similar to: nls problem

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

I apologize for posting this in essence the second time (no light at the end of the tunnel yet..): is there a way to enforce that "nls" takes both, the data *and* the model definition from the parent environment? the following fragment shows the problem. #======== cut here========== wrapper <- function (choose=0) { x <- seq(0,2*pi,len=100) y <- sin(1.5*x); y <-

If I source in the function (see below) calib(), I get: > source("papers/helle/threshold.r") > calib() Error in eval(expr, envir, enclos) : Object "energy" not found But if I cut and paste the code for calib() one line at a time into the R window it works fine. calib<-function() { contrast<-c(.01,.02,.0325,.055,.0775,.1,.125,.15,.175,.2)

Dear all Nobody responded to my previous post so far so I try with more offending subject. I just encountered a strange problem with nls formula. I tried to use nls in cycle but I was not successful. I traced the problem to some parse command. Here is an example DF<-data.frame(x=1:10, y=3*(1:10)^.5+rnorm(10)) coef(lm(log(DF[,2])~log(DF[,1]))) (Intercept) log(DF[, 1]) 0.7437320

Hi All, I've hit a problem using nls. I think it may be a restriction in the applicability of nls and I may have found a fix, but I've been wrong before. This example is simplified to the essentials. My real application is much more complicated. Take a function of matrix 'x' with additional arguments: matrix 'aMat' whose values are _not_ to be determined by nls vector

dear R experts: I am trying to orient myself using nls(). so, I am just trying to copy and adapt an example in the nls() function: > d= data.frame( y= runif(10), x= runif(10) ) > nls( y ~ 1/(1+x), data = d, start= list(x=0.5,y=0.5), trace=TRUE) Error in n%%respLength : non-numeric argument to binary operator the error message seems internal, so it would be nicer if there was a

Dear all I have tried to estimate the confidence intervals for predicted values of a nonlinear model fitted with nls. The function predict gives the predicted values and the lower and upper limits of the prediction, when the class of the object is lm or glm. When the object is derived from nls, the function predict (or predict.nls) gives only the predicted values. The se.fit and interval aguments

I have the following code, from which I get the following error message: Error in dim(data) <- dim : attempt to set an attribute on NULL I think the error is coming from the part of my code in BOLD RED. The script works fine until then. #Load libraries source("http://bioconductor.org/biocLite.R") biocLite() library(limma) library(Biobase) #change directory to folder where

Dear R users: Recently, I learn to use penalized logistic regression. Two packages (penalized and glmnet) have the function of lasso. So I write these code. However, I got different results of coef. Can someone kindly explain. # lasso using penalized library(penalized) pena.fit2<-penalized(HRLNM,penalized=~CN+NoSus,lambda1=1,model="logistic",standardize=TRUE) pena.fit2

Hello - I am using glmnet to generate a model for multiple cohorts i. For each i, I run 5 separate models, each with a different x variable. I want to compare the fit statistic for each i and x combination. When I use auc, the output is in some cases is < .5 (.49). In addition, if I compare mean MSE (with upper and lower bounds) ... there is no difference across my various x variables, but

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

Hello, I am reading in a file called fit2.txt (Limma). fit2.txt has 38 columns but when I dim(fit2) I only get 6 columns. The first column that it does not read in is df.residual. fit2<-read.table(fit2, file="fit2.txt",sep="\t",quote="",comment.char="",as.is=TRUE) The first few lines of fit2.txt (does not include all 38 columns) looks like this:

Hi everyone, I have a data frame (d), wich has the results of mosquitoes trapping in three different places. I suspect that one of these places (Local=='Palm') is biased by low numbers and will yield slower slopes in the variance-mean regression over the areas. I wonder if these slopes are diferents. I've looked trought the support list for methods for comparing slopes and found the

Hey, I have stacked a couple of garchFit objects in a list with names $fit1, $fit2, ..., $fiti assigning objects names using a loop, i.e. after running the loop modelStack = list($fit1, $fit2,...,$fiti). Thus the following apply; a = modelStack$fit2, then a is the second garchFit object of formal class 'fGarch' with 11 slots, @call, @formula... etc. I then want to extract information in

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Hey all, I seem to be having trouble fitting a spline to a large set of data using PSpline. It seems to work fine for a data set of size n=4476, but not for anything larger (say, n=4477). For example: THIS WORKS: ----------------------------- random = array(0,c(4476,2)) random[,1] = runif(4476,0,1) random[,2] = runif(4476,0,1) random = random[order(random[,1]),] plot(random[,1],random[,2])

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Hi, I have a technical question about rpart: according to Breiman et al. 1984, different costs for misclassification in CART can be modelled either by means of modifying the loss matrix or by means of using different prior probabilities for the classes, which again should have the same effect as using different weights for the response classes. What I tried was this: library(rpart)