Java 8 - customised sort based on specific order












11














I would like to sort the user list based on their status but the order must be based on the order that I set.



I want to set the order of list,



The order should be 1, 0 , 5. We should also keep in mind to order the username as well.



List<User> users = new ArrayList();
users.add(new User("A", 1));
users.add(new User("B", 5));
users.add(new User("C", 0));
users.add(new User("D", 1));
users.add(new User("E", 5));
users.add(new User("F", 0));


Here's the user class



public class User {
private String username;
private Integer status;
}


It should look like this



[
{
"username": "A",
"status": 1
},
{
"username": "D",
"status": 1
},
{
"username": "C",
"status": 0
},
{
"username": "F",
"status": 0
},
{
"username": "B",
"status": 5
},
{
"username": "E",
"status": 5
}
]


I not sure if it's possible to use Comparator.comparing, since this one is neither ascending nor descending order.










share|improve this question




















  • 2




    Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
    – MadProgrammer
    Nov 12 '18 at 4:02










  • if 1,0,5 is the only status numbers, then you can easily order it using comparator.
    – Abhi
    Nov 12 '18 at 4:04






  • 1




    Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
    – AJNeufeld
    Nov 12 '18 at 4:06






  • 1




    @KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
    – MadProgrammer
    Nov 12 '18 at 4:16






  • 1




    @KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
    – Slaw
    Nov 12 '18 at 4:17


















11














I would like to sort the user list based on their status but the order must be based on the order that I set.



I want to set the order of list,



The order should be 1, 0 , 5. We should also keep in mind to order the username as well.



List<User> users = new ArrayList();
users.add(new User("A", 1));
users.add(new User("B", 5));
users.add(new User("C", 0));
users.add(new User("D", 1));
users.add(new User("E", 5));
users.add(new User("F", 0));


Here's the user class



public class User {
private String username;
private Integer status;
}


It should look like this



[
{
"username": "A",
"status": 1
},
{
"username": "D",
"status": 1
},
{
"username": "C",
"status": 0
},
{
"username": "F",
"status": 0
},
{
"username": "B",
"status": 5
},
{
"username": "E",
"status": 5
}
]


I not sure if it's possible to use Comparator.comparing, since this one is neither ascending nor descending order.










share|improve this question




















  • 2




    Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
    – MadProgrammer
    Nov 12 '18 at 4:02










  • if 1,0,5 is the only status numbers, then you can easily order it using comparator.
    – Abhi
    Nov 12 '18 at 4:04






  • 1




    Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
    – AJNeufeld
    Nov 12 '18 at 4:06






  • 1




    @KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
    – MadProgrammer
    Nov 12 '18 at 4:16






  • 1




    @KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
    – Slaw
    Nov 12 '18 at 4:17
















11












11








11







I would like to sort the user list based on their status but the order must be based on the order that I set.



I want to set the order of list,



The order should be 1, 0 , 5. We should also keep in mind to order the username as well.



List<User> users = new ArrayList();
users.add(new User("A", 1));
users.add(new User("B", 5));
users.add(new User("C", 0));
users.add(new User("D", 1));
users.add(new User("E", 5));
users.add(new User("F", 0));


Here's the user class



public class User {
private String username;
private Integer status;
}


It should look like this



[
{
"username": "A",
"status": 1
},
{
"username": "D",
"status": 1
},
{
"username": "C",
"status": 0
},
{
"username": "F",
"status": 0
},
{
"username": "B",
"status": 5
},
{
"username": "E",
"status": 5
}
]


I not sure if it's possible to use Comparator.comparing, since this one is neither ascending nor descending order.










share|improve this question















I would like to sort the user list based on their status but the order must be based on the order that I set.



I want to set the order of list,



The order should be 1, 0 , 5. We should also keep in mind to order the username as well.



