similar to: smooth.spline vs loess

Hello R'users, I have a quick question. I wanted to know if there exist a function in R to compute the rank of a matrix. I could not find anything about it. Thank you, Raphael -------------- next part -------------- A non-text attachment was scrubbed... Name: raph.vcf Type: text/x-vcard Size: 303 bytes Desc: Card for Raphael Gottardo Url :

I'm updating the loess routines to allow for, among other things, arbitrary local polynomial degree and number of predictors. For now, I've given the updated package its own namespace. The trouble is, panel.loess still calls the original code in package:stats instead of the new loess package, regardless of whether package:loess or package:lattice comes first in the search list. If I

Hello R-users, I am currently trying to learn how to use the function gls of the nlme library. I fitted the following model: Generalized least squares fit by REML Model: response ~ array + dye + genes + variety + variety * genes + array * genes + dye * genes Data: data I have 11 arrays, 2 dyes, 2 varieties, 3200 genes, and 2 replications for each. Therefore I should have the corresponding

Has the loess.smooth() function changed? It used to work, but now it causes R to abort with a segmentation fault. I stole the function points.lines() from V&R 1st ed. pp. 67--68, but now it only works if I remove the line with loess.smooth. Here's the function I'm using: points.lines <- function(x, y, ...) { cor1 <-round(cor(x, y, use="pairwise"), digits=2)

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

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.

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

loess() should be able to do robust 2D smoothing. There's no natural ordering in 2D, so defining running medians can be tricky. I seem to recall Prof. Koenker talked about some robust 2D smoothing method at useR! 2004, but can't remember if it's available in some packages. Andy From: Vladislav Petyuk > > Hi, > Are there any multidimenstional versions of runmed() and >

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

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

It has bothered me for quite some time that a smoothing spline fit doesn't allow access to residuals or fitted values in general, since after fit <- smooth.spline(x,y, *) the resulting fit$x is really equal to the unique (up to 1e-6 precision) sorted original x values and fit$yin (and $y) is accordingly. There are several possible ways to implement the missing feature. My current

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

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 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)

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

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 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 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