similar to: package spline - default value of Boundary.knots of ns

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

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

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

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

Dear Sir, I am taking about the choosing knots of B-splines. Can you tel me the R docomentation for choosing the knots of B-splines automatically? It is possible to choose the knots by eye, but which is time consuming and a little arbitrary. Thanks for your reply. Shahid PhD Student -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

In the following case, library(splines) tt <- c(55, 251, 380, 289, 210, 385, 361, 669) nn <- rep(0:7,tt) ns(nn,4) ## knots are located at (0.25,0.5,0.75); quantiles = (2,5,7) we get the error Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call (arg 1) because the 75th quantile (the location of the last "interior" knot) ends up on the boundary. As a

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

dear all, I apologize for my delay in replying you. Here my contribution, maybe just for completeness: Similar to "earth", "segmented" also fits piecewise linear relationships with the number of breakpoints being selected by the AIC or BIC (recommended). #code (example and code from Martin Maechler previous email) library(segmented) o<-selgmented(y, ~x, Kmax=20,

>>>>> Anupam Tyagi >>>>> on Tue, 9 Jul 2024 16:16:43 +0530 writes: > How can I do automatic knot selection while fitting piecewise linear > splines to two variables x and y? Which package to use to do it simply? I > also want to visualize the splines (and the scatter plot) with a graph. > Anupam NB: linear splines, i.e. piecewise

Hello,        Dr. Simon Wood told me how to force a cubic spline passing through a point. The code is as following. Anyone  who knows how I can change the code to force the first derivative to be certain value. For example, the first derivative of the constrained cubic spline equals 2 at point (0, 0.6).        I really appreciate your help!        Thanks!                 Best             Victor   

--- 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 All, I am trying to figure out how to get the position of the knots in a pspline used in a cox model. my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being continuous How do I find out where the knot of the spline are? I would like to know to figure out how many cases are there between each knot. Best, Federico -- Federico C. F. Calboli Department of Epidemiology

Dear List, I'm using GAMs in a multiple imputation project, and I want to be able to combine the parameter estimates and covariance matrices from each completed dataset's fitted model in the end. In order to do this, I need the knots to be uniform for each model with partially-imputed data. I want to specify these knots based on the quantiles of the unique values of the non-missing

Dear all, I have a problem with specifying the no. of knots in our function which include gam(). I last worked with this in mid September but since then I have reinstalled R and Simon Wood's library(mgcv), which he has changed since then. The statistician (and good R-coder) with whom I co-operate is now unfortunately overloaded with teaching, and I'm in the sprut of my thesis.... I

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

Hi, all I want to choose same knots in GAM for 10 different studies so that they has the same basis function. Even though I choose same knots and same dimensions of basis smoothing, the basis representations are still not same. My command is as follows: data.gam<-gam(y~s(age,bs='cr',k=10)+male,family=binomial,knots=list(age=seq(45,64,length=10))) What is my mistake for choice of

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

Hi, When you use the default knot choice in spm() (library(SemiPar)) a figure showing the knot locations is sent to the screen and you have to accept the knots to move on. I am trying to run simulations using this function. Is there a way to get spm() to use the default "REML" knots without needing to approve each set of knots? Here is an example: library(SemiPar) data(scallop)

Hello, I am using R and the smooth.Pspline function in the pspline package to smooth some data by using natural cubic splines. After fitting a sufficiently smooth spline using the following call: (ps=smooth.Pspline(x,y,norder=2,spar=0.8,method=1) [the values of x are age in years from 1 to 100] I tried to check that R in fact had fitted a natural cubic spline by checking that the resulting

Dear list, I'm currently analyzing some count data using a hurdle model. I've used the rcspline.eval function in the Hmisc-library to contruct the spline terms for the regression model, and what I want in the end is the ability to compute coefficients and confidence intervals for different changes in the smooth function as well as plotting the smooth function along with the