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 so    gt  gt  gt  x   range 10   gt  gt  gt  x  0  1  2  3  4  5  6  7  8  9   gt  gt  gt  x -5    5  6  7  8  9   gt  gt  gt    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    gt  x  lt - 0 9  gt  x   1  0 1 2 3 4 5 6 7 8 9  gt  x  length x  - 4  length x    1  5 6 7 8 9  gt     The question is related to time series analysis btw where it is often useful to work only on recent data

User · Answer

Here is a function to do it and seems reasonably fast.

endv<-function(vec,val) 
{
if(val>length(vec))
{
stop("Length of value greater than length of vector")
}else
{
vec[((length(vec)-val)+1):length(vec)]
}
}

USAGE:

test<-c(0,1,1,0,0,1,1,NA,1,1)
endv(test,5)
endv(LETTERS,5)

BENCHMARK:

                                                    test replications elapsed relative
1                                 expression(tail(x, 5))       100000    5.24    6.469
2 expression(x[seq.int(to = length(x), length.out = 5)])       100000    0.98    1.210
3                       expression(x[length(x) - (4:0)])       100000    0.81    1.000
4                                 expression(endv(x, 5))       100000    1.37    1.691

User · Answer

I just add here something related  I was wanted to access a vector with backend indices  ie writting something like tail x  i  but to return x length x  - i   1  and not the whole tail   Following commentaries I benchmarked two solutions   accessRevTail  lt - function x  n        tail x n  1     accessRevLen  lt - function x  n      x length x  - n   1     microbenchmark  microbenchmark accessRevLen 1 100  87   accessRevTail 1 100  87   Unit  microseconds                      expr    min      lq     mean median      uq     max neval   accessRevLen 1 100  87   1 860  2 3775  2 84976  2 803  3 2740   6 755   100  accessRevTail 1 100  87  22 214 23 5295 28 54027 25 112 28 4705 110 833   100   So it appears in this case that even for small vectors  tail is very slow comparing to direct access

User · Answer

see  tail and  head for some convenient functions    gt  x  lt - 1 10  gt  tail x 5   1   6  7  8  9 10   For the argument s sake   everything but the last five elements would be     gt  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  lt - 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

User · Answer

How about rev x  1 5    x lt -1 10 system time replicate 10e6 tail x 5     user  system elapsed   138 85    0 26  139 28   system time replicate 10e6 rev x  1 5     user  system elapsed   61 97    0 25   62 23

User · Answer

You can do exactly the same thing in R with two more characters   x  lt - 0 9 x -5 -1   1  5 6 7 8 9   or  x - 1 5

User · Answer

The disapproval of tail here based on speed alone doesn t really seem to emphasize that part of the slower speed comes from the fact that tail is safer to work with  if you don t for sure that the length of x will exceed n  the number of elements you want to subset out   x  lt - 1 10 tail x  20     1   1  2  3  4  5  6  7  8  9 10 x length x  -  0 19    Error in x length x  -  0 19         only 0 s may be mixed with negative subscripts   Tail will simply return the max number of elements instead of generating an error  so you don t need to do any error checking yourself   A great reason to use it   Safer cleaner code  if extra microseconds milliseconds don t matter much to you in its use

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

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