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)]
To get the last element in an iterator loop you can use std::next() (from C++11). The loop is generally terminated by iterator != container. end() , where end() returns an iterator that points to the past-the-end element.
To get the last value of a vector , you can simply call std::vector::back() .
First of all, create a list. Then, use tail function with sapply function to extract the last value of all elements in the list.
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.
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)
, where mylast
is 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)]
.
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