Remove all duplicate rows including the “reference” row [duplicate]

Answer 1

Here's one way:

a[!(duplicated(a) | rev(duplicated(rev(a))))]
# [1] 1 2 4 5 6 8

Answer 2

I asked myself the same question (and i needed to do it quickly), so i came up with these solutions :

u =sample(x=1:10E6, size = 1000000, replace=T)
s1 <- function() setdiff(u, u[duplicated(u)])
s2 <- function() u[!duplicated(u) & !duplicated(u, fromLast=T)]
s3 <- function() u[!(duplicated(u) | rev(duplicated(rev(u))))]
s4 <- function() u[!u %in% u[duplicated(u)]]
s5 <- function() u[!match(u, u[duplicated(u)], nomatch = 0)]
s6 <- function() u[!is.element(u, u[duplicated(u)])]
s7 <- function() u[!duplicated2(u)]
library(rbenchmark)
benchmark(s1(), s2(), s3(), s4(), s5(), s6(), s7(),
          replications = 10,
          columns = c("test", "elapsed", "relative"),
          order = "elapsed")
     test elapsed relative
5 s5()    1.95    1.000
4 s4()    1.98    1.015
6 s6()    1.98    1.015
2 s2()    2.49    1.277
3 s3()    2.92    1.497
7 s7()    3.04    1.559
1 s1()    3.06    1.569

The choice is yours.

Answer 3

here is a solution to find the duplicated occurences with and their "original" occurences (and not only the duplicated occurences as with duplicated).

duplicated2 <- function(x){
  dup <- duplicated(x)
  if (sum(dup) == 0)
    return(dup)
  duplicated(c(x[dup], x))[-(1:sum(dup))]
}


a <- c(1,2,3,3,4,5,6,7,7,8)

> a[!duplicated2(a)]
[1] 1 2 4 5 6 8