List<User> users = new ArrayList();
users.add(new User("A", 1));
users.add(new User("B", 5));
users.add(new User("C", 0));
users.add(new User("D", 1));
users.add(new User("E", 5));
users.add(new User("F", 0));


Here's the user class



public class User {
private String username;
private Integer status;
}


It should look like this



[
{
"username": "A",
"status": 1
},
{
"username": "D",
"status": 1
},
{
"username": "C",
"status": 0
},
{
"username": "F",
"status": 0
},
{
"username": "B",
"status": 5
},
{
"username": "E",
"status": 5
}
]


I not sure if it's possible to use Comparator.comparing, since this one is neither ascending nor descending order.







java sorting java-8 comparator






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 '18 at 4:22









Mureinik

180k22130198




180k22130198










asked Nov 12 '18 at 3:59









KennethC

271215




271215








  • 2




    Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
    – MadProgrammer
    Nov 12 '18 at 4:02










  • if 1,0,5 is the only status numbers, then you can easily order it using comparator.
    – Abhi
    Nov 12 '18 at 4:04






  • 1




    Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
    – AJNeufeld
    Nov 12 '18 at 4:06






  • 1




    @KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
    – MadProgrammer
    Nov 12 '18 at 4:16






  • 1




    @KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
    – Slaw
    Nov 12 '18 at 4:17
















  • 2




    Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
    – MadProgrammer
    Nov 12 '18 at 4:02










  • if 1,0,5 is the only status numbers, then you can easily order it using comparator.
    – Abhi
    Nov 12 '18 at 4:04






  • 1




    Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
    – AJNeufeld
    Nov 12 '18 at 4:06






  • 1




    @KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
    – MadProgrammer
    Nov 12 '18 at 4:16






  • 1




    @KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
    – Slaw
    Nov 12 '18 at 4:17










2




2




Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
– MadProgrammer
Nov 12 '18 at 4:02




Yes, Comparator can deal with, it just needs to return +1/-1/0 based on the comparison of the two values supplied. Best bet is to just give it a try
– MadProgrammer
Nov 12 '18 at 4:02












if 1,0,5 is the only status numbers, then you can easily order it using comparator.
– Abhi
Nov 12 '18 at 4:04




if 1,0,5 is the only status numbers, then you can easily order it using comparator.
– Abhi
Nov 12 '18 at 4:04




1




1




Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
– AJNeufeld
Nov 12 '18 at 4:06




Subtract 1, then square. 1,0,5 -> 0, 1, 16 ... which is easy to sort
– AJNeufeld
Nov 12 '18 at 4:06




1




1




@KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
– MadProgrammer
Nov 12 '18 at 4:16




@KennethC Compator just returns a negative or positive or zero value based on the result of the comparison and the order you want the values to be sorted
– MadProgrammer
Nov 12 '18 at 4:16




1




1




@KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
– Slaw
Nov 12 '18 at 4:17






@KennethC Take a look at the documentation of Comparator.compare to see what the different return values mean (negative means less than, zero means equal to, positive means greater than).
– Slaw
Nov 12 '18 at 4:17














5 Answers
5






active

oldest

votes


















9














One approach could be to hold a list with the order you want and sort the users according to its index:



final List<Integer> order = Arrays.asList(1, 0, 5);
users.sort(
Comparator.comparing((User u) -> order.indexOf(u.getStatus()))
.thenComparing(User::getUsername));


Note that while this approach should be reasonable for a small number of statuses (like you currently have), it may slow down sorting if there are a large number of statuses and you need to do perform an O(n) search each time. A better performing approach (albeit arguably not as sleek), would be to use a map:



final Map<Integer, Integer> order = new HashMap<>();
order.put(1, 0);
order.put(0, 1);
order.put(5 ,2);
users.sort(Comparator.comparing((User u) -> order.get(u.getStatus()))
.thenComparing(User::getUsername));





share|improve this answer































    6














    If you don't mind using Guava in your project, you can use Ordering.explicit:



    users.sort(Ordering.explicit(1, 0, 5).onResultOf(User::getStatus));


    If you want to sort by name also, then add thenComparing:



    users.sort(Ordering
    .explicit(1, 0, 5)
    .onResultOf(User::getStatus)
    .thenComparing(User::getUsername));





    share|improve this answer





















    • this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
      – Eugene
      Nov 12 '18 at 4:26



















    1














    Assuming 1, 0, and 5 will be the only values of status, AJNeufeld made an excellent point in their comment; they stated that you can use an equation to map each value into an ascending order. In this case, the equation would be (x - 1)^2 where x is the value of status:



    users.sort(Comparator.comparingDouble(user -> Math.pow(user.getStatus() - 1, 2)));


