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I have a vector and I'm able to return highest and lowest value, but how to return 5 topmost values? Is there a simple one-line solution for this?

pixel
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4 Answers4

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> a <- c(1:100)
> tail(sort(a),5)
[1]  96  97  98  99 100
chrisamiller
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    Or `head(sort(a, decreasing=TRUE), 5)` – Marek Sep 11 '10 at 21:35
  • tail is slightly faster than head and decreasing = TRUE > x <- rnorm(50000000) > system.time(tail(sort(x), 5)) user system elapsed 22.64 0.25 22.95 > system.time(head(sort(x, decreasing = TRUE), 5)) user system elapsed 23.26 0.20 23.51 – Thierry Sep 11 '10 at 22:28
  • @Thierry You should run this more then once and take average time. Cause I think there is no difference (statistically speaking), based on my simulations. – Marek Sep 13 '10 at 08:06
  • I get on average 2% faster times for user.self and elapse. The gain on sys.self is 8%. But the relevance on the gain depends on the application. – Thierry Sep 19 '10 at 21:59
  • Using sort(x, method='quick') is significantly faster, but David's solution below using the partial argument is even faster. – Tommy Apr 08 '11 at 01:48
25
x[order(x)[1:5]]
hadley
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4

Yes, head( X, 5) where X is your sorted vector.

Dirk Eddelbuettel
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3
tail(sort.int(x, partial=length(x) - 4), 5)

Using sort.int with partial has the advantage of being (potentially) faster by (potentially) not doing a full sort. But in reality, my implementation appears a little slower. Maybe this is because with parameter partial != NULL, shell sort is used rather than quick sort?

> x <- 1:1e6
> system.time(replicate(100, tail(sort.int(x, partial=length(x) - 4), 5)))
   user  system elapsed 
  4.782   0.846   5.668
> system.time(replicate(100, tail(sort(x), 5)))
   user  system elapsed 
  3.643   0.879   4.854
David F
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  • If you instead use x<-runif(1e6) you'd see the benefit. Note that those 5 values you get back would indeed be the highest 5, but not necessarily in a sorted order. – Tommy Apr 08 '11 at 00:20