# [R] inserting elements in a list

Ray Brownrigg ray at mcs.vuw.ac.nz
Thu Feb 20 03:13:02 CET 2003

```> From: Peter Wolf <pwolf at wiwi.uni-bielefeld.de>
>
> "insert.values" <- function(x,pos.insert,x.insert){
> :

Which when reduced to its bare minimum to solve the stated problem,
becomes:
threes <- which(a==3)
a <- c(a, rep(7, length(threes)))[order(c(seq(a), threes))]

This is by far the overall fastest solution so far proposed, and
certainly is reasonably easy to understand.  However an even faster
solution is based on rep():

N <- which(threes <- a == 3)		# which ones
a <- rep(a, times = 1 + threes)	# duplicate them
a[N + 1:length(N)] <- 7		# replace the duplicates

Relative timing for vector lengths n=1000 and n=10000 with 1000 repeats
(on a 1.7GHz P4 running R-1.6.1 under NetBSD) is as follows:

a <- sample(1:9, n, TRUE)			n=1000	n=10000
lapply (Stephen Upton and Peter Dalgaard)	  9.2s	 117.8s
recursion (John Fox) [options(expressions=n)]	 59.9s	 stack overflow
offset <- (Thomas Lumley)			  1.4s	 12.7s
offset <- (Peter Dalgaard)			  1.3s	 10.9s
order() (Peter Wolf)				  1.1s	  8.7s
rep() (me)					  0.9s	  7.6s

Hope this helps,
Ray Brownrigg <ray at mcs.vuw.ac.nz>	http://www.mcs.vuw.ac.nz/~ray

```