Hi Nathan,
I am not sure that I understood what you need, and
also I know that it is not a elegant solution, but may
do the job.
n <- 1991
work <- vector()
for(x in 1:n) {
work[x] <- sum(1:(n-x+1))
}
plot(work)
number.groups <- 5
last.i<-0
number.groups.list<-NULL
for (i in 1:(number.groups-1))
{
number.groups.list<-c(number.groups.list, rep(i,
round(length(work)/number.groups,0)))
}
number.groups.list<-c(number.groups.list, rep(number.groups,
(length(work)-length(number.groups.list)) ))
aggregate(work, list(number.groups.list), sum)
plot(work, col=number.groups.list)
Regards a lot,
miltinho
brazil
On Wed, Mar 25, 2009 at 9:48 PM, Nathan S. Watson-Haigh
<nathan.watson-ha...@csiro.au> wrote:
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>
> I have some data generated as follows:
>
> <code>
> n <- 2000
> work <- vector()
> for(x in 1:n) {
> work[x] <- sum(1:(n-x+1))
> }
> plot(work)
> </code>
>
> What I want to do
> - -----------------
> I want to split work into a number of unequal chunks such that the sum of
> the
> values in each chunk is approximately equal.
>
>
>
> The numbers in "work" are proportional to the amount of work to be
> performed for
> each value of x by a function I've written. i.e. For each value of x, there
> are
> work[x] * y calculations to be performed (where y is a constant).
>
> I've written a parallel version of my function where I simply assign z
> number of
> x values to each slave. This is not ideal, since a slave that gets the 1:z
> smallest values of x will take longer to compute than the (n-z+1):n set of
> x
> values. For example, if I have 4 slaves available:
>
> slave 1 processes x in 1:500
> slave 2 processes x in 501:1000
> slave 3 processes x in 1001:1500
> slave 4 processes x in 1501:2000
>
> This means the total work performed by each slave is:
>
> slave 1 sum(work[1:500]) = 771708500
> slave 2 sum(work[501:1000]) = 396458500
> slave 3 sum(work[1001:1500]) = 146208500
> slave 4 sum(work[1501:2000]) = 20958500
>
> Manually plitting work into chunks where the sum of the values for the
> chunks is
> approximately equal, I get the following:
>
> sum(work[1:184])
> [1] 335533384
> > sum(work[185:415])
> [1] 334897871
> > sum(work[416:745])
> [1] 334672085
> > sum(work[746:2000])
> [1] 330230660
>
> I need to be able to do this automatically for any value of n and I think I
> should be able to do this by calculating the area under the curve and
> slicing it
> into equally sized regions, but don't really know how to get there from
> what
> I've said above!
>
> Cheers,
> Nathan
>
> - --
> - --------------------------------------------------------
> Dr. Nathan S. Watson-Haigh
> OCE Post Doctoral Fellow
> CSIRO Livestock Industries
> Queensland Bioscience Precinct
> St Lucia, QLD 4067
> Australia
>
> Tel: +61 (0)7 3214 2922
> Fax: +61 (0)7 3214 2900
> Web: http://www.csiro.au/people/Nathan.Watson-Haigh.html
> - --------------------------------------------------------
>
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