Message-ID: <20010113094608.R29175@jimmy.harvard.edu>
Date: 2001-01-13T14:46:08Z
From: Robert Gentleman
Subject: Histogram for factors
In-Reply-To: <14944.8292.121386.126936@gargle.gargle.HOWL>; from maechler@stat.math.ethz.ch on Sat, Jan 13, 2001 at 10:31:16AM +0100
On Sat, Jan 13, 2001 at 10:31:16AM +0100, Martin Maechler wrote:
> >>>>> "PD" == Peter Dalgaard BSA <p.dalgaard at biostat.ku.dk> writes:
>
> PD> Thomas Vogels <tov at ece.cmu.edu> writes:
> >> Hi,
> >>
> >> I keep running into this:
> R> hist (f)
> >> Error in hist.default(f) : `x' must be numeric
> >>
> >> To which of course something like (simplified but not beyond
> >> repair):
> R> hist.factor <- function (ff) {
> >> jj <- table (ff) jb <- barplot (jj, ylab="Frequency",
> >> xlab=deparse(substitute(ff))) axis (1, jb, names (jj)) }
> R> hist (f)
> >> is a possible solution.
> >>
> >> Why is a 'hist.factor' not part of the base distribution? Is this
> >> kind of thing discouraged? Usually not needed? Am I missing sth
> >> else?
>
> PD> Well histograms are density estimates defined for continuous
> PD> variables. What you're drawing is a barplot. barplot.factor might
> PD> be a good idea, though.
>
> Well,
> plot(f)
> already draws a barplot of a factor f (by the plot.factor method).
Which many of us would think of as being the right "default" plot
for a factor. However, I also think that the right thing for histogram
to do is to inform you that this is not a continuous variable.
I don't think of a barplot (barchart) as a specialized histogram.
>
> Martin
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