similar to: smooth.spline

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

Hola a todos: quiero consultarles para estar seguro de que estoy entendiendo bien el funcionamiento de la función smoothSpline() del paquete 'timeSeries'. Tengo una serie temporal con datos mensuales a la cual quiero suavizar usando splines para, por ejemplo, comparar con otras series temporales. Por lo que estuve viendo, me conviene usar la función smoothSpline() que se basa en

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 some new

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

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

Hi, there, I have 5 datasets. I would like to choose a basis spline with same knots in GAM function in order to obtain same basis function for 5 datasets. Moreover, the basis spline is used to for an interaction of two covarites. I used "cr" in one covariate, but it can only smooth w.r.t 1 covariate. Can anyone give me some suggestion about how to choose a proper smoothing spline

Goodmorning, This is a documentation related question about the B-spline function in R. In the help file it is stated that: "df degrees of freedom; one can specify df rather than knots; bs() then chooses df-degree-1 knots at suitable quantiles of x (which will ignore missing values)." So if one were to specify a spline with 6 degrees of freedom (and no intercept) then a basis

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,          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 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) +

Dear R listers, When using smooth.spline to interpolate data, results are generally good. However, some cases produce totally unreasonable results. The data are values of pressure, temperature, and salinity from a probe that is lowered into the ocean, and the objective is to interpolate temperature and salinity to specified pressures. While smooth.spline provides excellent values at the

Dear R-hep, I'm trying to get the first derivative of a smooth from a gam model like: model<-gam(y~s(x,bs="cr", k=5)+z) and need the derivative: ds(x)/dx. Since coef(model) give me all the parameters, including the parameters of the basis, I just need the 1st derivative of the basis s(x).1, s(x).2, s(x).3, s(x).4. If the basis were generated with the function

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) { a[i] = (x[i] -

Hi, I am trying to fit cubic spline to a data on mortality rate by age and year (1900-2008). The data is noisy and hence I would like to smooth using spline and also extrapolate beyond 2008. Data from 1900 to 1948 are very unreliable while data from 1948 to 2008 are reliable. I would like to have a higher weight for data between 1948 to 2008. I am not sure how to do this. When I smooth data from

Hi all, I have a vector of proportions (post_op_prw) such that >summary(amb$post_op_prw) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's 0.0000 0.0000 0.0000 0.3985 0.9134 0.9962 1.0000 > summary(cut2(amb$post_op_prw,0.0001)) [0.0000,0.0001) [0.0001,0.9962] NA's 1904 1672 1

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, 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

Apologies if this is this too obscure for R-help. In package splines, ns(x,,knots,intercept=TRUE) produces an n by K+2 matrix N, the values of K+2 basis functions for the natural splines with K (internal) knots, evaluated at x. It does this by first generating an n by K+4 matrix B of unconstrained splines, then postmultiplying B by H, a K+4 by K+2 representation of the nullspace of C (2 by K+4),

Suppose I have two variables, x and y. For a fixed number of knots, I want to create a spline transformation of x such that a loss function is minimized. Presumably, this loss function would be least squares, i.e. sum (f(x)-y)^2. The spline transformations would be linear, quadratic or cubic. I know I can solve this problem using some optimization function in R, but I was wondering if anyone

I would like to fit a logistic regression using a smothing spline, where the spline is a piecewise cubic polynomial. Is the knots option used to define the subintervals for each piece of the cubic spline? If yes and there are k knots, then why does the coefficients field in the returned object from gam only list k coefficients? Shouldn't there be 4k -4 coefficients? Sincerely, Bill