similar to: lme(y ~ ns(x, df=splineDF)) error

I would like to fit regression models of the form y ~ ns(x, df=splineDF) where splineDF is passed as an argument to a wrapper function. This works fine if the regression function is lm(). But with lme(), I get two different errors, depending on how I handle splineDF inside the wrapper function. A workaround is to turn the lme() command, along with the appropriate value of splineDF, into a text

I want to fit a series of lme() regression models that differ only in the degrees of freedom of a ns() spline. I want to use a wrapper function to do this. The models will be of the form y ~ ns(x, df=splineDF) where splineDF is passed as an argument to a wrapper function. This works fine if the regression function is lm(). But with lme(), I get an error. fitfunction() below demonstrates this.

Using a more stable nonlinear modeling tool will also help, but key is to get the periodicity right. y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156, 161, 169, 176, 182, 190, 197, 209, 218, 232, 240) lidata <- data.frame(y=y, t=t) #I use the

A maybe trivial and stupid question: In the case of a lm or glm fit, it is quite informative (to me) to have a look to the null deviance and the residual deviance of a model. This is generally provided in the print method or the summary, eg: Null Deviance: 658.8 Residual Deviance: 507.3 and (a bit simpled minded) I like to think that the proportion of deviance 'explained' by the

Hi lily, You can get fairly good starting values just by eyeballing the curves: plot(y) lines(supsmu(1:20,y)) lines(0.6*cos((1:20)/3+0.6*pi)+17.2) Jim On Wed, Jun 21, 2017 at 9:17 AM, lily li <chocold12 at gmail.com> wrote: > Hi R users, > > I have a question about fitting a cosine curve. I don't know how to set the > approximate starting values. Besides, does the method

Hi everybody, I have an experiment examining risky choice behavior where two groups of subjects were unevenly divided across two different MRI scanners while they performed a task. Each subject's data was recorded once and only once on a particular scanner. The table describing the distribution of subjects across the scanner (3TE and 3TW) and groups is below. 3TE 3TW Group1 10

Hi R users, I have a question about fitting a cosine curve. I don't know how to set the approximate starting values. Besides, does the method work for sine curve as well? Thanks. Part of the dataset is in the following: y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96,

Hi everybody, I've written a function to run an LME model on data derived from functional magnetic resonance images. When I run the function with contrasts included I get the following error Error in eval(expr, envir, enclos) : object 'inModelFormula' not found I think it has something do do with the way contrast evaluates arguments, but I've got no idea how to fix it. The code

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

What I did was to plot your initial values, then plot the smoothed values and guess the constants. That is, I got an "eyeball" fit to the smoothed values. As I have described this as "gross cheating" in the past, you should either split your data, estimate on one subset and then test on another, or estimate on your data and test on a replication. If you get pretty much the same

Hello, I have a simple question for you: making: mylm<-lm(y~x) summary(mylm) I get the following results: ****************************************************** Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 16.54087 0.19952 82.91 <2e-16 *** x[1:19] -2.32337 0.04251 -54.66 <2e-16 *** ******************************************************

I'd trying to subclass the "lm" class to produce a "mylm" class whose instances behave like lm objects (are accepted by methods like summary.lm) but have additional data or slots of my own design. For starters: setClass("mylm", "lm") produces the somewhat cryptic: Warning message: Old-style (``S3'') class "mylm" supplied as a

Hi folks, I am running a very simple regression using mylm <- lm(mass ~ tarsus, na.action=na.exclude) I would like the use the residuals from this analysis for more regression but I'm running into a snag when I try cbind(mylm$residuals, mydata) # where my data is the original data set The error tells me that it cannot use cbind because the length of mylm$residuals is

I am trying to learn how to use nlme by working on a simple example. I attach the data from a toy example I made up which is similar to my real problem. (My grasp of fixed/random effects is still a bit tenuous) It is a longitudinal study of the effect of two treatments: A and B. The data were created by: A: y<-12/(1+exp((2-time)/.5)),y<-8/(1+exp((2-time)/.5)) B:

Hello list, I have a linear regression: mylm = lm(y~x-1) I've been reading old mail postings as well as the plotmath demo and I came up with a way to print an equation resulting from a linear regression: model = substitute(list("y"==slope%*%"x", R^2==rsq), list(slope=round(mylm$coefficients[[1]],2),rsq=round(summary(mylm)$adj.r.squared, 2))) I have four models and I

Dear R users I am new R user, execuse me I bother you, but I worked hard to find a solution: # data ID <- c(1:100) set.seed(21) y <- rnorm(100, 10,2) x1 <- rnorm(100, 10,2) x2 <- rnorm(100, 10,2) x3 <- rnorm(100, 10,2) x4 <- rnorm(100, 10,2) x5 <- rnorm(100, 10,2) x6 <- rnorm(100, 10,2) mydf <- data.frame(ID,y, x1,x2, x3, x4, x5, x6) # just seperate analyis

Dear all - We perform some measurements with a machine that needs to be recalibrated. The best calibration we get with polynomial regression. The data might look like follows: > true_y <- c(1:50)*.8 > # the real values > m_y <- c((1:21)*1.1, 21.1, 22.2, 23.3 ,c(25:50)*.9)/0.3-5.2 > # the measured data > x <- c(1:50) > # and the x-axes > > # Now I do the following:

Hi Martin, When I attended the LinuxWorld Expo in NYC back in January, I chatted with some folks at the AMD booth, as well as guys from Penguin Computing (where we bought our Opteron box). I was told that the Operton has this somewhat strange setup that the memory is controlled by one CPU. The net effect of this being that when both CPUs are running, one might only be running at around 90%

Hello, I am randomly generating values and then using an ANOVA table to find the mean square value. I would like to form a loop that extracts the mean square value from ANOVA in each iteration. Below is an example of what I am doing. a<-rnorm(10) b<-factor(c(1,1,2,2,3,3,4,4,5,5)) c<-factor(c(1,2,1,2,1,2,1,2,1,2)) mylm<-lm(a~b+c) anova(mylm) Since I would like to use a loop to

DeaR list I would like to predict the values of each column of a matrix A by regressing it on all other columns of the same matrix A. I do this with a for loop: A <- B <- matrix(round(runif(10*3,1,10),0),10) A for (i in 1:length(A[1,])) B[,i] <- as.matrix(predict(lm( A[,i] ~ A[,-i] ))) B It works fine, but I need it to be faster. I've looked at *apply but just can't