similar to: typo in ts detrending implementation in spec.pgram?

Hello, I propose two small changes to spec.pgram to get modest speedup when dealing with input (x) having multiple columns. With plot = FALSE, I commonly see ~10-20% speedup, for a two column input matrix and the speedup increases for more columns with a maximum close to 45%. In the function as it currently exists, only the upper right triangle of pgram is necessary and pgram is not returned by

Hi, I am far from experienced in both R and time series hence the question. The code for spec.pgram() seems to involve a circularity of the kernel (see below) yielding new power estimates to all frequencies computed by FFT. " if (!is.null(kernel)) { for (i in 1:ncol(x)) for (j in 1:ncol(x)) pgram[, i, j] <- kernapply(pgram[, i, j], kernel, circular = TRUE)

Dear all, This is potentially a bug in spec.pgram, when the number of samples is odd,spec.pgramreturns a different result withfast = TRUE, the example below contains the two varieties with a reference spectrum calculated manually. the number of returned spectra is also larger (50 compared to 49) whenfast = TRUE x <- rnorm( 99 ) plot(spec.pgram(x, taper = 0 , detrend = FALSE , plot =

Hopefully this will not seem too ignorant of a question. I am having a hard time picking out the sources of the differences between: abs(fft(x))^2/length(x) and spec.pgram(x, taper=0, log="no", plot=FALSE) Also from the limited testing that I have done since the DC "frequency" is not returned from spec.pgram how can I tell what has happened to the series when I specify

Dear list, What on earth is spec.pgram() normalized too? If you would like to skip my proof as to why it's not normed too the mean squared or sum squared amplitude of the discrete function a[], feel free too skip the rest of the message. If it is, but you know why it's not exact in spec.pgram() when it should be, skip the rest of this message. The issue I refer herein refers only too a

hello, I have been using the per function in package longmemo to obtain a simple raw periodogram. I am considering to switch to the function spec.pgram since I want to be able to do tapering. To compare both I used spec.pgram with the options as suggested in the documentation of per {longmemo} to make them correspond. Now I have found on a variety of examples that there is a shift between

Hi everyone, first of all, I would like to say that I am a newbie in R, so I apologize in advance if my questions seem to be too easy for you. Well, I'm looking for periodicity in histograms. I have histograms of certain phenomenons and I'm asking whether a periodicity exists in these data. So, I make a periodogram with the function spec.pgram. For instance, if I have a histogram h, I

Dear list, I had some confusion regarding what function too use in order too relate results from spec.pgram() too a chi-square distribution. The documentation indicates that the PSD estimate can be approximated by a chi-square distribution with 2 degrees of freedom, but I am having trouble figuring out how to do it in R, and figuring out what specifically that statement in the documentation

Dear list, I've recently become interested in comparing the spectral estimates using the different methods ("pgram" and "ar") in the spectrum() function in the stats package. With many thanks to the authors of these complicated functions, I would like to point out what looks to me like a bit of an inconsistency -- but I would not be surprised if there is good reasoning

Can someone explain me this spec.pgram effect? Code: period.6<-c(0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10 ,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10) period.5<-c(0,0,0,0,0,10,0,0,0,0,10,0,0,0,0,0,0,10,0,0,0,0,10,0,0,0,0,0,0,10 ,0,0,0,0,10,0,0,0,0,0,0,10,0,0,0,0,10,0,0,0,0,0,0,10,0,0,0,0,10,0) par(mfrow=c(2,1))

I am trying to use the "kernel" function. To understand how it works I tried out some of the examples. None of them works as shown in the following: > kernel("daniell", 50) # Error in kernel("daniell", 50) : unused argument(s) (50) > kernel("daniell", 10) # Error in kernel("daniell", 10) : unused argument(s) (10) >

Hello! I tried to contact author of the package, but I got no reply. That is why I write it here. This might be useful for those who were using cts for spectral analysis of non-uniformly spaced data. In file spec.ls.R from cts_1.0-1.tar.gz lines 59-60 are written as pgram[k, i, j] <- 0.5 * ((sum(x[1:length(ti)]* cos(2 * pi * freq.temp[k] * (ti - tao))))^2/sum((cos(2 * pi * freq.temp[k] *

Dear R users I have a really simple question (hoping for a really simple answer :-): Having estimated the spectral density of a time series "x" (heart rate data) with: x.pgram <- spectrum(x,method="pgram") I would like to compute the power in a specific energy band. Assuming that frequency(x)=4 (Hz), and that I am interested in the band between f1 and f2, is the

Any Ideas to get an interactive periodogram? -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis

Hi, Are there any functions in R that could be used to estimate the phase-shift between two semi-sinusoidal vectors? Here is what I have tried so far, using the spectrum() function -- possibly incorrectly: # generate some fake data, normalized to unit circle x <- jitter(seq(-2*pi, 2*pi, by=0.1), amount=pi/8) # functions defining two out-of-phase phenomena f1 <- function(x)

Hello world, I am actually trying to transfer a lecture from Statistica to R and I ran into problems with spectral analysis, I think I just don't get it 8-( (The posting from "FFT, frequs, magnitudes, phases" from 2005 did not enlighten me) As a starter for the students I have a 10year data set of air temperature with daily values and I try to get a periodogram where the annual

Hi all, as I am trying to move slowly from just "working" to "good" code, I'd like to ask if there's a smarter way than using a for-loop in tasks like the example below. I need to obtain the extrema of the cumulated sum of a detrended time series. The following code is currently used, please have a look at the comments for my questions and remarks: system.time({ X

Dear R users I have two questions about estimating the spectral power of a time series: 1) I came across a funny thing with the following code: data(co2) par(mfrow=c(2,1)) co2.sp1<-spectrum(co2,detrend=T,demean=T,span=3) co2.sp2<-spectrum(co2[1:468],detrend=T,demean=T,span=3) The first plot displays the frequencies ranging from 0 to 6 whearas the second plot displays the same curve but

Dear R-Help, I wish to simulate a process so that it has certain properties in the frequency domain. What I attempted was to generate a random time-series signal, use spec-pgram(), apply a function in the frequency domain, and then inverse transform back to the time-domain. This idea does not seem as straight forward in practice as I anticipated. e.g. x<-ts(rnorm(1000, 0,1), frequency=256)

I need some real help on this, really stuck how are the coefficients for ur.ers(y, type = c("DF-GLS", "P-test"), model = c("constant", "trend"), lag.max = 0) The max lag is set at zero, so the regression should simply be Diff(zt) = a*z(t-1) where a is the value i'm trying to find and z(t)'s are the detrended values. but through performing