similar to: order of points in spline

I'm trying to use R to determine the quality of a cubic Bezier curve approximation of an elliptical arc. I know the four control points and I want to compute (x,y) coordinates of many points on the curve. I can't find anything in either the base distribution or CRAN that does this; all the spline-related packages seem to be about *fitting* piecewise Bezier curves to a data set.

Hi, I?m trying to get smooth curves connecting points in a plot using "spline" but I don?t get what I whant. Eg.: x<-1:5 y <- c(0.31, 0.45, 0.84, 0.43, 0.25) plot(x,y) lines(spline(x,y)) Creates a valley between the first and second points, then peaks at 3rd, and another valley between 4th and 5th. I?m trying to get a consistently growing curve up to the 3rth point and then a

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.

Hi, I am a total beginner with this whole thing so please have patience! I am trying to run an S-plus program with a certain line: spline(1:nrow(y), y[,1],n=100); This crashes with: Error: NAs in foreign function call (arg 8) Apparently, this is caused by the last command of spline: u <- seq(xmin, xmax, length.out = n) .C("spline_eval", z$method, length(u), x = u, y =

Dear R People: Could someone recommend a small example for spline regression, please? I would like to show this to my students, step by step. Thanks in advance, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com

Hi, I am using GAMs (package mgcv) to smooth event rates in a penalized regression setting and I was wondering if/how one can select the order of the derivative penalty. For my particular problem the order of the penalty (parameter "m" inside the "s" terms of the formula argument) appears to have a larger effect on the AIC/deviance of the estimated model than the

Hi Ruser I'm trying to replicate some SAS code. I have to add a spline to my longitudinal spaghetti plot. I have the plot, but I can't add the spline, a overall trend line. In the SAS code they use the command 'I=SM50S' and I would prefer something similar. I?m using R 2.10.1 on windows XP? I have made this working example. tolerance.pp <-

Dear all, I'm a bit stunned by the behaviour of a gam model using cyclic P-spline smoothers. I cannot provide the data, as I have about 61.000 observations from a time series. I use the following model : testgam <- gam(NO~s(x)+s(y,bs="cs")+s(DD,bs="cs")+s(TT),data=Final) The problem lies with the cyclic smoother I use for seasonal trends. The variable Final$y is a

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 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

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.

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

BUG IN SPLINE()? Version R-1.0.1, system i486,linux If the spline(x,y,method="natural") function is given values outside the range of the data, it does not give a warning. Moreover, the extrapolated value reported is not the ordinate of the natural spline defined by (x,y). Example. Let x <- c(2,5,8,10) and y <- c(1.2266,-1.7606,-0.5051,1.0390). Then interpolate/extrapolate with

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 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 ...

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

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