From: Petr Savicky <savicky_at_praha1.ff.cuni.cz>

Date: Tue, 29 Mar 2011 18:34:39 +0200

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Tue 29 Mar 2011 - 16:37:47 GMT

Date: Tue, 29 Mar 2011 18:34:39 +0200

On Tue, Mar 29, 2011 at 11:20:13AM -0500, Christopher Desjardins wrote:

> I have 3 vectors: p1, p2, and p3. I would like each vector to be any

*> possible value between 0 and 1 and p1 + p2 + p3 = 1. I want to graph these
**> and I've thought about using scatterplot3d(). Here's what I have so far.
**>
**> library(scatterplot3d)
**> p1 <- c(1,0,0,.5,.5,0,.5,.25,.25,.34,.33,.33,.8,.1,.1,.9,.05,.05)
**> p2 <- c(0,1,0,.5,0,.5,.25,.5,.25,.33,.34,.33,.1,.8,.1,.05,.9,.05)
**> p3 <- c(0,0,1,0,.5,.5,.25,.25,.5,.33,.33,.34,.1,.1,.8,.05,.05,.9)
**> scatterplot3d(p1,p2,p3)
**>
**>
**> However, I wonder if there is an easy way to create vectors p1, p2, and p3.
*

Hi.

The vectors p1, p2 and p3 are not uniquely determined. Try, for example, the following

n <- 16

pp <- expand.grid(p1=0:n, p2=0:n, p3=0:n) pp <- subset(pp, p1 + p2 + p3 == n) p1 <- pp$p1/n p2 <- pp$p2/n p3 <- pp$p3/n

If n is a power of 2, then p1 + p2 + p3 will be exactly all ones vector. Otherwise, there may be differences within machine rounding error.

Hope this helps.

Petr Savicky.

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