How to access the last value in a vector?
RDataframeVectorR Problem Overview
Suppose I have a vector that is nested in a dataframe one or two levels. Is there a quick and dirty way to access the last value, without using the length() function? Something ala PERL's $# special var?
So I would like something like:
dat$vec1$vec2[$#]
instead of
dat$vec1$vec2[length(dat$vec1$vec2)]
R Solutions
Solution 1 - R
I use the tail function:
tail(vector, n=1)
The nice thing with tail is that it works on dataframes too, unlike the x[length(x)] idiom.
Solution 2 - R
To answer this not from an aesthetical but performance-oriented point of view, I've put all of the above suggestions through a benchmark. To be precise, I've considered the suggestions
x[length(x)]mylast(x), wheremylastis a C++ function implemented through Rcpp,tail(x, n=1)dplyr::last(x)x[end(x)[1]]]rev(x)[1]
and applied them to random vectors of various sizes (10^3, 10^4, 10^5, 10^6, and 10^7). Before we look at the numbers, I think it should be clear that anything that becomes noticeably slower with greater input size (i.e., anything that is not O(1)) is not an option. Here's the code that I used:
Rcpp::cppFunction('double mylast(NumericVector x) { int n = x.size(); return x[n-1]; }')
options(width=100)
for (n in c(1e3,1e4,1e5,1e6,1e7)) {
x <- runif(n);
print(microbenchmark::microbenchmark(x[length(x)],
mylast(x),
tail(x, n=1),
dplyr::last(x),
x[end(x)[1]],
rev(x)[1]))}
It gives me
Unit: nanoseconds
expr min lq mean median uq max neval
x[length(x)] 171 291.5 388.91 337.5 390.0 3233 100
mylast(x) 1291 1832.0 2329.11 2063.0 2276.0 19053 100
tail(x, n = 1) 7718 9589.5 11236.27 10683.0 12149.0 32711 100
dplyr::last(x) 16341 19049.5 22080.23 21673.0 23485.5 70047 100
x[end(x)[1]] 7688 10434.0 13288.05 11889.5 13166.5 78536 100
rev(x)[1] 7829 8951.5 10995.59 9883.0 10890.0 45763 100
Unit: nanoseconds
expr min lq mean median uq max neval
x[length(x)] 204 323.0 475.76 386.5 459.5 6029 100
mylast(x) 1469 2102.5 2708.50 2462.0 2995.0 9723 100
tail(x, n = 1) 7671 9504.5 12470.82 10986.5 12748.0 62320 100
dplyr::last(x) 15703 19933.5 26352.66 22469.5 25356.5 126314 100
x[end(x)[1]] 13766 18800.5 27137.17 21677.5 26207.5 95982 100
rev(x)[1] 52785 58624.0 78640.93 60213.0 72778.0 851113 100
Unit: nanoseconds
expr min lq mean median uq max neval
x[length(x)] 214 346.0 583.40 529.5 720.0 1512 100
mylast(x) 1393 2126.0 4872.60 4905.5 7338.0 9806 100
tail(x, n = 1) 8343 10384.0 19558.05 18121.0 25417.0 69608 100
dplyr::last(x) 16065 22960.0 36671.13 37212.0 48071.5 75946 100
x[end(x)[1]] 360176 404965.5 432528.84 424798.0 450996.0 710501 100
rev(x)[1] 1060547 1140149.0 1189297.38 1180997.5 1225849.0 1383479 100
Unit: nanoseconds
expr min lq mean median uq max neval
x[length(x)] 327 584.0 1150.75 996.5 1652.5 3974 100
mylast(x) 2060 3128.5 7541.51 8899.0 9958.0 16175 100
tail(x, n = 1) 10484 16936.0 30250.11 34030.0 39355.0 52689 100
dplyr::last(x) 19133 47444.5 55280.09 61205.5 66312.5 105851 100
x[end(x)[1]] 1110956 2298408.0 3670360.45 2334753.0 4475915.0 19235341 100
rev(x)[1] 6536063 7969103.0 11004418.46 9973664.5 12340089.5 28447454 100
Unit: nanoseconds
expr min lq mean median uq max neval
x[length(x)] 327 722.0 1644.16 1133.5 2055.5 13724 100
mylast(x) 1962 3727.5 9578.21 9951.5 12887.5 41773 100
tail(x, n = 1) 9829 21038.0 36623.67 43710.0 48883.0 66289 100
dplyr::last(x) 21832 35269.0 60523.40 63726.0 75539.5 200064 100
x[end(x)[1]] 21008128 23004594.5 37356132.43 30006737.0 47839917.0 105430564 100
rev(x)[1] 74317382 92985054.0 108618154.55 102328667.5 112443834.0 187925942 100
This immediately rules out anything involving rev or end since they're clearly not O(1) (and the resulting expressions are evaluated in a non-lazy fashion). tail and dplyr::last are not far from being O(1) but they're also considerably slower than mylast(x) and x[length(x)]. Since mylast(x) is slower than x[length(x)] and provides no benefits (rather, it's custom and does not handle an empty vector gracefully), I think the answer is clear: Please use x[length(x)].
Solution 3 - R
If you're looking for something as nice as Python's x[-1] notation, I think you're out of luck. The standard idiom is
x[length(x)]
but it's easy enough to write a function to do this:
last <- function(x) { return( x[length(x)] ) }
This missing feature in R annoys me too!
Solution 4 - R
Combining lindelof's and Gregg Lind's ideas:
last <- function(x) { tail(x, n = 1) }
Working at the prompt, I usually omit the n=, i.e. tail(x, 1).
Unlike last from the pastecs package, head and tail (from utils) work not only on vectors but also on data frames etc., and also can return data "without first/last n elements", e.g.
but.last <- function(x) { head(x, n = -1) }
(Note that you have to use head for this, instead of tail.)
Solution 5 - R
The dplyr package includes a function last():
last(mtcars$mpg)
# [1] 21.4
Solution 6 - R
I just benchmarked these two approaches on data frame with 663,552 rows using the following code:
system.time(
resultsByLevel$subject <- sapply(resultsByLevel$variable, function(x) {
s <- strsplit(x, ".", fixed=TRUE)[[1]]
s[length(s)]
})
)
user system elapsed
3.722 0.000 3.594
and
system.time(
resultsByLevel$subject <- sapply(resultsByLevel$variable, function(x) {
s <- strsplit(x, ".", fixed=TRUE)[[1]]
tail(s, n=1)
})
)
user system elapsed
28.174 0.000 27.662
So, assuming you're working with vectors, accessing the length position is significantly faster.
Solution 7 - R
Another way is to take the first element of the reversed vector:
rev(dat$vect1$vec2)[1]
Solution 8 - R
I have another method for finding the last element in a vector.
Say the vector is a.
> a<-c(1:100,555)
> end(a) #Gives indices of last and first positions
[1] 101 1
> a[end(a)[1]] #Gives last element in a vector
[1] 555
There you go!
Solution 9 - R
Package data.table includes last function
library(data.table)
last(c(1:10))
# [1] 10
Solution 10 - R
Whats about
> a <- c(1:100,555)
> a[NROW(a)]
[1] 555
Solution 11 - R
The xts package provides a last function:
library(xts)
a <- 1:100
last(a)
[1] 100