    If you were to print the contents of user after calling the above snippet, you'd get:




    [User [username=A, status=1], User [username=D, status=1], User [username=C, status=0], User [username=F, status=0], User [username=B, status=5], User [username=E, status=5]]







    share|improve this answer

















    • 1




      Why not just |x - 1|?
      – ZhekaKozlov
      Nov 12 '18 at 5:58






    • 1




      @ZhekaKozlov v^1
      – Holger
      Nov 12 '18 at 10:34










    • @Holger Excuse me?
      – ZhekaKozlov
      Nov 12 '18 at 12:26










    • @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
      – Holger
      Nov 12 '18 at 12:27












    • @Holger 0^1 < 1^1 (should be greater)
      – ZhekaKozlov
      Nov 12 '18 at 12:29





















    0














    You can try to do this step by step



    //order define here
    List<Integer> statusOrder= Arrays.asList(1,0,5,2);

    //define sort by status
    Comparator<User> byStatus = (u1, u2) -> {
    return Integer.compare(statusOrder.indexOf(u1.getStatus()), statusOrder.indexOf(u2.getStatus()));
    };

    //define sort by name
    Comparator<User> byName = Comparator.comparing(User::getUsername);

    //actualy sort
    users.sort(byStatus.thenComparing(byName));





    share|improve this answer





























      0














      As you have mentioned you need custom ordering and that means you need to define somewhere that ordering either in HashMap<Status,Rank>> or one simple way add one more attribute say Integer rank; and you can define the rank based on your ordering for status attribute like say users.add(new User("A", 1,0)); here status 1 is sortest in order and its rank=0. And then you can use Comparator on rank attribute.



      For e.g :



      public class User {
      public String username;
      public Integer status;
      public Integer rank;

      public User(String username, Integer status, Integer rank)
      {
      this.username = username;
      this.status = status;
      this.rank = rank;
      }
      }


      Comparator class :



      class SortByRank implements Comparator<User> 
      {
      // Used for sorting in ascending order of
      // rank number
      public int compare(User a, User b)
      {
      return a.rank - b.rank;
      }
      }


      Main Class :



      class Main 
      {
      public static void main (String args)
      {
      List<User> users = new ArrayList();
      users.add(new User("A", 1, 0));
      users.add(new User("B", 5, 2));
      users.add(new User("C", 0, 1));
      users.add(new User("D", 1, 0));
      users.add(new User("E", 5, 2));
      users.add(new User("F", 0, 1));

      System.out.println("Unsorted");
      for (int i=0; i<users.size(); i++)
      System.out.print(users.get(i).username);

      Collections.sort(users, new SortByRank());

      System.out.println("nSorted by Rank");
      for (int i=0; i<users.size(); i++)
      System.out.print(users.get(i).username);
      }
      }





      share|improve this answer























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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9














        One approach could be to hold a list with the order you want and sort the users according to its index:



        final List<Integer> order = Arrays.asList(1, 0, 5);
        users.sort(
        Comparator.comparing((User u) -> order.indexOf(u.getStatus()))
        .thenComparing(User::getUsername));


        Note that while this approach should be reasonable for a small number of statuses (like you currently have), it may slow down sorting if there are a large number of statuses and you need to do perform an O(n) search each time. A better performing approach (albeit arguably not as sleek), would be to use a map:



        final Map<Integer, Integer> order = new HashMap<>();
        order.put(1, 0);
        order.put(0, 1);
        order.put(5 ,2);
        users.sort(Comparator.comparing((User u) -> order.get(u.getStatus()))
        .thenComparing(User::getUsername));





        share|improve this answer




























          9














          One approach could be to hold a list with the order you want and sort the users according to its index:



          final List<Integer> order = Arrays.asList(1, 0, 5);
          users.sort(
          Comparator.comparing((User u) -> order.indexOf(u.getStatus()))
          .thenComparing(User::getUsername));


