[R] Impaired boxplot functionality - mean instead of median
matthew_wiener at merck.com
Thu Dec 1 21:58:10 CET 2005
interaction(A, B) will create a single factor made up of the combinations of
the two factors A and B. Perhaps that would let you use plotmeans.
Hope this helps,
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Evgeniy Kachalin
Sent: Thursday, December 01, 2005 3:37 PM
To: r-help at stat.math.ethz.ch
Subject: Re: [R] Impaired boxplot functionality - mean instead of median
Marc Schwartz (via MN) │п│и│ш│е│т:
>>Marc Schwartz (via MN) │п│и│ш│е│т:
>>So plotmeans is incapable of: boxplot(numerical~fact1+fact2). Is there
>>any way further?
> I think that somehow we are talking past each other here.
> plotmeans() does what it is designed to do, which is to simplify the
> process of plotting group-wise point estimates and user defined error
> bars/intervals around the point estimates.
> In your case, these intervals would be standard deviations around each
> of the group means as you have indicated.
> Review the examples in ?plotmeans.
> As Martin and others have pointed out, you need to remove boxplots from
> the equation here, as they were not designed to plot means and standard
Again, what I'm talking about: plotmeans is incapable of analyzing the
formula. For example, I have two factors: A - a, b, c, and B - d, e, f.
If i plot: boxplot(num~A+B) what do I get? Eight boxes: ad, ae, af, ba,
be, bf, cd, ce, cf. If I plot: plotmeans(num~A+B) - what do I get?
Nothing. Because plotmeans cannot combine two factors in various
combination. Is there a simple way to do it?
Anyway... That's wrong way, all what is neccessary is to have a boxplot
with mean istead of median. Is there simple way to do it?
Statistical software like Statistica 7.0 offers any possible combination
of what "Boxplot" could mean. Is it possible to have only one
modification to R's boxplot?
Thank you for kind answers.
Also please tell me, where should I send replies: to conference adress
or to those who answer me directly.
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