similar to: obscure error message from splines::ns

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

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 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

Dear Maintainer, Thanks for the excellent package splines. I am writing this email to request you to consider a suggestion I have with regards to the function ns. While trying to rework an example from a textbook, I couldn't call ns with appropriate arguments to reproduce the results. The package documentation also couldn't help me find the problem. Finally, I found a stack exchange

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

Hi - How can I 'manually' reproduce the results in 'pred1' below? My attempt is pred_manual, but is not correct. Any help is much appreciated. library(splines) set.seed(12345) y <- rgamma(1000, shape =0.5) age <- rnorm(1000, 45, 10) glm1 <- glm(y ~ ns(age, 4), family=Gamma(link=log)) dd <- data.frame(age = 16:80) mm <- model.matrix( ~ ns(dd$age, 4)) pred1 <-

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

--AZaLVt6Pw+ Content-Type: text/plain; charset=us-ascii Content-Description: message body text Content-Transfer-Encoding: 7bit Dear all, I was looking at "splinefun" and the underlying C code and believe that there is a memory access error in the C routine "spline_eval". Specifically, on line 368 and following the following code appears: if(ul < x[i] || x[i+1] < ul)

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

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 [[alternative HTML version deleted]]

I've got a problem with the R function spline.des. I use the following arguments: x_seq(0,1,length=200) knot_quantile(x,(1:8)/9) ord_3 k_sort(c(rep(range(x),ord,knot))) derivs_rep(2,200) When I do spline.des(k,x,ord,derivs)$design on Splus and on R, I don't have the same results. However, if I take an order ord=4 (ie a spline of degree 3), then, I have the same

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   

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

Dear all, I need to compute tensor product of B-spline defined over equi-spaced break-points. I wrote my own program (it works in a 2-dimensional setting) library(splines) # set the break-points Knots = seq(-1,1,length=10) # number of splines M = (length(Knots)-4)^2 # short cut to splineDesign function bspline = function(x) splineDesign(Knots,x,outer.ok = T) # bivariate tensor product of

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

Hello. I seem to be having the same problem that Paul had in the thread titled "[Rd] R 2.15.2 make check failure on 32-bit --with-blas="-lgoto2"" from October of last year <https://stat.ethz.ch/pipermail/r-devel/2012-October/065103.html> Unfortunately, that thread ended without an answer to his last question. Briefly, I am trying to compile an Rblas for Windows NT 32bit

Laurent 14 juin 12:02 afficher les options De : "Laurent" <eddy_l... at hotmail.com> - Rechercher les messages de cet auteur Date : Tue, 14 Jun 2005 03:02:37 -0700 Local : Mar 14 juin 2005 12:02 Objet : bs() function of the splines package R??pondre | R??pondre ?? l'auteur | Transf??rer | Imprimer | Message individuel | Afficher l'original | Retirer | Signaler un

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