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.../2015-May/071205.html https://stat.ethz.ch/pipermail/r-devel/2015-April/070982.html Were these actually published to the list? If so - big apology. Regards, Ivan On Fri, May 29, 2015 at 12:43 AM David Winsemius <dwinsemius at comcast.net> wrote: > > On May 28, 2015, at 9:11 PM, Ivan Popivanov wrote: > > > Hello, > > > > Over the last two months I have sent two messages (same topic) to the > list. > > None of them showed on the list. For the first, I got a message that it > is > > in some queue and waiting for an administrator to look at it. Is the...

Hello, This turned out to be surprisingly hard for me: Let's say I have mm = matrix(as.character(seq(1,30, 1)), nrow=3); mm [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] "1" "4" "7" "10" "13" "16" "19" "22" "25" "28" [2,] "2" "5"

Let me start with the code: library(quantmod) library(rugarch) getSymbols("SPY", from="1900-01-01") rets=na.trim(diff(log(Cl(SPY)))) tt = tail(rets["/2004-10-29"], 1000) spec = ugarchspec(variance.model=list(garchOrder=c(1,1)), mean.model=list(armaOrder=c(2,5)), distribution.model="sged") for(ii in 1:10) { ttFit = ugarchfit( spec=spec,

Hello, Over the last two months I have sent two messages (same topic) to the list. None of them showed on the list. For the first, I got a message that it is in some queue and waiting for an administrator to look at it. Is the queue THAT long?! [[alternative HTML version deleted]]

Hello, I have some data, and I want to generate random numbers following the distribution of this data (in other words, to generate a synthetic data set sharing the same stats as a given data set). Reading an old thread I found the following text: >If you can compute the quantile function of the distribution (i.e., the >inverse of the integral of the pdf), then you can use the

On May 28, 2015, at 9:11 PM, Ivan Popivanov wrote: > Hello, > > Over the last two months I have sent two messages (same topic) to the list. > None of them showed on the list. For the first, I got a message that it is > in some queue and waiting for an administrator to look at it. Is the queue > THAT long?! The moderators...

library(xts) library(TTR) ndx = getYahooData("^NDX") aa = ndx$Close bb = aggregate(aa, as.yearweek, tail, 1) The last operation takes forever, and then the bb dates are messed up. The following produces the desired result: time(aa) = as.Date(time(aa)) bb = aggregate(aa, as.yearweek, tail, 1) The index of ndx and aa is of POSIXct (as reported by is(time(ndx))) , which apparently

Given a dataset x, the ecdf is ecdf(x). Then I can use ecdf(x)(y) to find the percentile of y. Given the ecdf is there a way to determine what is the value of y that is the boundary of let's say 95 percentile? In other words, is there a function I can call on the ecdf like: fomeFunc( ecdf( x ), 0.95 ) Which will return the highest value of y, for which ecdf( y ) < 0.95? The only solution

Hello, Have been thinking that sometimes users may want each process to initialize its random seed with a specific value rather then the current seed. This could be keyed off depending whether mc.set.seed is logical, preserving the current behaviour, or numerical, using the value in a call to set.seed. Does this make sense? If you wonder how I came up with the idea: I spent a couple of hours

Hello, I have been struggling with this for a while, tried a few things, but no clean solution so far. Here is an example from the documentation for geom_line: ========================================== # Summarise number of movie ratings by year of movie mry <- do.call(rbind, by(movies, round(movies$rating), function(df) { nums <- tapply(df$length, df$year, length)