search for: interpspline

...checking examples ... SKIPPED * checking tests ... Running 'spline-tst.R' ERROR Running the tests in 'tests/spline-tst.R' failed. Last 13 lines of output: > proc.time() user system elapsed 2.272 0.020 2.291 > > ###----------------- sparse / dense interpSpline() --------------------------- > > ## from help(interpSpline) -- ../man/interpSpline.Rd > ispl <- interpSpline( women$height, women$weight) > isp. <- interpSpline( women$height, women$weight, sparse=TRUE) Error in splineDesign(knots, x, ord, derivs, sparse = sparse) :...

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...

...checking tests ... > Running 'spline-tst.R' > ERROR > Running the tests in 'tests/spline-tst.R' failed. > Last 13 lines of output: >> proc.time() > user system elapsed > 2.272 0.020 2.291 >> >> ###----------------- sparse / dense interpSpline() --------------------------- >> >> ## from help(interpSpline) -- ../man/interpSpline.Rd >> ispl <- interpSpline( women$height, women$weight) >> isp. <- interpSpline( women$height, women$weight, sparse=TRUE) > Error in splineDesign(knots, x, ord, derivs, sparse =...

...| > Running 'spline-tst.R' | > ERROR | > Running the tests in 'tests/spline-tst.R' failed. | > Last 13 lines of output: | >> proc.time() | > user system elapsed | > 2.272 0.020 2.291 | >> | >> ###----------------- sparse / dense interpSpline() --------------------------- | >> | >> ## from help(interpSpline) -- ../man/interpSpline.Rd | >> ispl <- interpSpline( women$height, women$weight) | >> isp. <- interpSpline( women$height, women$weight, sparse=TRUE) | > Error in splineDesign(knots, x, ord, derivs...

...6, 2.22717964214416, 1.87754657382891, 1.55678176465949, 1.26649764837839, 1.00770187223770, 0.780805622450771, 0.585650849661306, 0.421553364080296, 0.287358347766713, 0.18150469048976, 0.102094654969619 ) plot(d2,r2,type="b") require(splines) sp <- interpSpline(r2,d2) psp <- predict(sp) points(psp$y,psp$x,col=5) bsp <- backSpline(sp) lines(predict(bsp,seq(0,6,length=101)),col=2) The prediction from the spline (cyan dots) looks perfectly reasonable. The prediction from the inverted spline matches the curve over part of the range but goes...

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] -

Dear R-Users, I am using 'contour' to plot a graph using values x,y and a matrix z. I would like to obtain 'smooth lines' instead of no-smooth contour lines. I tried with filled.contour too. In a Post I found yy <-predict(interpSpline(x, y)) I could use that method for each range that I want to plot but I hope to find a easier method. Thank you Tommaso Letterio DIAF - UniFi Via San Bonaventura 13 Firenze (IT) - 50145 +39-055-3288605

...enough to allow breaking current R-devel's code freeze. I hope I will have corrected it for 1.5.1.. ## Here is code reproducing the problems; ## I use try(.) whenever I know the current versions of R would ## give an error: library(splines) x <- c(1:3,5:6) y <- c(3:1,5:6) (isP <- interpSpline(x,y))# poly-spline representation (isB <- interpSpline(x,y, bSpl = TRUE))# B-spline repr. xo <- c(0, x, 10)# x + outside points options(digits = 4) for(der in 0:3) # deriv=3 fails! print(formatC(try(predict(isP, xo, deriv = der)$y), wid=7,format="f"), quote = FALSE)...

...stats::dpois(y, lambda=ymax/(1+x/xhalf), log=TRUE)) (fit <- mle(ll)) plot(profile(fit), absVal=FALSE) everything works fine. Now run (fit <- mle(ll, method="BFGS", control=list(ndeps=c(1e-3, 1e-3)))) plot(profile(fit), absVal=FALSE) and you will get Error in interpSpline.default(obj[[i]]$par.vals[, i], obj[[i]]$z, na.action = na.omit) : only 0's may mix with negative subscripts This happens because optim gets the old ndeps but just one variable to optimize. The optim error is not reported, so the problem is rather hard to trace. Would you agree that t...

...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, and interpSpline but not to any predicting function. Example: > library(splines) > predict.bs Erreur : objet "predict.bs" non trouv? # in english: object "predict.bs" not found Also, I cannot track the package splines on the CRAN website. Why is that? 2) Package fda: it does not handle...

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

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

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, Having a set of values (non-time series data), what are the approximation functions that could determine the trend of the values? Cheers, Carol [[alternative HTML version deleted]]

Hello, I have 15 data points (weight at birth) by age which I want to interpolate back in time (to 5 more age points). There are many functions in R to do this and I wonder if anyone has experience in using these -- any preference/caveat etc? I am trying to find an alternative to linear interpolation. Using R 1.6.0 under windows TIA Marwan

Is there a way of calculating the derivative of a function returned by splinefun()? Such a function is a cubic spline, whence it has a calculable derivative, but is there a (simple) way of getting at it? One workaround that I have thought of is to take a fine grid of points, evaluate the function returned by splinefun() at these points, put an interpolating spline through these points using

...to profile within that range only (this is a very crude adjustment, see the code below). The new profile works, and I can increase the number of points profiled (by changing delta.t), but for some reason I cannot plot it. I get the following error message: Loading required package: splines Error in interpSpline.default(obj[[i]]$par.vals[,i], obj[[i]]$tau Thank you for your info Katarina My code in full is: (Note: I'm just using simulated data in this example, and the profile will sometimes work with this) # parameter assignments Nmin <- 0.8852 Nopt1 <- 16.78...

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

R-1.7.0 built on NetBSD 1.6, but the validation test suite failed: Machinetype: Intel Pentium III (600 MHz); NetBSD 1.6 (GENERIC) Remote gcc version: gcc (GCC) 3.2.2 Remote g++ version: g++ (GCC) 3.2.2 Configure environment: CC=gcc CXX=g++ LDFLAGS=-Wl,-rpath,/usr/local/lib make[5]: Entering directory `/local/build/R-1.7.0/src/library' >>> Building/Updating