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How to access the last value in a vector?

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)] 
like image 602
user14008 Avatar asked Sep 16 '08 21:09

user14008


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2 Answers

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.

like image 83
lindelof Avatar answered Oct 17 '22 01:10

lindelof


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)].

like image 29
anonymous Avatar answered Oct 17 '22 01:10

anonymous