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





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5















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







share|improve this question


















  • 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


















5















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







share|improve this question


















  • 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














5












5








5


1






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







share|improve this question














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




r sorting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 16 '18 at 12:46









vonjdvonjd

2,13732844




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














  • 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








3




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












2 Answers
2






active

oldest

votes


















6














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.






share|improve this answer





















  • 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



















1














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





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    2 Answers
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    2 Answers
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    active

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    6














    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.






    share|improve this answer





















    • 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
















    6














    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.






    share|improve this answer





















    • 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














    6












    6








    6







    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.






    share|improve this answer















    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 16 '18 at 14:10

























    answered Nov 16 '18 at 12:55









    RLaveRLave

    5,28911226




    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














    • 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








    1




    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





    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







    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





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

    – Csd
    Nov 16 '18 at 17:15













    1














    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





    share|improve this answer




























      1














      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





      share|improve this answer


























        1












        1








        1







        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





        share|improve this answer













        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






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 16 '18 at 17:33









        CsdCsd

        31819




        31819






























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            App crashed after uploaded to heroku server

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I am uploading my code to heroku with the command: git push heroku master But whenever i push the code, whole code pushed on the server, but there is nothing there but application error. 2018-11-17T07:14:17.465955+00:00 heroku[router]: at=error code=H10 desc="App crashed" method=GET path="/" host=ecodrivebot.herokuapp.com request_id=47dc6aff-19dc-45e3-8a86-2d41eeef86b0 fwd="122.180.29.179" dyno= connect= service= status=503 bytes= protocol=https 2018-11-17T07:14:18.030853+00:00 heroku[router]: at=error code=H10 desc="App crashed" method=GET path="/favicon.ico" host=ecodrive...
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