I've generalised Duncan's code:
seats <- function(N,M, r0=2.5){
radii <- seq(r0, 1, len=M)
counts <- numeric(M)
pts = do.call(rbind,
lapply(1:M, function(i){
counts[i] <<- round(N*radii[i]/sum(radii[i:M]))
theta <- seq(0, pi, len = counts[i])
N <<- N - counts[i]
data.frame(x=radii[i]*cos(theta), y=radii[i]*sin(theta), r=i,
theta=theta)
} )
)
pts = pts[order(-pts$theta,-pts$r),]
pts
}
and written this:
election <- function(seats, counts){
stopifnot(sum(counts)==nrow(seats))
seats$party = rep(1:length(counts),counts)
seats
}
sample usage:
> layout = seats(449,16)
> result = election(layout, c(200,200,49)) # no overall majority!!!
> plot(result$x, result$y, col=result$party,pch=19, asp=1)
Looks like a start...
On Fri, Sep 12, 2014 at 12:41 PM, Duncan Murdoch
<murdoch.dun...@gmail.com> wrote:
> On 12/09/2014, 7:18 AM, Barry Rowlingson wrote:> On Fri, Sep 12, 2014 at
> 11:25 AM, Jim Lemon <j...@bitwrit.com.au> wrote:
>>
>>> I can see how you would plot the points going from right to left (the
> easy
>>> way), by plotting the next point on the arc with the least increase in
>>> angle from the last point plotted. If this is the way you have worked
> out, I
>>> think all that you have to do is to turn the party affiliation into a
> factor (if
>>> it is not already) and plot the points by the sorted numeric value of the
>>> factor. You will probably want to adjust the levels to some political
>>> dimensions before doing the sort.
>>
>> I'm interested in how you get exactly N seats in M rows that look as
>> neat as that. My eyes are going funny trying to count the dots in each
>> arc but there must be some nice algorithm for generating a sequence
>> that sums to N, has M elements, and has a small variable difference
>> between the row sizes to constrain the sum...
>>
>> Or am I overthinking this?
>
> I would guess it's something like this:
>
> 1. Set the radii of each arc. Since the spacing of the dots looks
> even, the number of dots in each arc should be proportional to the radius.
>
> 2. Using the proportions above find the number of dots in the largest
> arc by rounding the proportion times total to an integer.
>
> 3. Repeat for each arc moving inwards, subtracting the number of dots
> already shown from the grand total.
>
> The same scheme can be used to set the number of dots of each colour in
> each arc.
>
> For example:
>
> radii <- seq(2.5, 1, len=11)
> N <- 449
> counts <- numeric(11)
> plot(c(-2.5, 2.5), c(0, 2.5), type="n", axes=FALSE, asp=1)
> for (i in 1:11) {
> counts[i] <- round(N*radii[i]/sum(radii[i:11]))
> theta <- seq(0, pi, len = counts[i])
> points(radii[i]*cos(theta), radii[i]*sin(theta))
> N <- N - counts[i]
> }
>
> Duncan Murdoch
>
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