          Note that while this approach should be reasonable for a small number of statuses (like you currently have), it may slow down sorting if there are a large number of statuses and you need to do perform an O(n) search each time. A better performing approach (albeit arguably not as sleek), would be to use a map:



          final Map<Integer, Integer> order = new HashMap<>();
          order.put(1, 0);
          order.put(0, 1);
          order.put(5 ,2);
          users.sort(Comparator.comparing((User u) -> order.get(u.getStatus()))
          .thenComparing(User::getUsername));





          share|improve this answer


























            9












            9








            9






            One approach could be to hold a list with the order you want and sort the users according to its index:



            final List<Integer> order = Arrays.asList(1, 0, 5);
            users.sort(
            Comparator.comparing((User u) -> order.indexOf(u.getStatus()))
            .thenComparing(User::getUsername));


            Note that while this approach should be reasonable for a small number of statuses (like you currently have), it may slow down sorting if there are a large number of statuses and you need to do perform an O(n) search each time. A better performing approach (albeit arguably not as sleek), would be to use a map:



            final Map<Integer, Integer> order = new HashMap<>();
            order.put(1, 0);
            order.put(0, 1);
            order.put(5 ,2);
            users.sort(Comparator.comparing((User u) -> order.get(u.getStatus()))
            .thenComparing(User::getUsername));





            share|improve this answer














            One approach could be to hold a list with the order you want and sort the users according to its index:



            final List<Integer> order = Arrays.asList(1, 0, 5);
            users.sort(
            Comparator.comparing((User u) -> order.indexOf(u.getStatus()))
            .thenComparing(User::getUsername));


            Note that while this approach should be reasonable for a small number of statuses (like you currently have), it may slow down sorting if there are a large number of statuses and you need to do perform an O(n) search each time. A better performing approach (albeit arguably not as sleek), would be to use a map:



            final Map<Integer, Integer> order = new HashMap<>();
            order.put(1, 0);
            order.put(0, 1);
            order.put(5 ,2);
            users.sort(Comparator.comparing((User u) -> order.get(u.getStatus()))
            .thenComparing(User::getUsername));






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 10 '18 at 6:37

























            answered Nov 12 '18 at 4:20









            Mureinik

            180k22130198




            180k22130198

























                6














                If you don't mind using Guava in your project, you can use Ordering.explicit:



                users.sort(Ordering.explicit(1, 0, 5).onResultOf(User::getStatus));


                If you want to sort by name also, then add thenComparing:



                users.sort(Ordering
                .explicit(1, 0, 5)
                .onResultOf(User::getStatus)
                .thenComparing(User::getUsername));





                share|improve this answer





















                • this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                  – Eugene
                  Nov 12 '18 at 4:26
















                6














                If you don't mind using Guava in your project, you can use Ordering.explicit:



                users.sort(Ordering.explicit(1, 0, 5).onResultOf(User::getStatus));


                If you want to sort by name also, then add thenComparing:



                users.sort(Ordering
                .explicit(1, 0, 5)
                .onResultOf(User::getStatus)
                .thenComparing(User::getUsername));





                share|improve this answer





















                • this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                  – Eugene
                  Nov 12 '18 at 4:26














                6












                6








                6






                If you don't mind using Guava in your project, you can use Ordering.explicit:



                users.sort(Ordering.explicit(1, 0, 5).onResultOf(User::getStatus));


                If you want to sort by name also, then add thenComparing:



                users.sort(Ordering
                .explicit(1, 0, 5)
                .onResultOf(User::getStatus)
                .thenComparing(User::getUsername));





                share|improve this answer












                If you don't mind using Guava in your project, you can use Ordering.explicit:



                users.sort(Ordering.explicit(1, 0, 5).onResultOf(User::getStatus));


                If you want to sort by name also, then add thenComparing:



                users.sort(Ordering
                .explicit(1, 0, 5)
                .onResultOf(User::getStatus)
                .thenComparing(User::getUsername));






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 '18 at 4:09









                ZhekaKozlov

                14.7k866111




                14.7k866111












                • this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                  – Eugene
                  Nov 12 '18 at 4:26


















                • this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                  – Eugene
                  Nov 12 '18 at 4:26
















                this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                – Eugene
                Nov 12 '18 at 4:26




                this is how I would do it, mainly because that explicit is well such a clear and "explicit" name
                – Eugene
                Nov 12 '18 at 4:26











