similar to: Application of results from smooth.spline outside R

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 ...

Here I think S3 dispatch is very natural. Try the following: page <- function(x, method = c("dput", "print"), ...) UseMethod("page") page.getAnywhere <- function(x, ..., idx=NULL) { name <- x$name; objects <- x$obj; if (length(objects) == 0) stop("no object named '", name, "' was found"); if (is.null(idx)) {

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

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

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.

I use smooth.spline(x, y) in package modreg and I would like to get value of penalized log likelihood and preferable also its two parts. To make clear what I am asking for (and make sure that I am asking for the right thing) I clarify my problem trying to use the same notation as in help(smooth.spline): I want to find the natural cubic spline f(x) such that L(f) = \sum_{k=1}{n} w[k](y[k] -

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)

--- 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 > > > > -------- ???????????? ????????? > -------- > ?? ????:

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. -------- >

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

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

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

Hello, I've noticed that SPLUS seems to have a function for evaluating derivative matrices of splines. I've found the R function that evaluates matrices from 'smooth.spline'; maybe someone has written something to do the same with smooth.basis? regards, s

Can anyone tell me the trick for obtaining the smoother matrix from smooth.spline when there are non-unique values for x. I have the following code but, of course, it only works when all values of x are unique. ## get the smoother matrix (x having unique values smooth.matrix = function(x, df){ n = length(x); A = matrix(0, n, n); for(i in 1:n){ y = rep(0, n); y[i]=1; yi =

'getAnywhere' is not reporting methods when there are periods in the class name or the generic name (in R-devel). > getAnywhere( 'predict.loess') A single object matching 'predict.loess' was found It was found in the following places registered S3 method for predict from namespace modreg namespace:modreg with value <<...>> > getAnywhere(

Can someone please show me how to smooth time series data that I have in the form of a zoo object? I have a monthly economies series and all I really need is to see a less jagged line when I plot it. If I do something like s <- smooth.spline(d.zoo$Y, spar = 0.2) plot(predict(s,index(d.zoo)), xlab = "Year") # not defined for Date objects and if I do something like

Hello All I have a very long vector of unique predictor values and 6 significant digits setting for the smooth.spline rounds them off. Is there any way of increasing the significant digits withour recompiling a lot if code (simple editing and tham sourcing of "smooth.spline.r" function does not work, probably due to presence of Fortan functional calls)? Thank you very much in advance

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

I've had a play with this and, due to my own short-comings, remain none the wiser. In particular, I'm not sure what value of 'spar' is consistent with the magic lambda=1/1600 for quarterly data. I initially interpreted spar as lambda and tried setting spar=1/1600. This results in almost no smoothing while spar=1600 causes an error. The smooth.spline function seems to want