[R] finding derivative of a data series in R

[nandan amar]

Thanks Ravi,
It worked.
trying to understand the features.R  :)
Regards

On 27 May 2011 21:21, Ravi Varadhan <rvaradhan at jhmi.edu> wrote:
> I am attaching a function that I have written sometime ago for extracting some of the features of a noisy, discretely sampled times series.  In order to extract the features, I first estimate the smoothed function and smoothed (first and second order) derivatives.
>
> Here is a simple demo of how to use it:
>
> source("h:/features.R")
> n <- 200
> x <- sort(runif(n))
> y <- exp(-0.2 * sin(2*pi*x)) + rnorm(n, sd=0.05)
>
> ans <- features(x, y, fits.return=TRUE, control=list(plot.it=TRUE))
>
> fits <- attr(ans, "fits")
>
> plot(fits$x, fits$fits1, type="l", xlab="x", ylab="First derivative")
>
> yexact <- -0.2 * 2*pi * cos(2*pi*x) * exp(-0.2 * sin(2*pi*x))
>
> lines(x, yexact, col=2)
>
> # Similarly, you can also look at the smoothed second derivative
>
> Hope this is helpful,
> Ravi.
> -------------------------------------------------------
> Ravi Varadhan, Ph.D.
> Assistant Professor,
> Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University
>
> Ph. (410) 502-2619
> email: rvaradhan at jhmi.edu
>
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of nandan amar
> Sent: Friday, May 27, 2011 10:44 AM
> To: r-help; Dennis Murphy
> Subject: Re: [R] finding derivative of a data series in R
>
> However if we have some discrete data set like daily temperature etc
> how can we can differentiate it
> because as.expression(D()) will not work then
>
> On 27 May 2011 16:52, nandan amar <nandan.amar at gmail.com> wrote:
>> Thanks Dennis.
>> I got you.
>> My main consern was how to differentiate a series correctly.
>> I think the first procedure is more accurate.
>> Regards.
>>
>> On 27 May 2011 16:27, Dennis Murphy <djmuser at gmail.com> wrote:
>>> Hi:
>>>
>>> A function and its spline approximation are not equivalent functions,
>>> hence neither are their corresponding derivatives. I modified an
>>> example from the splinefun() help page to illustrate this.
>>>
>>> op <- par(mfrow = c(2, 1))
>>> x <- seq(1, 9, by = 0.01)
>>> # u is the true function, u1-u3 are its symbolic derivatives wrt x
>>> # u1 = f', u2 = f'', u3 = f'''
>>> u <- expression(sin(pi * (x - 0.5)))
>>> u1 <- as.expression(D(u, 'x'))
>>> u2 <- as.expression(D(u1, 'x'))
>>> u3 <- as.expression(D(u2, 'x'))
>>> plot(x, eval(u), type = 'l', ylim = c(-30, 30),
>>>     ylab = "", xlab = "",
>>>     main = expression(f(x) == sin(pi * (x - 0.5))))
>>> lines(x, eval(u1), type = 'l', col = 'red')
>>> lines(x, eval(u2), type = 'l', col = 'blue')
>>> lines(x, eval(u3), type = 'l', col = 'green')
>>> # legend('topright', legend = c("f", "f'", "f''", "f'''"),
>>> #        col = c('black', 'red', 'blue', 'green'), lwd = 2)
>>>
>>> # y2 is an evaluation of u at each x point
>>> y2 <- sin((x-0.5)*pi)
>>> # Construct the spline function approximation to u
>>> f <- splinefun(x, y2)
>>> # Plot the interpolation function and its first three derivatives
>>> curve(f(x), 1, 10, col = "black", lwd = 1.5, ylim = c(-30, 30),
>>>      ylab = "", main = 'Spline interpolation of f')
>>> curve(f(x, deriv=1), 1, 10, col = 'red', lwd = 1.5, add = TRUE)
>>> curve(f(x, deriv=2), 1, 10, col = 'blue', lwd = 1.5, add = TRUE)
>>> curve(f(x, deriv=3), 1, 10, col = 'green', lwd = 1.5, add = TRUE)
>>> par(op)
>>>
>>> Notice that the peaks and troughs of the derivatives of the spline
>>> approximation are not at the same x locations as in the original
>>> function. Also notice the linearity in the derivatives when x is
>>> between 9 and 10.
>>>
>>> I suppose you could improve the approximations by setting some knot
>>> points, but I don't have the time to chase down that hypothesis for
>>> you right now. I'll leave that as homework :)
>>>
>>> Your example is simpler since it is polynomial, but the concept is the
>>> same: the derivative of the interpolator shouldn't necessarily match
>>> the derivative of the function exactly. Obviously, though, you want
>>> them to be close.
>>>
>>> HTH,
>>> Dennis
>>>
>>> On Fri, May 27, 2011 at 1:23 AM, nandan amar <nandan.amar at gmail.com> wrote:
>>>> Dear All,
>>>> I tried following for getting derivative of a polynomial in R
>>>>
>>>> i<- -10:10
>>>> x<-i*i*i+3*i*i+2
>>>> fun_spline<-splinefun(i,x)
>>>> plot(x,type="l")
>>>> lines(x,fx_spline(x, deriv=1), col='green')
>>>> lines(x,fx_spline(x, deriv=2), col='green')
>>>>
>>>> Now when I plot
>>>> 3*i*i + 6*i and 6*i + 6
>>>> the plot was not same for first deivative.
>>>> where as the 2nd derivative was same
>>>>
>>>> Is this a correct method for getting derivative.
>>>> where am I doing wrong as first derivative  lines(x,fx_spline(x,
>>>> deriv=1), col='green') is not correct.
>>>>
>>>> amar
>>>> --
>>>> Amar Kumar Nandan
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>>
>>
>>
>> --
>> Amar Kumar Nandan
>> Karnataka, India, 560100
>> http://aknandan.co.nr
>>
>
>
>
> --
> Amar Kumar Nandan
> Karnataka, India, 560100
> http://aknandan.co.nr
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Amar Kumar Nandan
Karnataka, India, 560100
http://aknandan.co.nr