Fastest way to find *the index* of the second (third...) highest/lowest value in vector or column

Fastest way to find the index of the second (third...) highest/lowest value in vector or column ?

i.e. what

sort(x,partial=n-1)[n-1]
开发者_如何学Go

is to

max()

but for

which.max()

Best,

Fastest way to find second (third...) highest/lowest value in vector or column

One possible route is to use the index.return argument to sort. I'm not sure if this is fastest though.

set.seed(21)
x <- rnorm(10)
ind <- 2
sapply(sort(x, index.return=TRUE), `[`, length(x)-ind+1)
#        x       ix 
# 1.746222 3.000000

EDIT 2 :

As Joshua pointed out, none of the given solutions actually performs correct when you have a tie on the maxima, so :

X <- c(11:19,19)

n <- length(unique(X))
which(X == sort(unique(X),partial=n-1)[n-1])

fastest way of doing it correctly then. I deleted the order way, as that one doesn't work and is a lot slower, so not a good answer according to OP.

To point to the issue we ran into :

> X <- c(11:19,19)    
> n <- length(X)
> which(X == sort(X,partial=n-1)[n-1])
[1]  9 10 #which is the indices of the double maximum 19

> n <- length(unique(X))
> which(X == sort(unique(X),partial=n-1)[n-1])
[1] 8 # which is the correct index of 18

The timings of the valid solutions :

> x <- runif(1000000)

> ind <- 2

> n <- length(unique(x))

> system.time(which(x == sort(unique(x),partial=n-ind+1)[n-ind+1]))
   user  system elapsed 
   0.11    0.00    0.11 

> system.time(sapply(sort(unique(x), index.return=TRUE), `[`, n-ind+1))
   user  system elapsed 
   0.69    0.00    0.69 

library Rfast has implemented the nth element function with return index option.

UPDATE (28/FEB/21) package kit offers a faster implementation (topn) as shown in the simulations below.

x <- runif(1e+6)

n <- 2

which_nth_highest_richie <- function(x, n)
{
  for(i in seq_len(n - 1L)) x[x == max(x)] <- -Inf
  which(x == max(x))
}

which_nth_highest_joris <- function(x, n)
{
  ux <- unique(x)
  nux <- length(ux)
  which(x == sort(ux, partial = nux - n + 1)[nux - n + 1])
} 

microbenchmark::microbenchmark(
        topn = kit::topn(x, n,decreasing = T)[n],
        Rfast = Rfast::nth(x,n,descending = T,index.return = T),
        order = order(x, decreasing = TRUE)[n],
        richie = which_nth_highest_richie(x,n),
        joris = which_nth_highest_joris(x,n))

Unit: milliseconds
          expr    min     lq      mean     median       uq      max   neval 
          topn  3.741101 3.7917  4.517201  4.060752  5.108901 7.403901 100
         Rfast  15.8121 16.7586  20.64204  17.73010  20.7083  47.6832  100
         order 110.5416 113.4774 120.45807 116.84005 121.2291 164.5618 100
        richie  22.7846 24.1552  39.35303  27.10075  42.0132  179.289  100
         joris 131.7838 140.4611 158.20704 156.61610 165.1735 243.9258 100

Topn is the clear winner in finding the index of the 2nd biggest value in 1 million numbers.

Futher, simulations where run to estimate running times of finding the nth biggest number for varying n. Variable x was repopulated for each n but it's size was always 1 million numbers.

As shown topn is the best option for finding the nth biggest element and it's index, given that n is not too big. In the plot we can observe that topn becomes slower than Rfast's nth for bigger n. It is worthy to note that topn has not been implemented for n > 1000 and will throw an error in such cases.

Method: Set all max values to -Inf, then find the indices of the max. No sorting required.

X <- runif(1e7)
system.time(
{
  X[X == max(X)] <- -Inf
  which(X == max(X))
})

Works with ties and is very fast.

