similar to: generalized least squares with empirical error covariance matrix

Hi, I'm trying to use the kalman filter to estimate the variable drift of a random walk, given that I have a vector of time series data. Anyone have any thoughts on how to do this in R? Thanks, Alex [[alternative HTML version deleted]]

> structSSM Is no longer part of KFAS. All you needed to do was: library(KFAS) ?KFAS and you would have seen that if you went to the index. A structural state space model is now built up from its components, much like in LM. Look at; ?SSModel -Roy > On Jul 29, 2017, at 9:26 PM, Staff <rbertematti at gmail.com> wrote: > > I found an example at >

I found an example at http://www.bearcave.com/finance/random_r_hacks/kalman_smooth.html shown below. But it seems the structSSM function has been removed from KFAS library so it won't run. Does anyone know how to fix the code so that it runs? library(KFAS) library(tseries) library(timeSeries) library(zoo) library(quantmod) getDailyPrices = function( tickerSym, startDate, endDate ) {

To whomever it may concern, I'm a young Industrial Engineer working on Senior Design at Georgia Tech and have found the StructTS method to be excellent for the training set for my forecasting project. There's only one problem: I don't actually understand what a Structural Time Series IS. I've looked up resources on it, and get that essentially you're dividing the Time

Hi All: I have been looking for an elegant way to do the following, but haven't found it, I have never had a good understanding of any of the "apply" functions. A simplified idea is I have an array, say: junk(5, 10, 3) where (5, 10, 3) give the dimension sizes, and I want to reverse the second dimension, so I could do: junk1 <- junk[, rev(seq_len(10), ] but what I am

How about this: f <- function(a,wh){ ## a is the array; wh is the index to be reversed l<- lapply(dim(a),seq_len) l[[wh]]<- rev(l[[wh]]) do.call(`[`,c(list(a),l)) } ## test z <- array(1:120,dim=2:5) ## I omit the printouts f(z,2) f(z,3) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

Hi All: I am using httr to download files form a service, in this case a .csv file. When I use httr::content on the result, I get a message. Since this will be in a package. I want to suppress the message, but haven't figured out how to do so. The following should reproduce the result: myURL <-

yes, thanks, and I was getting close to that. One thing I found is the manual says the height is the distance above the y-line, which should be, but doesn't have to be positive. In fact, the time series are estimates of a cycle, and has negative values, which unfortunately are not included in my sub-sample. And the negative values are not handled properly (the series disappears for

Hi All: In creating a R Notebook I know that in the text I can link to a (sub) section by using the command: [Header 1](#anchor) and putting the appropriate anchor name at the appropriate header. But can the same be done for code chunks, if the code chunk is named? What I want to do is say that such and such code chunk is an example of how to do something, and have that link to the

I have tried: ggplot(plotFrame, aes(x = time, y = cycle, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, group = depth)) + geom_density_ridges() none are producing a plot that was a ridgeline for each depth showing the time series at that depth. The plot

The min_height = -0.25 is there to make it show cycle values down to -1/4. You may want to change it to -1 so it shows more of the cycle values. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 1:26 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > yes, thanks, and I was getting close to that. One thing I found is the > manual says the

Does the following work for you? ggplot2::ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + ggridges::geom_ridgeline(fill="red", min_height=-0.25) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 12:43 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > I have tried: > > ggplot(plotFrame, aes(x =

Hi All: I am just not understanding ggridges. The data I have are time series at different depths in the ocean. I want to make a joy plot of the time series by depth. If I was just doing a ggplot2 line plot I would be doing: ggplot(plotFrame, aes(x = time, y = cycle, group = depth)) + geom_line() but translating that to ggridges has not worked right. Below is the result from dput() of a