What is the difference between sorted list vs list sort

Question

list sort   sorts the list and replaces the original list  whereas sorted list  returns a sorted copy of the list  without changing the original list    When is one preferred over the other  Which is more efficient  By how much  Can a list be reverted to the unsorted state after list sort   has been performed

User · Answer

Note  Simplest difference between sort   and sorted   is  sort     doesn t return any value while  sorted   returns an iterable list    sort   doesn t return any value   The sort   method just sorts the elements of a given list in a specific order - Ascending or Descending without returning any value   The syntax of sort   method is   list sort key      reverse           Alternatively  you can also use Python s in-built function sorted     for the same purpose  sorted function return sorted list    list sorted list  key      reverse

User · Answer

The main difference is that sorted some list  returns a new list   a    3  2  1  print sorted a    new list print a           is not modified   and some list sort    sorts the list in place   a    3  2  1  print a sort     in place print a           it s modified   Note that since a sort   doesn t return anything  print a sort   will print None     Can a list original positions be retrieved after list sort     No  because it modifies the original list

User · Answer

What is the difference between sorted list  vs list sort       list sort mutates the list in-place  amp  returns None sorted takes any iterable  amp  returns a new list  sorted    sorted is equivalent to this Python implementation  but the CPython builtin function should run measurably faster as it is written in C   def sorted iterable  key None       new list   list iterable       make a new list     new list sort key key          sort it     return new list                return it      when to use which     Use list sort when you do not wish to retain the original sort order  Thus you will be able to reuse the list in-place in memory   and when  you are the sole owner of the list  if the list is shared by other code and you mutate it  you could introduce bugs where that list is used   Use sorted when you want to retain the original sort order or when you  wish to create a new list that only your local code owns       Can a list s original positions be retrieved after list sort      No - unless you made a copy yourself  that information is lost because the sort is done in-place       And which is faster  And how much faster     To illustrate the penalty of creating a new list  use the timeit module  here s our setup   import timeit setup       import random lists    list range 10000   for   in range 1000      list of lists for l in lists      random shuffle l    shuffle each list shuffled iter   iter lists    wrap as iterator so next   yields one at a time       And here s our results for a list of randomly arranged 10000 integers  as we can see here  we ve disproven an older list creation expense myth   Python 2 7   gt  gt  gt  timeit repeat  next shuffled iter  sort     setup setup  number   1000   3 75168503401801  3 7473005310166627  3 753129180986434   gt  gt  gt  timeit repeat  sorted next shuffled iter     setup setup  number   1000   3 702025591977872  3 709248117986135  3 71071034099441    Python 3   gt  gt  gt  timeit repeat  next shuffled iter  sort     setup setup  number   1000   2 797430992126465  2 796825885772705  2 7744789123535156   gt  gt  gt  timeit repeat  sorted next shuffled iter     setup setup  number   1000   2 675589084625244  2 8019039630889893  2 849375009536743      After some feedback  I decided another test would be desirable with different characteristics  Here I provide the same randomly ordered list of 100 000 in length for each iteration 1 000 times   import timeit setup       import random random seed 0  lst   list range 100000   random shuffle lst        I interpret this larger sort s difference coming from the copying mentioned by Martijn  but it does not dominate to the point stated in the older more popular answer here  here the increase in time is only about 10    gt  gt  gt  timeit repeat  lst    sort     setup setup  number   10000   572 919036605  573 1384446719999  568 5923951   gt  gt  gt  timeit repeat  sorted lst       setup setup  number   10000   647 0584738299999  653 4040515829997  657 9457361929999    I also ran the above on a much smaller sort  and saw that the new sorted copy version still takes about 2  longer running time on a sort of 1000 length   Poke ran his own code as well  here s the code   setup       import random random seed 12122353453462456  lst   list range  length    random shuffle lst  lists    lst    for   in range  repeats    it   iter lists      t1    l   next it   l sort    t2    l   next it   sorted l   length   10    7 repeats   10    2 print length  repeats  for t in t1  t2      print t      print timeit t  setup setup format length length  repeats repeats   number repeats     He found for 1000000 length sort   ran 100 times  a similar result  but only about a 5  increase in time  here s the output   10000000 100 l   next it   l sort   610 5015971539542 l   next it   sorted l  646 7786222379655   Conclusion   A large sized list being sorted with sorted making a copy will likely dominate differences  but the sorting itself dominates the operation  and organizing your code around these differences would be premature optimization  I would use sorted when I need a new sorted list of the data  and I would use list sort when I need to sort a list in-place  and let that determine my usage

User · Answer

Here are a few simple examples to see the difference in action   See the list of numbers here   nums    1  9  -3  4  8  5  7  14    When calling sorted on this list  sorted will make a copy of the list   Meaning your original list will remain unchanged    Let s see   sorted nums    returns   -3  1  4  5  7  8  9  14    Looking at the nums again  nums   We see the original list  unaltered and NOT sorted    sorted did not change the original list   1  2  -3  4  8  5  7  14    Taking the same nums list and applying the sort function on it  will change the actual list    Let s see   Starting with our nums list to make sure  the content is still the same   nums   -3  1  4  5  7  8  9  14   nums sort     Now the original nums list is changed and looking at nums we see our original list has changed and is now sorted    nums  -3  1  2  4  5  7  8  14

User · Answer

The  sort   function stores the value of new list directly in the list variable  so answer for your third question would be NO  Also if you do this using sorted list   then you can get it use because it is not stored in the list variable  Also sometimes  sort   method acts as function  or say that it takes arguments in it   You have to store the value of sorted list  in a variable explicitly   Also for short data processing the speed will have no difference  but for long lists  you should directly use  sort   method for fast work  but again you will face irreversible actions

User · Answer

sorted   returns a new sorted list  leaving the original list unaffected  list sort   sorts the list in-place  mutating the list indices  and returns None  like all in-place operations    sorted   works on any iterable  not just lists  Strings  tuples  dictionaries  you ll get the keys   generators  etc   returning a list containing all elements  sorted    Use list sort   when you want to mutate the list  sorted   when you want a new sorted object back  Use sorted   when you want to sort something that is an iterable  not a list yet  For lists  list sort   is faster than sorted   because it doesn t have to create a copy  For any other iterable  you have no choice  No  you cannot retrieve the original positions  Once you called list sort   the original order is gone

[python] What is the difference between `sorted(list)` vs `list.sort()`?

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