Getting the last n elements of a vector. Is there a better way than using the length() function?

Question

If, for argument's sake, I want the last five elements of a 10-length vector in Python, I can use the - operator in the range index like so:

>>> x = range(10) >>> x [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> x[-5:] [5, 6, 7, 8, 9] >>> 

What is the best way to do this in R? Is there a cleaner way than my current technique, which is to use the length() function?

> x <- 0:9 > x  [1] 0 1 2 3 4 5 6 7 8 9 > x[(length(x) - 4):length(x)] [1] 5 6 7 8 9 >  

The question is related to time series analysis btw where it is often useful to work only on recent data.

Answer 1

see ?tail and ?head for some convenient functions:

> x <- 1:10 > tail(x,5) [1]  6  7  8  9 10 

For the argument's sake : everything but the last five elements would be :

> head(x,n=-5) [1] 1 2 3 4 5 

As @Martin Morgan says in the comments, there are two other possibilities which are faster than the tail solution, in case you have to carry this out a million times on a vector of 100 million values. For readibility, I'd go with tail.

test                                        elapsed    relative  tail(x, 5)                                    38.70     5.724852      x[length(x) - (4:0)]                           6.76     1.000000      x[seq.int(to = length(x), length.out = 5)]     7.53     1.113905      

benchmarking code :

require(rbenchmark) x <- 1:1e8 do.call(   benchmark,   c(list(     expression(tail(x,5)),     expression(x[seq.int(to=length(x), length.out=5)]),     expression(x[length(x)-(4:0)])   ),  replications=1e6) )