similar to: In-sample / Out-of-sample using R

I have a situation where lagged values of a time-series are used to predict future values. I have packed together the time-series and the lagged values into a data frame: > str(D) 'data.frame': 191 obs. of 13 variables: $ y : num -0.21 -2.28 -2.71 2.26 -1.11 1.71 2.63 -0.45 -0.11 4.79 ... $ y.l1 : num NA -0.21 -2.28 -2.71 2.26 -1.11 1.71 2.63 -0.45 -0.11 ... $ y.l2 : num

I am doing > library(rpart) > m <- rpart("y ~ x", D[insample,]) > D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 > p <- predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA

Hi, See below I reply your message for <https://stat.ethz.ch/pipermail/r-help/2008-April/160966.html>[R] predict.glm & newdata posted on Fri Apr 4 21:02:24 CEST 2008 You say it ##works fine but it does not: if you look at the length of yhat2, you will find 100 and not 200 as expected. In fact predict(reg1, data=x2) gives the same results as predict(reg1). So I am still looking for

I think I have isolated a problem with integration between Sweave and beamer. Could you please see the file: http://www.mayin.org/ajayshah/tmp/bugdemo.Rnw Unfortunately, it uses some of my internal libraries, so you can't run it. When I put it through Sweave, I get: http://www.mayin.org/ajayshah/tmp/bugdemo.tex which is, of course, a generic latex file which you can read and

I learned R & MLE in the last few days. It is great! I wrote up my explorations as http://www.mayin.org/ajayshah/KB/R/mle/mle.html I will be most happy if R gurus will look at this and comment on how it can be improved. I have a few specific questions: * Should one use optim() or should one use stats4::mle()? I felt that mle() wasn't adding much value compared with optim, and

Folks, I have this code fragment: set.seed(1001) x <- c(0.79211363702017, 0.940536712079832, 0.859757602692931, 0.82529998629531, 0.973451006822, 0.92378802164835, 0.996679563355802, 0.943347739494445, 0.992873542980045, 0.870624707845108, 0.935917364493788) range(x) # from 0.79 to 0.996 e <- function(x,d) { median(x[d]) } b <- boot(x, e, R=1000)

I have two data frames. Suppose the first has rows r1 r2 r3 and the second has rows R1 R2 R3 I'd like to generate the data frame: r1 R1 r1 R2 r1 R3 r2 R1 r2 R2 r2 R3 r3 R1 r3 R2 r3 R3 How would I go about doing this? I'm sure there's a clean way to do it but I find myself thinking in loops. -- Ajay Shah

How do I parse a date "yyyymmdd"? I tried asking chron(s, "ymd") but that didn't work. Would the date parsing routines of the Date class of 1.9 grok this? -- Ajay Shah Consultant ajayshah at mayin.org Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi

I wondered was people on this list felt about this article: http://www.voxeu.org/index.php?q=node/2363 which talks about the problems of obtaining sound answers in numerical optimisation in settings such as MLE or NLS. -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org http://ajayshahblog.blogspot.com <*(:-? -

I have : raw <- read.table("monthly.text", skip=3, sep="|", col.names=c("junk", "junk2", "wpi", "g.wpi", "wpi.primary", "g.wpi.primary", "wpi.fuel", "g.wpi.fuel", "wpi.manuf", "g.wpi.manuf",

How does one convert objects c("a","b","c") and "d" into "abcd"? > paste(c("a","b","c"), "d") of course yields [1] "a d" "b d" "c d" -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org

Folks, I have a personal package which used to build fine. Today when I tried to build the package again, some errors popped up. Could you please help? When I paste the offending function into an R it works correctly. But the package build breaks. $ R CMD check ansecon * checking for working pdflatex ... OK * using log directory '/Users/ajayshah/L/R/packages/ansecon.Rcheck' * using R

Folks, I have a time-series of 875 readings of the weekly returns of a stock market index (India's Nifty). I am interested in the AR(1) coefficient. When I do arima(r, order=c(1,0,0)) I get a statistically significant AR1 coefficient. If we apply the ordinary bootstrap to this problem, this involves sampling with replacement, which destroys the time-series structure. Hence, if we do

Hi Ajay: A casual search of google using terms like "MCMC", "Markov Chains", "Monte Carlo", and "tutorial" pulled up over 4000 hits. A few links that may interest you, I thought, like: http://csep1.phy.ornl.gov/CSEP/MC/MC.html http://www.stat.fi/isi99/proceedings/arkisto/varasto/gree0167.pdf http://www.maths.soton.ac.uk/staff/Sahu/utrecht/ ... and so

See http://mayin.nfshost.com/ajayshah/KB/R/documents/mle/mle.html -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org http://ajayshahblog.blogspot.com <*(:-? - wizard who doesn't know the answer.

Folks, I asked a question on this mailing list about the subset support of lm(). In a flash, I got three helpful responses from Rajarshi Guha <rxg218 at psu.edu> <http://jijo.cjb.net> Erin Hodgess <hodgess at gator.uhd.edu> and Peter Dalgaard <p.dalgaard at biostat.ku.dk> :-) and it was just great. The mistake I was making was in not understanding the notion of a

Folks, A nice new data resource has come up -- http://data.un.org/ I thought it would be wonderful to setup an R function like tseries::get.hist.quote() which would be able to pull in some or all of this data. I walked around a bit of it and I'm not able to map the resources to predictable URLs which can then be wget. There's some javascript going on that I'm not understanding.

How would I do something like this: f <- function(x, g) { s <- as.character(g) # THIS DOES NOT WORK sprintf("The %s of x is %.0f\n", s, g(x)) } f(c(2,3,4), "median") f(c(2,3,4), "mean") and get the results "The median of x is 3" "The mean of x is 3" -- Ajay Shah

Folks, I'm doing fine with using orthogonal polynomials in a regression context: # We will deal with noisy data from the d.g.p. y = sin(x) + e x <- seq(0, 3.141592654, length.out=20) y <- sin(x) + 0.1*rnorm(10) d <- lm(y ~ poly(x, 4)) plot(x, y, type="l"); lines(x, d$fitted.values, col="blue") # Fits great! all.equal(as.numeric(d$coefficients[1] + m

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals