similar to: How to get the penalized log likelihood from smooth.spline()?

Hi, I got two questions and would really appreciate any help from here. First, is the penalized maximum likelihood estimation(Firth Type Estimation) only fit for binary response (0,1 or TRUE, FALSE)? Can it be applied to multinomial logistic regression? If yes, what's the formula for LL and U(beta_i)? Can someone point me to the right reference? Second, when I used *logistf *on a dataset with

Dear list, if I do smooth.spline(tmpSec, tmpT, all.knots=T) with the attached data, I get this error-message: Error in smooth.spline(tmpSec, tmpT, all.knots = T) : smoothing parameter value too small If I do smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary number], all.knots=T) it works! I just don't see it. It works for hundrets other datasets, but not for

Hi there, How may I smooth spline two vectors with the smoothing parameter selected by generalized maximum likelihood (GML) .? Thanks a lot. -- View this message in context: http://r.789695.n4.nabble.com/Smoothing-spline-with-smoothing-parameters-selected-by-generalized-maximum-likelihood-tp4210679p4210679.html Sent from the R help mailing list archive at Nabble.com.

Hi r-helpers. Please forgive my ignorance, but I would like to plot a smoothing spline (smooth.spline) from package "stats", and show the knots in the plot, and I can't seem to figure out where smooth.spline has located the knots (when I use nknots). Unfortunately, I don't know a lot about splines, but I know that they provide me an easy way to estimate the location of local

Hello everone,            Anyone who knows how to force a cubic smoothing spline to pass through a particular point?            I found on website  someone said that we can use "cobs package" to force the spline pass through certain points or impose shape           constraints (increasing, decreasing). However,  this package is using  B-spline and can only do linear and quadratic

Hello, I come from a non statistics background, but R is available to me, and I needed to test an implementation of smoothing spline that I have written in c++, so I would like to match the results with R (for my unit tests) I am following http://www.nabble.com/file/p25569553/SPLINES.PDF SPLINES.PDF where we have a list of points (xi, yi), the yi points are random such that: y_i = f(x_i) +

I have noticed a slightly puzzling behaviour exhibited by smooth.spline(). If I do sss <- smooth.spline(x,y) for a certain pair of data vectors x and y, and then do length(sss$x) I get the result ``18''. However if I do length(unique(x)) I get ``27''. Trying to force smooth.spline() to use more knots I tried sss <- smooth.spline(x,y,all.knots=TRUE) but again

I'm using R 1.7.0 on linux. With this version of R the package modreg is automatically loaded at start of session. However attempting to use predict.smooth.spline() produces Error: couldn't find function predict.smooth.spline. The function smooth.spline() is OK. What am I missing? ====================================== I.White ICAPB, University of Edinburgh Ashworth Laboratories, West

Dear R-list, I?m trying to use the lmer of the lme4 package to fit a linear mixed model of the form Y = Xb + Zu + e and I can?t figure out how to control the covariance structure of u. I want u ~ N(0,sigma^2*I). More precisely I?m trying to smooth a curve through data using the "Penalized Splines as BLUPs" method as described in Ruppert, Wand & Carroll (2003). So I have Z = [Z1

Hi, I want to use the result from smooth.spline outside R. I take my data ,which is 180 point stored in x and y s <- smooth(x,y) I can know use to e.g. find the interpolated value at e.g. x=500 predict (s,500) My problem is, that i don't know how to implement the predict function. I have looked at literature, but i cannot connect the output of the smooth.spline() to an actual spline

Hi all, I'm using the ssanova function from the gss package to fit smoothing spline anovas, and am running into some difficulty. For my data, I have measurements at 2 milisecond intervals for every observation. Every observation does not have the same duration, so I have scaled the times for each observation to a scale between 0 and 1. I would like to smooth over time, and the following

Is there any way to identify or infer the inflection points in a smooth spline object? I am doing a comparison of various methods of time-series analysis (polynomial regression, spline smoothing, recursive partitioning) and I am specifically interested in obtaining the julian dates associated with the inflection points inferred by the various models. Tyler e.g.

Hello. I'm getting an unexpected result when running smooth.spline(). Here is a simple example that replicates the error I'm getting: > aa <- c(1, 2, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 13, 14) > bb <- 1:length(aa) > plot(aa, bb) > smooth.spline(aa, bb) Error in smooth.spline(aa, bb) : need at least four unique 'x' values As you can see from the example, my

Hi, I have three original curves as follows, n<-seq(20,200,by=10) t<-c(0.1138, 0.1639, 0.2051, 0.2473, 0.2890, 0.3304, 0.3827, 0.4075, 0.4618, 0.4944, 0.5209, 0.5562, 0.5935, 0.6197, 0.6523, 0.6771, 0.6984, 0.7209, 0.7453) es<-c(0.3682, 0.4268, 0.5585, 0.6095, 0.7023, 0.7534, 0.8225, 0.8471, 0.8964, 0.9098, 0.9371, 0.9514, 0.9685, 0.9747, 0.9812, 0.9859, 0.9905, 0.9923, 0.9940)

I like what smooth.spline does but I am unclear on the output. I can see from the documentation that there are fit.coef but I am unclear what those coeficients are applied to.With spline I understand the "noraml" coefficients applied to a cubic polynomial. But these coefficients I am not sure how to interpret. If I had a description of the algorithm maybe I could figure it out but as it

Dear R listers, When using smooth.spline to interpolate data, results are generally good. However, some cases produce totally unreasonable results. The data are values of pressure, temperature, and salinity from a probe that is lowered into the ocean, and the objective is to interpolate temperature and salinity to specified pressures. While smooth.spline provides excellent values at the

Hi, forget about the below details. It is not related to the fact that the function is returned from a function. Sorry about that. I've been troubleshooting soo much I've been shoting over the target. Here is a much smaller reproducible example: x <- 1:10 y <- 1:10 + rnorm(length(x)) sp <- smooth.spline(x=x, y=y) ypred <- predict(sp$fit, x) # [1] 2.325181 2.756166 ...

Hello I have installed R-0.90.1 on my Linux (Redhat 6.2) machine, unfortunately I am not able to use a number of commands like e.g. smooth.spline and predict.smooth.spline. The error messages being given by is: Error: Object "smooth.spline" not found With the command library() I have checked or the libraries for the smoothing functions are there, as shown below. -------- >

--- On Mon, 3/12/12, aleksandr shfets <a_shfets at mail.ru> wrote: > From: aleksandr shfets <a_shfets at mail.ru> > Subject: Fwd: Re[2]: [R] B-spline/smooth.basis derivative matrices > To: "Vassily Shvets" <shv736 at yahoo.com> > Received: Monday, March 12, 2012, 5:15 PM > > > > -------- ???????????? ????????? > -------- > ?? ????:

I have some data fit to a smooth.spline object as follows: (x=vector of data for the predictor variable, y=vector of data for the response variable) fit <- smooth.spline(x,y) Now, given a spline fit point y_new, I want to be able to find out what value of x_new yielded this fit value. How to do so? (This problem is the inverse of the predict.smooth.spline function, which takes x_new as input