test function which I will optimize using the nmmin function which underlies R's optim() with Nelder-Mead. I guess I wonder what library/header files I should be using. I was using R_ext/Applic.h and linking with libR but I

I would like to produce a complimentary log-log survival plot with only the points appearing on the graph. I am using the code below, taken from the plot.survfit page of help for the the survival package (version 2.35-7). I am running in R 2.10.0 on Windows XP, and the list of packages following the error is loaded. Is there some specific 'type= ' syntax, or an additional parameter that

I am trying to use the R API to call optim functions (nmmin, vmmin, lbfgsb, etc.) through a C program but I couldn't find the shared library to link under the R-2.6.0 build which is compiled under Linux (REL5). main.cpp:35: undefined reference to `Rf_initEmbeddedR(int, char**)' main.cpp:41: undefined reference to `nmmin' Thanks in advance for any help. ------------------------

Hi there I am new user of r, i would need some help to translate som code for vectors in matlab to r. I have managed to translate the first 7 rows, but not the rest. Could anyone give me any suggestions for this problem?? Matlab code: tempo=[]; temps=[]; tempn=[]; tempao=[]; tempas=[]; tempan=[]; for k=1:5 tempo = [tempo n_o(k,:)]; temps = [temps n_s(k,:)]; tempn = [tempn

Dear all, I have been trying to use R's optimization routines from C and I have some questions. Specifically, I am testing the usage of nmmin which performs Nelder-Mead optimization in order to learn how to do the above exercise. I understand from the notes in Chapter 5 that the declaration for both optimfn as well as nmmin are included in the header file R_ext/Applic.h so I have included

Sorry, that's a cut and paste error. It should be either tempx or xtemp throughout. Terry T On 06/11/2015 12:59 PM, John Nolan wrote: > Is there a misprint in your example? The first line of code uses tempx, but the rest uses a different variable xtemp? > > John > > -----"R-devel" <r-devel-bounces at r-project.org> wrote: ----- > To: "r-devel at

Dear list, I was trying to use the R API for optimization method "Simulated annealing" void samin(int n, double *x, double *Fmin, optimfn fn, int maxit, int tmax, double temp, int trace, void *ex); but I encountered the following problem: The implementation of the function samin (as seen in src/main/optim.c) passes its void * argument "ex" into the function

Hola. Estoy empezando a ver temas de series temporales y me gustaría saber como obtener con R los IVES (índices de variación estacionales). Por el momento estoy viendo cosas muy simples, como descomponer una serie en tendencia y componente estacional, usando la función decompose y también la función stl. Por ejemplo # generamos 48 datos de una normal por ejemplo x <- rnorm(48) # creamos el

A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related problem. The basic problem is that when predicting a categorical using a subset, the subset may not have all the categories

(Apparently, I posted this to the wrong place. I am hopefully posting this is the correct place now. If not, please advise.) A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related

Hi, I am trying to create what are called "spider diagrams" in the geochemical literature using R. A spider diagram is basically a plot of the atomic number versus the concentration on a log scale. Lines are drawn from each atomic number for each sample. Right now, my data frame looks like: SAMPLE SITE V3 LA.NASC LAATNUM CE.NASC CEATMNUM 1 1A:001

Dear Akshay, The best response was given by Andrew. "{...}" is not a closure. This is unusual for someone used to C-type languages. But I will try to explain some of the rationale. In the case that "{...}" was a closure, then external variables would need to be explicitly declared before the closure (in order to reuse those values): intermediate = c() { ??? intermediate

i want to estimate parameters with maximum likelihood method with contraints (contant numbers). for example sum(Ai)=0 and sum(Bi)=0 i have done it without the constraints but i realised that i have to use the contraints. Without constraints(just a part-not complete): skellamreg_LL=function(parameters,z,design) { n=length(z); mu=parameters[1]; H=parameters[2]; Apar=parameters[3:10];

The following code snippet taken from plot.survfit now gives me a surprising result: tempx <- c(0, 400, 0, 0) ytemp <- c(.0657, .98, 1, 0) plot(range(xtemp), range(ytemp), log='y', type='n') In that the max value for the x axis is now 10. The 0 on the end of ytemp was actually a mistake and shouldn't be there, if you remove it then the upper limit is sensible.

Naresh, I am not sure how you are creating your data.frame so it has no, I think, column names. There are two scenarios including one where it is not really a valid data.frame and one where it can be handled before any other use as shown below. If it cannot be used, you might need to modify how your SQL or the function you call creates it so it includes either names it chooses or that you supply.

There you go, once again helping strengthen ;) John Get Outlook for iOS<https://aka.ms/o0ukef> ________________________________ From: R-help <r-help-bounces at r-project.org> on behalf of avi.e.gross at gmail.com <avi.e.gross at gmail.com> Sent: Tuesday, January 28, 2025 12:01:25 AM To: 'Naresh Gurbuxani' <naresh_gurbuxani at hotmail.com>; r-help at r-project.org

Hi all, I have a little problem. I make an weibull survival analysis using the survival package. It,s OK, them I have the functions. I plot this funcions with curve(). I want to make a plot with the real survival points (proportion of alive x time) and them add the curves to points. I have the time to dead, the censor data and my trataments. To analysis the model is: model1 <-

On 2025-01-28 1:55 p.m., Naresh Gurbuxani wrote: > Data.frame is returned by SQL query. It does have column names. In the function, I make small changes to some columns. > > Something like: > > Myquery <- ?SELECT date, price, stock FROM stocktab WHERE stock = ?ABC? AND date > ?2025-01-01?;? > > Prices <- dbGetQuery(con, myquery) > SetDT(Prices) > Prices[,