Is there a good reason to use `sort` with `index.return = TRUE` instead of `order`?

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sort has the argument index.return which is by default FALSE. If you set it to TRUE you get the ordering index... basically the same as when you use order.

My question

Are there cases where it makes sense to use sort with index.return = TRUE instead of order?

asked Nov 16 '18 at 12:46

vonjd

2,13732844

3

FWIW sort calls this x[order(x, na.last = na.last, decreasing = decreasing)] for non-integer inputs.

– hrbrmstr
Nov 16 '18 at 12:58

add a comment |

sort has the argument index.return which is by default FALSE. If you set it to TRUE you get the ordering index... basically the same as when you use order.

My question

Are there cases where it makes sense to use sort with index.return = TRUE instead of order?

asked Nov 16 '18 at 12:46

vonjd

2,13732844

3

FWIW sort calls this x[order(x, na.last = na.last, decreasing = decreasing)] for non-integer inputs.

– hrbrmstr
Nov 16 '18 at 12:58

add a comment |

sort has the argument index.return which is by default FALSE. If you set it to TRUE you get the ordering index... basically the same as when you use order.

My question

Are there cases where it makes sense to use sort with index.return = TRUE instead of order?

asked Nov 16 '18 at 12:46

vonjd

2,13732844

sort has the argument index.return which is by default FALSE. If you set it to TRUE you get the ordering index... basically the same as when you use order.

My question

Are there cases where it makes sense to use sort with index.return = TRUE instead of order?

r sorting

asked Nov 16 '18 at 12:46

vonjd

2,13732844

asked Nov 16 '18 at 12:46

vonjd

2,13732844

asked Nov 16 '18 at 12:46

vonjd

2,13732844

asked Nov 16 '18 at 12:46

vonjd

2,13732844

asked Nov 16 '18 at 12:46

vonjd

2,13732844

3

FWIW sort calls this x[order(x, na.last = na.last, decreasing = decreasing)] for non-integer inputs.

– hrbrmstr
Nov 16 '18 at 12:58

add a comment |

3

FWIW sort calls this x[order(x, na.last = na.last, decreasing = decreasing)] for non-integer inputs.

– hrbrmstr
Nov 16 '18 at 12:58

FWIW sort calls this x[order(x, na.last = na.last, decreasing = decreasing)] for non-integer inputs.

– hrbrmstr
Nov 16 '18 at 12:58

add a comment |

2 Answers
2

active

oldest

votes

order simply gives the indexes, instead sort gives also the values (and with index.return=T a list):

x <- runif(10, 0, 100)

order(x)

# [1]  2  7  1  9  6  5  8 10  4  3

sort(x, index.return=T)

# $`x`

# [1] 0.08140348 0.18272011 0.23575252 0.51493537 0.64281259 0.92121388 0.93759670 0.96221375 0.97646916 0.97863369

# 

# $ix

# [1]  2  7  1  9  6  5  8 10  4  3

It seems that order is a little faster with big numbers (longer vector size):

x <- runif(10000000, 0, 100)



microbenchmark::microbenchmark(

  sort = {sort(x, index.return=T)},

  order = {x[order(x)]},

  times = 100

)

# Unit: milliseconds

# expr      min       lq     mean   median       uq      max neval

# sort 63.48221 67.79530 78.33724 70.74215 74.10109 173.1129   100

# order 56.46055 57.18649 60.88239 58.29462 62.13086 155.5481   100

So probably you should pick sort with index.return = TRUE only if you need a list object to be returned. I can't find an example where sort is better than the other.

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

1

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

add a comment |

My suggestions are based on RLave's answer.

You could use the argument method, sort(x,method="quick",index.return=TRUE), and the function might be a little faster than the default. Also if you want a faster (for large vectors) alternative method of this, you can use this function:

sort_order <- function(x){

    indices <- order(x) #you can choose a method also but leave default.

    list("x"=x[indices],"ix"=indices)

}

Here are some benchmarks.

microbenchmark::microbenchmark(

     sort=s<-sort(x,index.return=T),

     "quick sort"=sq<-sort(x,method="quick",index.return=T),

     "order sort"=so<-sort_order(x),times = 10

     times=10

)



Unit: seconds

         expr      min       lq     mean   median       uq      max neval

         sort 1.493714 1.662791 1.737854 1.708502 1.887993 1.960912    10

   quick sort 1.366938 1.374874 1.451778 1.444342 1.480122 1.668693    10

   order sort 1.181974 1.344398 1.359209 1.369108 1.424569 1.461862    10



all.equal(so,sq)

[1] TRUE

all.equal(s,so)

