Re: [R] Seeking a more efficient way to find partition maxima

From: Marc Schwartz <marc_schwartz_at_comcast.net>
Date: Mon, 07 Jan 2008 10:44:20 -0600

Talbot Katz wrote:
> Hi.
>
> Suppose I have a vector that I partition into disjoint, contiguous
> subvectors. For example, let v = c(1,4,2,6,7,5), partition it into
> three subvectors, v1 = v[1:3], v2 = v[4], v3 = v[5:6]. I want to
> find the maximum element of each subvector. In this example, max(v1)
> is 4, max(v2) is 6, max(v3) is 7. If I knew that the successive
> subvector maxima would never decrease, as in the example, I could do
> the following:
>
> partiCmax <- function( values, seriesIdx ) { # assume seriesIdx is
> increasing integer sequence beginning with 1, ending at less than or
> equal to length(values) parti <- cbind( seriesIdx, c( ( seriesIdx[ -1
> ] - 1 ), length( values ) ) ) return( cummax( values )[ parti[ , 2 ]
> ] ) }
>
>
> The use of cummax makes that pretty efficient, but if the subvector
> maxima are not non-decreasing, it doesn't work. The following
> function works (at least it did on the examples I tried):
>
> partiMax <- function( values, seriesIdx ) { # assume seriesIdx is
> increasing integer sequence beginning with 1, ending at less than or
> equal to length(values) parti <- cbind( seriesIdx, c( ( seriesIdx[ -1
> ] - 1 ), length( values ) ) ) return( sapply( ( 1:length(seriesIdx)
> ), function ( i ) {return( max( values[ parti[ i, 1 ]:parti[ i, 2 ] ]
> ) ) } ) ) }
>
>
> but I figured someone out there could come up with something
> cleverer. Thanks!

It is not clear how you are creating the partitions, but if you are (hopefully) ending up with a list, such as:

 > Vec.List
$v1
[1] 1 4 2

$v2
[1] 6

$v3
[1] 7 5

Then you can use:

 > sapply(Vec.List, max, na.rm = TRUE)
v1 v2 v3
  4 6 7

See ?sapply for more information.

HTH, Marc Schwartz



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