How to expand a list with NULLs up to some length?

Question

Given a list whose length <= N, what is the best / most efficient way to fill it up with trailing NULLs up to length (so that it has length N).

This is something which is a one-liner in any decent language, but I don't have a clue how to do it (efficiently) in a few lines in R so that it works for every corner case (zero length list etc.).

score 9 · Accepted Answer · answered Nov 11 '11 at 18:18

9

Let's keep it really simple:

tst<-1:10 #whatever, to get a vector of length 10
tst<-tst[1:15]

answered Nov 11 '11 at 18:18

Nick Sabbe

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That's it. This solution is 2x faster than than the one proposed by Matthew Dowle, and about 5x faster then my clumsy solution. – eold Nov 11 '11 at 18:40
1

@leden. Yet another way is simply `length(tst)=15`. Any faster? How did you test? – Matt Dowle Nov 11 '11 at 20:14
Yep, this is even faster for large lists. Thanks. – eold Nov 11 '11 at 23:32
@RomanLuštrik : ahem. Fortunately, it also works for `tst<-list(1,2,3)`. But of course, a list is just a special case of a vector, so I thought I'd provide the more general solution. – Nick Sabbe Nov 17 '11 at 10:27

score 6 · Answer 2 · answered Nov 11 '11 at 18:22

6

Try this :

> l = list("a",1:3)
> N = 5
> l[N+1]=NULL
> l
[[1]]
[1] "a"

[[2]]
[1] 1 2 3

[[3]]
NULL

[[4]]
NULL

[[5]]
NULL

>

answered Nov 11 '11 at 18:22

Matt Dowle

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score 6 · Answer 3 · answered Nov 11 '11 at 20:16

6

How about this ?

> l = list("a",1:3)
> length(l)=5
> l
[[1]]
[1] "a"

[[2]]
[1] 1 2 3

[[3]]
NULL

[[4]]
NULL

[[5]]
NULL

answered Nov 11 '11 at 20:16

Matt Dowle

58,872
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166
224

score 4 · Answer 4 · answered Nov 11 '11 at 22:32

Directly editing the list's length appears to be the fastest as far as I can tell:

tmp <- vector("list",5000)
sol1 <- function(x){
    x <- x[1:10000] 
}
sol2 <- function(x){
    x[10001] <- NULL
}
sol3 <- function(x){
    length(x) <- 10000
}

library(rbenchmark)
benchmark(sol1(tmp),sol2(tmp),sol3(tmp),replications = 5000)
       test replications elapsed relative user.self sys.self user.child sys.child
1 sol1(tmp)         5000   2.045 1.394952     1.327    0.727          0         0
2 sol2(tmp)         5000   2.849 1.943383     1.804    1.075          0         0
3 sol3(tmp)         5000   1.466 1.000000     0.937    0.548          0         0

But the differences aren't huge, unless you're doing this a lot on very long lists, I suppose.

score 1 · Answer 5 · answered Nov 11 '11 at 18:14

1

I'm sure there are shorter ways, but I would be inclined to do:

l <- as.list(1:10)
N <- 15
l <- c(l, as.list(rep(NA, N - length(l) )))

answered Nov 11 '11 at 18:14

JD Long

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score 0 · Answer 6 · edited Nov 11 '11 at 21:35

0

Hi: I'm not sure if you were talking about an actual list but, if you were, below will work. It works because, once you access the element of a vector ( which is a list is ) that is not there, R expands the vector to that length.

length <- 10
temp <- list("a","b")
print(temp)
temp[length] <- NULL
print(temp)

edited Nov 11 '11 at 21:35

Brian Diggs

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answered Nov 11 '11 at 18:17

mark leeds

131
3

How to expand a list with NULLs up to some length?

6 Answers6