[R] Sorting in trees problem
bgunter.4567 at gmail.com
Wed Feb 24 17:57:26 CET 2016
"%in%" is a wrapper for match(), which uses hashing I believe
(correction welcome!),and so is generally very fast.
See ?Rprof for profiling R code to get timings. (Haven't used it
myself, so not sure how useful it would be for this situation).
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Feb 24, 2016 at 8:45 AM, Axel Urbiz <axel.urbiz at gmail.com> wrote:
> As decision trees require sorting the variable used for splitting a given
> node, I'm trying to avoid having this recurrent sorting by only sorting all
> numeric variable first (and only once).
> My attempt in doing this is shown in "Solution 2" below, but although I get
> the desired result I think the %in% operation may be a costly one (and may
> even offset the benefits of pre-sorting).
> Any alternative solutions would be highly appreciated.
> ### Sample data
> df <- data.frame(x1 = rnorm(20), x2 = rnorm(20))
> w <- rep(1L, nrow(df))
> # w == 1L denote observation present in the current node
> w[c(1, 8, 10)] <- 0L
> ### The problem: sort x1 within observations present in the current node
> ### Solution 1: slow for repeated sorting
> nodeObsInd <- which(w == 1L)
> sol1 <- df[nodeObsInd, ]
> sol1 <- sol1[order(sol1$x1), ]$x1
> ### Solution 2: sort all variables initially only.
> sort_fun <- function(x)
> index <- order(x)
> x <- x[index]
> data.frame(x, index) # the index gives original position of the obs
> s_df <- lapply(df, function(x) sort_fun(x))
> sol2 <- s_df[][s_df$x1$index %in% nodeObsInd, ]
> ### check same result
> all.equal(sol1, sol2$x)
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