How to access the last value in a vector

Question

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

User · Answer

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.

User · Answer

The xts package provides a last function   library xts  a  lt - 1 100 last a   1  100

User · Answer

I have another method for finding the last element in a vector  Say the vector is a    gt  a lt -c 1 100 555   gt  end a        Gives indices of last and first positions  1  101   1  gt  a end a  1      Gives last element in a vector  1  555   There you go

User · Answer

Another way is to take the first element of the reversed vector   rev dat vect1 vec2  1

User · Answer

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  lt - 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

User · Answer

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

User · Answer

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  lt - 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

User · Answer

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  lt - function x    return  x length x         This missing feature in R annoys me too

User · Answer

I have another method for finding the last element in a vector  Say the vector is a    gt  a lt -c 1 100 555   gt  end a        Gives indices of last and first positions  1  101   1  gt  a end a  1      Gives last element in a vector  1  555   There you go

User · Answer

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  lt - function x    return  x length x         This missing feature in R annoys me too

User · Answer

Package data table includes last function  library data table  last c 1 10      1  10

User · Answer

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

User · Answer

Whats about   gt  a  lt - c 1 100 555   gt  a NROW a    1  555

User · Answer

Combining lindelof s and Gregg Lind s ideas   last  lt - 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  lt - function x    head x  n   -1       Note that you have to use head for this  instead of tail

User · Answer

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

User · Answer

Package data table includes last function  library data table  last c 1 10      1  10

User · Answer

Whats about   gt  a  lt - c 1 100 555   gt  a NROW a    1  555

User · Answer

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  lt - function x    return  x length x         This missing feature in R annoys me too

User · Answer

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  lt - function x    return  x length x         This missing feature in R annoys me too

User · Answer

Combining lindelof s and Gregg Lind s ideas   last  lt - 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  lt - function x    head x  n   -1       Note that you have to use head for this  instead of tail

User · Answer

The xts package provides a last function   library xts  a  lt - 1 100 last a   1  100

User · Answer

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

User · Answer

The dplyr package includes a function last     last mtcars mpg     1  21 4

User · Answer

Another way is to take the first element of the reversed vector   rev dat vect1 vec2  1

User · Answer

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.

User · Answer

The dplyr package includes a function last     last mtcars mpg     1  21 4

[r] How to access the last value in a vector?

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