[1] TRUE

answered Nov 16 '18 at 17:33

Csd

31819

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

order simply gives the indexes, instead sort gives also the values (and with index.return=T a list):

x <- runif(10, 0, 100)

order(x)

# [1]  2  7  1  9  6  5  8 10  4  3

sort(x, index.return=T)

# $`x`

# [1] 0.08140348 0.18272011 0.23575252 0.51493537 0.64281259 0.92121388 0.93759670 0.96221375 0.97646916 0.97863369

# 

# $ix

# [1]  2  7  1  9  6  5  8 10  4  3

It seems that order is a little faster with big numbers (longer vector size):

x <- runif(10000000, 0, 100)



microbenchmark::microbenchmark(

  sort = {sort(x, index.return=T)},

  order = {x[order(x)]},

  times = 100

)

# Unit: milliseconds

# expr      min       lq     mean   median       uq      max neval

# sort 63.48221 67.79530 78.33724 70.74215 74.10109 173.1129   100

# order 56.46055 57.18649 60.88239 58.29462 62.13086 155.5481   100

So probably you should pick sort with index.return = TRUE only if you need a list object to be returned. I can't find an example where sort is better than the other.

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

1

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

add a comment |

order simply gives the indexes, instead sort gives also the values (and with index.return=T a list):

x <- runif(10, 0, 100)

order(x)

# [1]  2  7  1  9  6  5  8 10  4  3

sort(x, index.return=T)

# $`x`

# [1] 0.08140348 0.18272011 0.23575252 0.51493537 0.64281259 0.92121388 0.93759670 0.96221375 0.97646916 0.97863369

# 

# $ix

# [1]  2  7  1  9  6  5  8 10  4  3

It seems that order is a little faster with big numbers (longer vector size):

x <- runif(10000000, 0, 100)



microbenchmark::microbenchmark(

  sort = {sort(x, index.return=T)},

  order = {x[order(x)]},

  times = 100

)

# Unit: milliseconds

# expr      min       lq     mean   median       uq      max neval

# sort 63.48221 67.79530 78.33724 70.74215 74.10109 173.1129   100

# order 56.46055 57.18649 60.88239 58.29462 62.13086 155.5481   100

So probably you should pick sort with index.return = TRUE only if you need a list object to be returned. I can't find an example where sort is better than the other.

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

1

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

add a comment |

order simply gives the indexes, instead sort gives also the values (and with index.return=T a list):

x <- runif(10, 0, 100)

order(x)

# [1]  2  7  1  9  6  5  8 10  4  3

sort(x, index.return=T)

# $`x`

# [1] 0.08140348 0.18272011 0.23575252 0.51493537 0.64281259 0.92121388 0.93759670 0.96221375 0.97646916 0.97863369

# 

# $ix

# [1]  2  7  1  9  6  5  8 10  4  3

It seems that order is a little faster with big numbers (longer vector size):

x <- runif(10000000, 0, 100)



microbenchmark::microbenchmark(

  sort = {sort(x, index.return=T)},

  order = {x[order(x)]},

  times = 100

)

# Unit: milliseconds

# expr      min       lq     mean   median       uq      max neval

# sort 63.48221 67.79530 78.33724 70.74215 74.10109 173.1129   100

# order 56.46055 57.18649 60.88239 58.29462 62.13086 155.5481   100

So probably you should pick sort with index.return = TRUE only if you need a list object to be returned. I can't find an example where sort is better than the other.

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

order simply gives the indexes, instead sort gives also the values (and with index.return=T a list):

x <- runif(10, 0, 100)

order(x)

# [1]  2  7  1  9  6  5  8 10  4  3

sort(x, index.return=T)

# $`x`

# [1] 0.08140348 0.18272011 0.23575252 0.51493537 0.64281259 0.92121388 0.93759670 0.96221375 0.97646916 0.97863369

# 

# $ix

# [1]  2  7  1  9  6  5  8 10  4  3

It seems that order is a little faster with big numbers (longer vector size):

x <- runif(10000000, 0, 100)



microbenchmark::microbenchmark(

  sort = {sort(x, index.return=T)},

  order = {x[order(x)]},

  times = 100

)

# Unit: milliseconds

# expr      min       lq     mean   median       uq      max neval

# sort 63.48221 67.79530 78.33724 70.74215 74.10109 173.1129   100

# order 56.46055 57.18649 60.88239 58.29462 62.13086 155.5481   100

So probably you should pick sort with index.return = TRUE only if you need a list object to be returned. I can't find an example where sort is better than the other.

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

edited Nov 16 '18 at 14:10

answered Nov 16 '18 at 12:55

RLave

5,28911226

answered Nov 16 '18 at 12:55

RLave

5,28911226

answered Nov 16 '18 at 12:55

RLave

5,28911226

1

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

add a comment |

1

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

To clarify, by "big numbers" you mean order is faster with sorting vectors with lots of numbers, not numbers that are large in magnitude, right?

