similar to: interpolation and numerical differentiation in R ?

I finally got (mostly) what I wanted. In an attempt to figure out how to get nls to deal with a non-differentiable function, I had (stupidly) 'simplified' the problem until it became singular. Can I do something to make optim() less sensitive to my initial guess? For this example, I get a lousy solution if I make the initial guess for t0 = min(t) = 0.05. Thanks again, -- Robert Merithew

First, thanks to all who helped me with my question about rescaling axes on the fly. Using unlist() and range() to set the axis ranges in advance worked well. I've since plotted about 300 datasets with relative ease. Now I'm trying to fit a lossy oscillator resonance to (the square root of) a lorentzian (testframe$y is oscillator amplitude, testframe$x is drive frequency): lorentz

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

Readers, I have looked at various documents hosted on the web site; I couldn't find anything on interpolation. So I started r and accessed the help (help.start()). (by the way is it possible to configure r to open help in opera instead of firefox?) Initially I read the help for the akima package but couldn't understand it. Next I tried the asplines package help. I tried to copy the

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) *

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 <-

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,

Hi, I would like to solve a double integral of the form \int_0^1 \int_0^1 x*y dx dy using Gauss Quadrature. I know that I can use R's integrate function to calculate it: integrate(function(y) { sapply(y, function(y) { integrate(function(x) x*y, 0, 1)$value }) }, 0, 1) but I would like to use Gauss Quadrature to do it. I have written the following code (using R's statmod package)

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

I'm trying to use nls() to fit an exponential decay with an unknown offset in the time (independent variable). (Perhaps this is inherently very difficult?). > decay.pl <- nls (amp ~ expn(b0,b1,tau,t0,t), data = decay, + start = c(b0=1, b1=7.5, tau=3.5, t0=0.1), trace=T) Error in nlsModel(formula, mf, start) : singular gradient matrix at initial parameter estimates

(I'm new to R) Is there a way to add data to an existing plot, and have the plot axes rescaled automatically (i.e. if the new data lie outside the current axes) ? If not, how can I specify multiple datasets at once, so the axes are scaled to accomodate all sets? Details: (Am I using R's data structures in a reasonable way?) I have many small datasets taken under different

I think this is a bug; at least this behavior is not documented in plot or plot.default. plot.default resets xaxp, and leaves xaxp reset when it exits: par(xaxp=c(0,1,4)) print(par("xaxp")) plot(c(0,1),c(0.2,0.3)) print(par("xaxp")) R. Woodrow Setzer, Jr. Phone: (919) 541-0128 Experimental Toxicology Division

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] -

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 would like like to compute first and second derivative from numerical data. I hoped I could compute a spline object from the data (that works), and then compute the derivative from that spline objects, but I couldn't find anything like that. Does somebody know how to get numerical derivatives? Oliver -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Dear R users, I am running a maximum likelihood model with optim. I chose the simulated annealing method (method="SANN"). SANN is not performing bad, but I guess it would be much more effecive if I could set the `parscale' parameter. The help sais: `parscale' A vector of scaling values for the parameters. Optimization is performed on `par/parscale' and these

Ah... I searched for half an hour for this function... you know, the help function in R could really be a lot better... But wait a minute... looking at this, it appears you have to pass in an expression. What if it is an unknown function, where you only have a handle to the function, but you cannot see it's implementation ? Will this work then ? -----Original Message----- From: Berton Gunter

Hi, I have been trying to do numerical differentiation using R. I found some old S code using Richardson Extrapolation which I managed to get to work. I am posting it here in case anyone needs it. ######################################################################## richardson.grad <- function(func, x, d=0.01, eps=1e-4, r=6, show=F){ # This function calculates a numerical approximation

Hi there, I've got this vector: -84 -87 -90 -90 -89 -86 NA NA NA NA NA NA NA NA NA NA NA NA -96 -99 -100 -99 -96 -92 -89 -87 -87 -88 -90 -92 -94 -95 -96 -97 -97 -97 -96 -95 Is there a function in R which replaces the NAs with "interpolated" values between -86 and -96? Thanks, Sven

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