search for: splines

Hi, I am having an issue with a custom panel for lattice. The problem comes when I try passing a groups argument. Here is the custom panel, a wrapper for smooth spline. I copied panel.loess and replaced the loess arguments with smooth.spline(). [Note: I would like to use the cross-validation fitting properties of smooth.spline.] library(lattice)

# I need to generate some data. I'm modeling some time series that follow a # negative exponential decay (mostly). I have 20 samples that can easily be fit with cubic splines. # What I want to do is generate many thousands of similar samples using the parameters from the splines # For instance one data sample looks not unlike this: foo.curve <- 1 * exp(-0.01 * 1:500) + 0.5 ts.plot(foo.curve, lwd = 2) # Another sample looks not unlike this: foo.curve2 <- 0.9 * e...

Hello, I am trying to estimate a multivariate regression of Y on X with regression splines. Y is (nx1), and X is (nxd), with d>1. I assume the data is generated by some unknown regression function f(X), as in Y = f(X) + u, where u is some well-behaved regression error. I want to estimate f(X) via regression splines (tensor product splines). Then, I want to get the predicted values for...

Hello, I'm trying to use na.spline (package zoo) to fill some missing data in a time series. this works fine, however, if I apply the 'maxgap' argument, I always get the error: <------ Error in na.spline.vec(x., coredata(object.), xout = xout., ...) : attempt to apply non-function ------> I couldn't find a similar error for this case in the mailing lists and zoo vignette,

Hi all, I'm looking for smooth monotone estimation packages, preferably using splines. I downloaded the 'cobs' package and intend to use it, but since it offers only quadratic splines based on L1 minimization, I'd like to compare its performance to that of a more 'mainstream' cubic-spline, L2-norm minimizing spline. Preferably a smoothing spline. Does anyone kn...

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] -

Suppose I have two var x and y,now I want to fits a natural cubic spline in x to y,at the same time create new var containing the smoothed values of y. How can I get it?

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) + e_i where e_i is normal with mean 0 and variance sigma_i^2 There is a smoothing parameter lambda between 0 and 1. .when lambda is 0, smoothness is all that matters, and the fitting func...

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

Greetings, A couple general questions regarding the use of splines to interpolate depth profile data. Here is an example of a set of depths, with associated attributes for a given soil profile, along with a function for calculating midpoints from a set of soil horizon boundaries: #calculate midpoints: mid <- function(x) { for( i in 1:length(x)) { if( i &...

...1:10 using either spline(x,y,n=10,method="natural",xmin=1,xmax=10) or fn <- splinefun(x,y,method="natural") fn(1:10) This gives c(2.5366,1.2266,-0.0834,...,1.0390). I agree with all values except that at x=1. The interpolation formula in Green and Silverman's book on splines gives the derivative of a natural spline at the first knot as f'(x1)=(f2-f1)/(x2-x1) -(1/6)(x2-x1)d, where d=second derivative at the second knot. For the above example, d=0.7071, f'(x1)=-1.3493, and the natural spline interpolant has value 1.2266-(-1.3493)=2.5759 at x=1, not 2.5366. PS Th...

Hi, I am looking for an R package that would calculate multivarite (mostly 2d and 3d, tensor) cubic interpolating splines, so that I could evaluate these splines (and their derivatives) at many points (unkown at the time of calculating the spline polynomials) repeatedly. To make things concrete, I have an array V with dim(V) = k and gridpoint vectors grid=list(...), length(grid[[i]])==k[i], and would like to get so...

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

Hello everyone, I was trying to fit a spline to some points and I was quite surprised to find out that the function spline does not take into account the order of the points themselves, but orders them by x. For instance, I have: x <- c(262, 275, 264, 250, 247, 242, 238, 233) y <- c(422, 389, 359, 308, 269, 229, 191, 176) plot(x, y, xlim=c(0, 500), ylim=c(0,500)) s <- spline(x,y)

Hi All, I want to fit a piecewise spline of degree 1, i.e. a spline consisting of a straight line over each piece. I downloaded the R package pspline, then I have following questions: 1. in the program, the degree of the spline is specified by 2*norder-1. Why do they adopt such scheme that we can only fit a spline with odd degree? 2. norder cannot be set to 1. Is there any specific reason

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

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

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 everyone,          Dr. wood told me that I can adapting his example to force cubic spline to pass through certain point.          I still have no idea how to achieve this. Suppose we want to force the cubic spline to pass (1,1), how can I achieve this by adapting the following code? # Penalized example: monotonic penalized regression spline ..... # Generate data from a monotonic truth.

Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit<-smooth.spline(cdf) xx<-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x