On Sun, Jan 16, 2011 at 11:58 AM, Hugo Mildenberger
<hugo.mildenber...@web.de> wrote:
> Thank you very much for your qualified answers, and also for the
> link to the Tukey paper. I appreciate Tukey's writings very much.
Yes, thanks to Hadley for the nice reference, I hadn't seen it before.
> Looking at the lattice code (below), a possible implementation might
> involve binning, not so?
>
> I see a problematic part here:
>
> xx <- sort(unique(x))
>
> Unique certainly works well with Poisson distributed data, but is
> essentially a no-op when confronted with continous floating-point
> numbers.
True, but as Achim said, rootogram() is intended to work with data
arising from discrete distributions, not continuous ones. I see now
that this is not as explicit as it could be in the help page (although
"frequency distribution" gives a hint), which I will try to improve.
I don't think automatic handling of continuous distributions is simple
(because it is not clear how you would specify the reference
distribution). However, a little preliminary work will get you close
with the current implementation:
xnorm <- rnorm(1000)
## 'discretize' by binning and replacing data by bin midpoints
h <- hist(xnorm, plot = FALSE) # add arguments for more control
xdisc <- with(h, rep(mids, counts))
## Option 1: Assume bin probabilities proportional to dnorm()
norm.factor <- sum(dnorm(h$mids, mean(xnorm), sd(xnorm)))
rootogram(~ xdisc,
dfun = function(x) {
dnorm(x, mean(xnorm), sd(xnorm)) / norm.factor
})
## Option 2: Compute probabilities explicitly using pnorm()
## pdisc <- diff(pnorm(h$breaks)) ## or estimated:
pdisc <- diff(pnorm(h$breaks, mean = mean(xnorm), sd = sd(xnorm)))
pdisc <- pdisc / sum(pdisc)
rootogram(~ xdisc,
dfun = function(x) {
f <- factor(x, levels = h$mids)
pdisc[f]
})
-Deepayan
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