finding unique vector elements in a list efficient

问题:

I have a list of numerical vectors, and I need to create a list containing only one copy of each vector. There isn't a list method for the identical function, so I wrote a function to apply to check every vector against every other.

F1 <- function(x){

    to_remove <- c()
    for(i in 1:length(x)){
        for(j in 1:length(x)){
            if(i!=j && identical(x[[i]], x[[j]]) to_remove <- c(to_remove,j)
        }
    }
    if(is.null(to_remove)) x else x[-c(to_remove)] 
} 

The problem is that this function becomes very slow as the size of the input list x increases, partly due to the assignment of two large vectors by the for loops. I'm hoping for a method that will run in under one minute for a list of length 1.5 million with vectors of length 15, but that might be optimistic.

Does anyone know a more efficient way of comparing each vector in a list with every other vector? The vectors themselves are guaranteed to be equal in length.

Sample output is shown below.

x = list(1:4, 1:4, 2:5, 3:6)
F1(x)
> list(1:4, 2:5, 3:6)

回答1:

As per @JoshuaUlrich and @thelatemail, ll[!duplicated(ll)] works just fine.
And thus, so should unique(ll) I previously suggested a method using sapply with the idea of not checking every element in the list (I deleted that answer, as I think using unique makes more sense)

Since efficiency is a goal, we should benchmark these.

# Let's create some sample data
xx <- lapply(rep(100,15), sample)
ll <- as.list(sample(xx, 1000, T))
ll

Putting it up against some becnhmarks

fun1 <- function(ll) {
  ll[c(TRUE, !sapply(2:length(ll), function(i) ll[i] %in% ll[1:(i-1)]))]
}

fun2 <- function(ll) {
  ll[!duplicated(sapply(ll, digest))]
}

fun3 <- function(ll)  {
  ll[!duplicated(ll)]
}

fun4 <- function(ll)  {
  unique(ll)
}

#Make sure all the same
all(identical(fun1(ll), fun2(ll)), identical(fun2(ll), fun3(ll)), 
    identical(fun3(ll), fun4(ll)), identical(fun4(ll), fun1(ll)))
# [1] TRUE


library(rbenchmark)

benchmark(digest=fun2(ll), duplicated=fun3(ll), unique=fun4(ll), replications=100, order="relative")[, c(1, 3:6)]

        test elapsed relative user.self sys.self
3     unique   0.048    1.000     0.049    0.000
2 duplicated   0.050    1.042     0.050    0.000
1     digest   8.427  175.563     8.415    0.038
# I took out fun1, since when ll is large, it ran extremely slow

Fastest Option:

unique(ll)

回答2:

You could hash each of the vectors and then use !duplicated() to identify unique elements of the resultant character vector:

library(digest)  

## Some example data
x <- 1:44
y <- 2:10
z <- rnorm(10)
ll <- list(x,y,x,x,x,z,y)

ll[!duplicated(sapply(ll, digest))]
# [[1]]
#  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
# [26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
# 
# [[2]]
# [1]  2  3  4  5  6  7  8  9 10
# 
# [[3]]
#  [1]  1.24573610 -0.48894189 -0.18799758 -1.30696395 -0.05052373  0.94088670
#  [7] -0.20254574 -1.08275938 -0.32937153  0.49454570

To see at a glance why this works, here's what the hashes look like:

sapply(ll, digest)
[1] "efe1bc7b6eca82ad78ac732d6f1507e7" "fd61b0fff79f76586ad840c9c0f497d1"
[3] "efe1bc7b6eca82ad78ac732d6f1507e7" "efe1bc7b6eca82ad78ac732d6f1507e7"
[5] "efe1bc7b6eca82ad78ac732d6f1507e7" "592e2e533582b2bbaf0bb460e558d0a5"
[7] "fd61b0fff79f76586ad840c9c0f497d1"