similar to: R 3.2.4 rc issue

Thanks for the info, Dirk. The tarball builds don't run make check (because of a policy decision that it is better to have the sources available on all platforms for testing than to have none if it breaks on a single platform). However the build as of tonight has no problem with make check on the build machine. Did you by any chance forget that Matrix is a recommended package and expected to

On 4 March 2016 at 09:11, peter dalgaard wrote: | Thanks for the info, Dirk. | | The tarball builds don't run make check (because of a policy decision that it is better to have the sources available on all platforms for testing than to have none if it breaks on a single platform). However the build as of tonight has no problem with make check on the build machine. Did you by any chance forget

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 R-helpers, can I use a POSIXct date as the x variable in interpSpline? The help page says x and y need to be numeric... is there a workaround? example: library(splines) testdfr <- data.frame(Date=seq(as.POSIXct("2008-08-01"),as.POSIXct("2008-09-01"), length=10)) testdfr$yvar <- rnorm(10) sp <- interpSpline(yvar ~ Date, testdfr) preddfr <-

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

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

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

I'm sure I'm doing something completely boneheaded here, but I've used this idiom (constructing an interpolation spline and using prediction from a backSpline to find an approximation profile confidence interval) many times before and haven't hit this particular problem: r2 <- c(1.04409027570601, 1.09953936543359, 1.15498845516117, 1.21043754488875,

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

Full_Name: David D Degras Version: 2.8.0 OS: Mac OS X Submission from: (NULL) (128.135.239.11) I have recently installed the version 2.8.0 of R along with package fda (v 2.0.2) and its dependencies (including package splines v. 2.8.0). Here are my problems: 1) The package splines should feature functions such a predict.bs, predict.bSpline and such and it does not! I can make calls to bs, ns,

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

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] -

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

Hi, I would like to fit an (interpolating) spline to data where the derivatives at the endpoints of the interval are nonzero, thus the natural spline endpoint-specification does not make sense. Books (de Boor, etc) suggest that in this case I use not-a-knot splines. I know what not-a-knot splines are (so if I were solving for the coefficients directly I knew how to do this), but I don't

BUG IN SPLINE()? Version R-1.0.1, system i486,linux If the spline(x,y,method="natural") function is given values outside the range of the data, it does not give a warning. Moreover, the extrapolated value reported is not the ordinate of the natural spline defined by (x,y). Example. Let x <- c(2,5,8,10) and y <- c(1.2266,-1.7606,-0.5051,1.0390). Then interpolate/extrapolate with

Hi, I am a total beginner with this whole thing so please have patience! I am trying to run an S-plus program with a certain line: spline(1:nrow(y), y[,1],n=100); This crashes with: Error: NAs in foreign function call (arg 8) Apparently, this is caused by the last command of spline: u <- seq(xmin, xmax, length.out = n) .C("spline_eval", z$method, length(u), x = u, y =

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

I like what smooth.spline does but I am unclear on the output. I can see from the documentation that there are fit.coef but I am unclear what those coeficients are applied to.With spline I understand the "noraml" coefficients applied to a cubic polynomial. But these coefficients I am not sure how to interpret. If I had a description of the algorithm maybe I could figure it out but as it

Hello, In the following I tried 3 versions of an example in R help. Only the two first predict command work. After : library(splines) require(stats) 1) fm1 <- lm(weight ~ bs(height, df = 5), data = women) ht1 <- seq(57, 73, len = 200) ph1 <- predict(fm1, data.frame(height=ht1)) # OK plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht1, ph1) 2)

Hi, I'm trying to determine if R is useful to me. I've browsed 'The Basics of S and S-Plus' (Krause & Olson), and like the logic of the language. However, I don't see an easy way to do things like this: * given a set of observations (x,y) (x and y equal-length vectors), and a 2nd set of abscissas x2, interpolate y at the new abscissas x2. (for now, I don't really