Andrei -
Looking inside the code for cut, it looks like you could retrieve
the breaks as follows:
getbreaks = function(x,nbreaks){
nb = nbreaks + 1
dx = diff(rx <- range(x,na.rm=TRUE))
seq.int(rx[1] - dx/1000,rx[2] + dx/1000,length.out=nb)
}
The dx/1000 is what makes cut()'s break different than
a simple call to seq().
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Fri, 15 Oct 2010, Andrei Zorine wrote:
Hello,
My question is assuming I have cut()'ed my sample and look at the
table() of it, how can I compute probabilities for the bins? Do I have
to parse table's names() to fetch bin endpoints to pass them to
p[distr-name] functions? i really don't want to input arguments to PDF
functions by hand (nor copy-and-paste way).
x.fr <- table(cut(x,10))
x.fr
(0.0617,0.549] (0.549,1.04] (1.04,1.52] (1.52,2.01] (2.01,2.5]
16 28 26 18 6
(2.5,2.99] (2.99,3.48] (3.48,3.96] (3.96,4.45] (4.45,4.94]
3 2 0 0 1
names(x.fr)
[1] "(0.0617,0.549]" "(0.549,1.04]" "(1.04,1.52]" "(1.52,2.01]"
[5] "(2.01,2.5]" "(2.5,2.99]" "(2.99,3.48]" "(3.48,3.96]"
[9] "(3.96,4.45]" "(4.45,4.94]"
--
Andrei Zorine
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and provide commented, minimal, self-contained, reproducible code.