If you can guarantee no ties, then an even faster version is

system.time(
{
  X[which.max(X)] <- -Inf
  which.max(X)
})

EDIT: As Joris mentioned, this method doesn't scale that well for finding third, fourth, etc., highest values.

which_nth_highest_richie <- function(x, n)
{
  for(i in seq_len(n - 1L)) x[x == max(x)] <- -Inf
  which(x == max(x))
}

which_nth_highest_joris <- function(x, n)
{
  ux <- unique(x)
  nux <- length(ux)
  which(x == sort(ux, partial = nux - n + 1)[nux - n + 1])
}

Using x <- runif(1e7) and n = 2, Richie wins

system.time(which_nth_highest_richie(x, 2))   #about half a second
system.time(which_nth_highest_joris(x, 2))    #about 2 seconds

For n = 100, Joris wins

system.time(which_nth_highest_richie(x, 100)) #about 20 seconds, ouch! 
system.time(which_nth_highest_joris(x, 100))  #still about 2 seconds

The balance point, where they take the same length of time, is about n = 10.

No ties which() is probably your friend here. Combine the output from the sort() solution with which() to find the index that matches the output from the sort() step.

> set.seed(1)
> x <- sample(1000, 250)
> sort(x,partial=n-1)[n-1]
[1] 992
> which(x == sort(x,partial=n-1)[n-1])
[1] 145

Ties handling The solution above doesn't work properly (and wasn't intended to) if there are ties and the ties are the values that are the ith largest or larger values. We need to take the unique values of the vector before sorting those values and then the above solution works:

> set.seed(1)
> x <- sample(1000, 1000, replace = TRUE)
> length(unique(x))
[1] 639
> n <- length(x)
> i <- which(x == sort(x,partial=n-1)[n-1])
> sum(x > x[i])
[1] 0
> x.uni <- unique(x)
> n.uni <- length(x.uni)
> i <- which(x == sort(x.uni, partial = n.uni-1)[n.uni-1])
> sum(x > x[i])
[1] 2
> tail(sort(x))
[1]  994  996  997  997 1000 1000

order() is also very useful here:

> head(ord <- order(x, decreasing = TRUE))
[1] 220 145 209 202 211 163

So the solution here is ord[2] for the index of the 2nd highest/largest element of x.

Some timings:

> set.seed(1)
> X <- sample(1e7, 1e7)
> system.time({n <- length(X); which(X == sort(X, partial = n-1)[n-1])})
   user  system elapsed 
  0.319   0.058   0.378 
> system.time({ord <- order(X, decreasing = TRUE); ord[2]})
   user  system elapsed 
 14.578   0.084  14.708 
> system.time({order(X, decreasing = TRUE)[2]})
   user  system elapsed 
 14.647   0.084  14.779

But as the linked post was getting at and the timings above show, order() is much slower, but both provide the same results:

> all.equal(which(X == sort(X, partial = n-1)[n-1]), 
+           order(X, decreasing = TRUE)[2])
[1] TRUE

And for the ties-handling version:

foo <- function(x, i) {
    X <- unique(x)
    N <- length(X)
    i <- i-1
    which(x == sort(X, partial = N-i)[N-i])
}

> system.time(foo(X, 2))
   user  system elapsed 
  1.249   0.176   1.454

So the extra steps slow this solution down a bit, but it is still very competitive with order().

Use maxN function given by Zach to find the next max value and use which() with arr.ind = TRUE.

which(x == maxN(x, 4), arr.ind = TRUE)

Using arr.ind will return index position in any of the above solutions as well and simplify the code.

This is my solution for finding the index of the top N highest values in a vector (not exactly what the OP wanted, but this might help other people)

index.top.N = function(xs, N=10){
    if(length(xs) > 0) {
    o = order(xs, na.last=FALSE)
    o.length = length(o)
    if (N > o.length) N = o.length
    o[((o.length-N+1):o.length)]
  }
  else {
    0
  }
}