similar to: smooth.spline(): residuals(), fitted(),...

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

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 ...

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.

--- 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 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] -

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

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 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

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)

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

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'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 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) +

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 Smoothing Splines, D.G. Pollock (available online) where we have a list of points (xi, yi), the yi points are random such that: y_i = f(x_i) + e_i

Dear all, I want to use smooth.spline to construct a cubic smoothing spline and its first derivative to my data. However, the predict.smooth.spline does not seem to provide a SE for both the fitted values and their derivatives. How should I calculate it? Thank you very much, Bingzhang

Could someone, please, write me, how to compute the spar-value for the smooth.spline-routine to get the same HP-filtered time-series with a parameter lambda for a function (see mail "Hodrick-Prescott-Filter example" from ggrothendieck at yifan.net): hpf <- function(y,lambda{ eye <- diag(length(y)) d <- diff(eye,d=2) z <- solve(eye+lambda*crossprod(d),y)} ? Second, is

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

Hi R Community. I would like to use smooth.spline to fit a set of data and constrain the endpoints of the fit to have specific derivatives. I know this is possible with cubic splines, but I can't figure out how to specify this with arguments to the smooth.spline function. In general, is it possible to specify a set of "knots" w/locations and derivatives to constrain the fit? I

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