search for: parzen

Displaying 9 results from an estimated 9 matches for "parzen".

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2002 Jul 26
0
Parzen Windows
I suspect Prof. Ripley's response suffices. However, there *is* a Parzen kernel for kernel smoothing: Parzen K(z) = { 4/3 - 8z^2 + 8|z|^3 if |z| <= 1/2 8(1 -|z|)^3/3 if 1/2 < |z| <= 1 0 otherwise If I'm not mistaken, this appeared in Parzen's original 1962 paper on kernel density estimation....
2001 Sep 25
1
parzen-window, tukey window
Dear R-user and -programmer, has one R-package the ability to compute smoothed periodograms of time series using the Tukey-window and/or the Parzen-window? In the ts- and tseries-packages I have found only Daniell-smoothers. With many thanks in advance for any hint Albrecht Kauffmann -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send...
2012 Jan 22
1
Problem with sapa package and spectral density function (SDF)
...subdat$yf : subdat$mf > sub.datm <- aggregate(subdat$DA,list(subdat$ymf),mean) > > serie.day <- ts(subdat$DA,start(sub.dat$Year[1],sub.dat$Day[1],frequency=365)) > n.d <- length(serie.day) > espec.day <- SDF(serie.day,method="lag window",window=taper(type="parzen",n.sample=n.d,cutoff=(2*sqrt(n.d))),npad=2*n.d) > > serie.month <- ts(sub.datm,start(sub.dat$Year[1],sub.dat$Month[1]),frequency=12) > n.m<-length(serie.month) > n.m [1] 322 > espec.month<-SDF(serie.month,method="lag window",window=taper(type="parzen&quot...
2012 Jan 26
2
Calculate a function repeatedly over sections of a ts object
Hi, I want to apply a function (in my case SDF; package ?sapa?) repeatedly over discrete sections of a daily time series object by sliding a time window of constant length (e.g. 10 consecutive years or 1825 days) over the entire ts at increments of 1 time unit (e.g. 1 year or 365 days). So for example, the first SDF would be calculated for the daily values of my variable recorded between years 1
2012 Mar 07
4
Difference in Kaplan-Meier estimates plus CI
...Imagine a simple case with two survival curves (e.g. treatment & control). I just want to calculate the difference in KM estimates at a specific time point (e.g. 1 year) plus the estimate's 95% CI. The former is straightforward, but the estimates not so much. I know methods exist such as Parzen, Wei, and Ying, but was surprised not to find a package that included this. Before I code it up, I thought I'd ask if I was just missing it somewhere. Thank you Jason [[alternative HTML version deleted]]
2009 Nov 23
1
Calibration score for survival probability
...produce what I need (ie a chi-sq type statistic with a table of expected vs observed probabilities). Any other functions I should be aware of? Also, has anybody come across an implementation of the statistic described in: "A global goodness of fit statistic for Cox regression models" by Parzen & Lpisitz, Biometrics 55, 1999 Many thanks in advance Eleni Rapsomaniki Research Associate Strangeways Research Laboratory Department of Public Health and Primary Care University of Cambridge ?
2009 Oct 11
2
spectral analysis
Dear all, I am searching the period of a time series usering R. Is there some lag window functions in R? Could you give me some books about spectral analysis usering R? best wishes, Wang [[alternative HTML version deleted]]
2002 Jul 26
5
Is there a function for finding local extrema.
I have a vector with about 100.000 values representing a quite regular function (sinusoid like). I would like to find all local maxima of this function (should be about 4000). Is there a native routine for R? Thanks in advance Eryk. -- _|_ \|/ \|/ Eryk Witold Wolski tel :0049-(0)30-8413-1543 w w ?v? 'v? \'v'/ MPI Moleculare Genetik fax :0049-(0)30-8413-1139 |
2012 Mar 25
2
avoiding for loops
I have data that looks like this: > df1 group id 1 red A 2 red B 3 red C 4 blue D 5 blue E 6 blue F I want a list of the groups containing vectors with the ids. I am avoiding subset(), as it is only recommended for interactive use. Here's what I have so far: df1 <- data.frame(group=c("red", "red", "red", "blue",