No, that's not what I meant, but maybe I didn't understand the question.
What I suggested would involve sorting y, not x: "sort the *distances*".
If you want to minimize the sd of a subset of numbers, you sort the numbers and find a subset that is clumped together.
If the numbers are a function of pairs, you compute the function for all pairs of numbers, and find a subset that's clumped together.
Anyway, it's an idea, not a theorem, so proof is left as an exercise for the esteemed reader.
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Hans W Borchers
Sent: Monday, February 28, 2011 2:17 PM
To: r-help at stat.math.ethz.ch
Subject: Re: [R] Finding pairs with least magnitude difference from mean
<rex.dwyer <at> syngenta.com> writes:
James,
It seems the 2*mean(x) term is irrelevant if you are seeking to
minimize sd. Then you want to sort the distances from smallest to
largest. Then it seems clear that your five values will be adjacent in
the list, since if you have a set of five adjacent values, exchanging
any of them for one further away in the list will increase the sd. The
only problem I see with this is that you can't use a number more than
once. In any case, you need to compute the best five pairs beginning
at position i in the sorted list, for 1<=i<=choose(n,2), then take the
max over all i.
There no R in my answer such as you'd notice, but I hope it helps just
the same.
Rex
You probably mean something like the following:
x <- rnorm(10)
y <- outer(x, x, "+") - (2 * mean(x))
o <- order(x)
sd(c(y[o[1],o[10]], y[o[2],o[9]], y[o[3],o[8]], y[o[4],o[7]], y[o[5],o[6]]))
This seems reasonable, though you would have to supply a more stringent
argument. I did two tests and it works alright.
--Hans Werner
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