– camille
Nov 16 '18 at 14:04

Yes I meant longer vector size. I checked at it seems that even with numbers of bigger magniture is still a little faster. It doesn't change much if you use runif(10000000, 0, 100) or runif(10000000, 0, 10000).

– RLave
Nov 16 '18 at 14:05

Sometimes sort(x,method-"quick",...) is faster so you could test also with this arguments. Nice answer.

– Csd
Nov 16 '18 at 17:15

add a comment |

My suggestions are based on RLave's answer.

sort_order <- function(x){

    indices <- order(x) #you can choose a method also but leave default.

    list("x"=x[indices],"ix"=indices)

}

Here are some benchmarks.

microbenchmark::microbenchmark(

     sort=s<-sort(x,index.return=T),

     "quick sort"=sq<-sort(x,method="quick",index.return=T),

     "order sort"=so<-sort_order(x),times = 10

     times=10

)



Unit: seconds

         expr      min       lq     mean   median       uq      max neval

         sort 1.493714 1.662791 1.737854 1.708502 1.887993 1.960912    10

   quick sort 1.366938 1.374874 1.451778 1.444342 1.480122 1.668693    10

   order sort 1.181974 1.344398 1.359209 1.369108 1.424569 1.461862    10



all.equal(so,sq)

[1] TRUE

all.equal(s,so)

[1] TRUE

answered Nov 16 '18 at 17:33

Csd

31819

add a comment |

My suggestions are based on RLave's answer.

sort_order <- function(x){

    indices <- order(x) #you can choose a method also but leave default.

    list("x"=x[indices],"ix"=indices)

}

Here are some benchmarks.

microbenchmark::microbenchmark(

     sort=s<-sort(x,index.return=T),

     "quick sort"=sq<-sort(x,method="quick",index.return=T),

     "order sort"=so<-sort_order(x),times = 10

     times=10

)



Unit: seconds

         expr      min       lq     mean   median       uq      max neval

         sort 1.493714 1.662791 1.737854 1.708502 1.887993 1.960912    10

   quick sort 1.366938 1.374874 1.451778 1.444342 1.480122 1.668693    10

   order sort 1.181974 1.344398 1.359209 1.369108 1.424569 1.461862    10



all.equal(so,sq)

[1] TRUE

all.equal(s,so)

[1] TRUE

answered Nov 16 '18 at 17:33

Csd

31819

add a comment |

My suggestions are based on RLave's answer.

sort_order <- function(x){

    indices <- order(x) #you can choose a method also but leave default.

    list("x"=x[indices],"ix"=indices)

}

Here are some benchmarks.

microbenchmark::microbenchmark(

     sort=s<-sort(x,index.return=T),

     "quick sort"=sq<-sort(x,method="quick",index.return=T),

     "order sort"=so<-sort_order(x),times = 10

     times=10

)



Unit: seconds

         expr      min       lq     mean   median       uq      max neval

         sort 1.493714 1.662791 1.737854 1.708502 1.887993 1.960912    10

   quick sort 1.366938 1.374874 1.451778 1.444342 1.480122 1.668693    10

   order sort 1.181974 1.344398 1.359209 1.369108 1.424569 1.461862    10



all.equal(so,sq)

[1] TRUE

all.equal(s,so)

[1] TRUE

answered Nov 16 '18 at 17:33

Csd

31819

My suggestions are based on RLave's answer.

sort_order <- function(x){

    indices <- order(x) #you can choose a method also but leave default.

    list("x"=x[indices],"ix"=indices)

}

Here are some benchmarks.

microbenchmark::microbenchmark(

     sort=s<-sort(x,index.return=T),

     "quick sort"=sq<-sort(x,method="quick",index.return=T),

     "order sort"=so<-sort_order(x),times = 10

     times=10

)



Unit: seconds

         expr      min       lq     mean   median       uq      max neval

         sort 1.493714 1.662791 1.737854 1.708502 1.887993 1.960912    10

   quick sort 1.366938 1.374874 1.451778 1.444342 1.480122 1.668693    10

   order sort 1.181974 1.344398 1.359209 1.369108 1.424569 1.461862    10



all.equal(so,sq)

[1] TRUE

all.equal(s,so)

[1] TRUE

answered Nov 16 '18 at 17:33

Csd

31819

answered Nov 16 '18 at 17:33

Csd

31819

answered Nov 16 '18 at 17:33

Csd

31819

answered Nov 16 '18 at 17:33

Csd

31819

add a comment |

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