similar to: splineDesign and not-a-knot conditions

Hi, I am trying to find an elegant way to compute and store some frequently used matrices "on demand". The Matrix package already uses something like this for storing decompositions, but I don't know how to do it. The actual context is the following: A list has information about a basis of a B-spline space (nodes, order) and gridpoints at which the basis functions would be

On 08/02/2012 05:00 AM, r-devel-request at r-project.org wrote: > Now I just have to grovel over the R code in ns() and bs() to figure > out how exactly they pick knots and handle boundary conditions, plus > there is some code that I don't understand in ns() that uses qr() to > postprocess the output from spline.des. I assume this is involved > somehow in imposing the boundary

Dear all, I'm trying to figure out the exact meaning of the B-spline coefficients generated by the R command bs(). After reading a lot of things, I still have no clue... Here's my data. > test time f0 1 1 94.76328 2 2 102.47954 3 3 105.01234 4 4 107.21387 5 5 108.63279 6 6 109.54507 7 7 113.87931 8 8 118.21356 9 9 121.08652 10

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'm using function gam from package mgcv to fit splines. ?When I try to make a prediction slightly beyond the original 'x' range, I get this error: > A = runif(50,1,149) > B = sqrt(A) + rnorm(50) > range(A) [1] 3.289136 145.342961 > > > fit1 = gam(B ~ s(A, bs="ps"), outer.ok=TRUE) > predict(fit1, newdata=data.frame(A=149.9), outer.ok=TRUE) Error

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 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

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

I generally run 'make; make check' (with more settings) when building the Debian package. Running 3.2.4 rc from last night, I see a lot of package loading issues during 'make check'. Here is splines as one examples: checking package 'splines' * using log directory '/build/r-base-3.2.3.20160303/tests/splines.Rcheck' * using R version 3.2.4 RC (2016-03-02 r70270) *

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

Windows Vista R 2.10 - I know it is old, I will update later today. How might I perform a piece-wise linear regression where two linear segments are separated by a single knot? In addition to estimating the slopes of the two segments (or the slope in one segment and the difference between the slope of the first and second segment), I would like the analysis to select the optimum knot. My first

How can I fit a piecewise continuous logistic regression with a single free knot (i.e. the knot is not specified; the model produce an estimate of the value of the knot). Thank you, John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center

Hello! So I am handling this problem with some arrays grp1-grp7, I want to write a loop to avoid tedious work, but I don't know how to transform string to boor? For example I used i=1 paste("grp",i, sep="") I only got "grp1" instead of grp1, which can't be manipulate using mean() or other function. I am not sure if I make myself clear... THANKS!!!!

Hi, I have fit a series of ols() models, by group, in this manner: l <- ols(y ~ rcs(x, 4)) ... where the series of 'x' values in each group is the same, however knots are not always identical between groups. The result is a table of 'coefs' derived from the ols objects, by group: group Intercept top top' top'' 1 6.864 0.01 2.241 -2.65

Dear R-help, In the rms package, I have fitted an ols model with a variable represented as a restricted cubic spline, with the knot locations specified as a previously defined vector. When I save the model object and open it in another workspace which does not contain the vector of knot locations, I get an error message if I try to predict with that model. This also happens if only one workspace

Dear R developers, The trace of the hat matrix H~(n,n) is computed as follows: tr(H) = tr(BS^-1B') = tr(S^-1B'B) := tr(X) = sum(diag(X)) with B~(n,p), S~(p,p). Since p is of the order 10^3 but S is sparse I would like to employ Taucs linear solver ( http://www.tau.ac.il/~stoledo/taucs/ ) on SX = B'B. (Further improvement by implying a looping over i=1,...,p, calling

Hi, I have a question about the usage of eval(). Wonder if any experienced user can help me out of it. I use eval() in the following function: semireg.pwl <- function(coef.s=rnorm(1),coef.a=rnorm(1),knots.pos=knots.x,knots.ini.val=knots.val){ knotn <- length(knots.pos) def.par.env <- sys.frame(1) print(def.par.env) print(environment(coef.s)) tg <- eval( (parse(text=

--- 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 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 am trying to understand idea for splint, but it use Fortran code. Does anyone know how to see that Fortran code? Is splint a not-knot spline method? Thanks, jfm [[alternative HTML version deleted]]