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celt
2008 Jan 07
3
Seeking a more efficient way to find partition maxima
..., 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!
-- TMK --212-460-5430 home917-656-5351 cellt o p k a t z @ m s n . c o m
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