[Rd] hist.default()$density
Martin Becker
martin.becker at mx.uni-saarland.de
Tue Mar 30 17:17:36 CEST 2010
Dear developers,
the current implementation of hist.default() calculates 'density' (and
'intensities') as
dens <- counts/(n*h)
where h has been calculated before as
h <- diff(fuzzybreaks)
which results in 'fuzzy' values for the density, see e.g.
> tmp <- hist(1:10,breaks=c(-2.5,2.5,7.5,12.5),plot=FALSE)
> print(tmp$density,digits=15)
[1] 0.0399999920000016 0.1000000000000000 0.0600000000000000
Since hist.default()$breaks are not the fuzzy breaks used for the
calculation of dens, the sum of the bins' area is significantly
different from 1 in many cases, see e.g.
> print(sum(tmp$density*diff(tmp$breaks)),digits=15)
[1] 0.999999960000008
Is this intended, or should the calculation of dens read
dens <- counts/(n*diff(breaks))
instead (or should hist.default()$breaks return the fuzzy breaks)?
Best wishes
Martin
--
Dr. Martin Becker
Statistics and Econometrics
Saarland University
Campus C3 1, Room 206
66123 Saarbruecken
Germany
More information about the R-devel
mailing list