                1














                Assuming 1, 0, and 5 will be the only values of status, AJNeufeld made an excellent point in their comment; they stated that you can use an equation to map each value into an ascending order. In this case, the equation would be (x - 1)^2 where x is the value of status:



                users.sort(Comparator.comparingDouble(user -> Math.pow(user.getStatus() - 1, 2)));


                If you were to print the contents of user after calling the above snippet, you'd get:




                [User [username=A, status=1], User [username=D, status=1], User [username=C, status=0], User [username=F, status=0], User [username=B, status=5], User [username=E, status=5]]







                share|improve this answer

















                • 1




                  Why not just |x - 1|?
                  – ZhekaKozlov
                  Nov 12 '18 at 5:58






                • 1




                  @ZhekaKozlov v^1
                  – Holger
                  Nov 12 '18 at 10:34










                • @Holger Excuse me?
                  – ZhekaKozlov
                  Nov 12 '18 at 12:26










                • @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                  – Holger
                  Nov 12 '18 at 12:27












                • @Holger 0^1 < 1^1 (should be greater)
                  – ZhekaKozlov
                  Nov 12 '18 at 12:29


















                1














                Assuming 1, 0, and 5 will be the only values of status, AJNeufeld made an excellent point in their comment; they stated that you can use an equation to map each value into an ascending order. In this case, the equation would be (x - 1)^2 where x is the value of status:



                users.sort(Comparator.comparingDouble(user -> Math.pow(user.getStatus() - 1, 2)));


                If you were to print the contents of user after calling the above snippet, you'd get:




                [User [username=A, status=1], User [username=D, status=1], User [username=C, status=0], User [username=F, status=0], User [username=B, status=5], User [username=E, status=5]]







                share|improve this answer

















                • 1




                  Why not just |x - 1|?
                  – ZhekaKozlov
                  Nov 12 '18 at 5:58






                • 1




                  @ZhekaKozlov v^1
                  – Holger
                  Nov 12 '18 at 10:34










                • @Holger Excuse me?
                  – ZhekaKozlov
                  Nov 12 '18 at 12:26










                • @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                  – Holger
                  Nov 12 '18 at 12:27












                • @Holger 0^1 < 1^1 (should be greater)
                  – ZhekaKozlov
                  Nov 12 '18 at 12:29
















                1












                1








                1






                Assuming 1, 0, and 5 will be the only values of status, AJNeufeld made an excellent point in their comment; they stated that you can use an equation to map each value into an ascending order. In this case, the equation would be (x - 1)^2 where x is the value of status:



                users.sort(Comparator.comparingDouble(user -> Math.pow(user.getStatus() - 1, 2)));


                If you were to print the contents of user after calling the above snippet, you'd get:




                [User [username=A, status=1], User [username=D, status=1], User [username=C, status=0], User [username=F, status=0], User [username=B, status=5], User [username=E, status=5]]







                share|improve this answer












                Assuming 1, 0, and 5 will be the only values of status, AJNeufeld made an excellent point in their comment; they stated that you can use an equation to map each value into an ascending order. In this case, the equation would be (x - 1)^2 where x is the value of status:



                users.sort(Comparator.comparingDouble(user -> Math.pow(user.getStatus() - 1, 2)));


                If you were to print the contents of user after calling the above snippet, you'd get:




                [User [username=A, status=1], User [username=D, status=1], User [username=C, status=0], User [username=F, status=0], User [username=B, status=5], User [username=E, status=5]]








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 '18 at 4:22









                Jacob G.

                15.2k52162




                15.2k52162








                • 1




                  Why not just |x - 1|?
                  – ZhekaKozlov
                  Nov 12 '18 at 5:58






                • 1




                  @ZhekaKozlov v^1
                  – Holger
                  Nov 12 '18 at 10:34










                • @Holger Excuse me?
                  – ZhekaKozlov
                  Nov 12 '18 at 12:26










                • @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                  – Holger
                  Nov 12 '18 at 12:27












                • @Holger 0^1 < 1^1 (should be greater)
                  – ZhekaKozlov
                  Nov 12 '18 at 12:29
















                • 1




                  Why not just |x - 1|?
                  – ZhekaKozlov
                  Nov 12 '18 at 5:58






                • 1




                  @ZhekaKozlov v^1
                  – Holger
                  Nov 12 '18 at 10:34










                • @Holger Excuse me?
                  – ZhekaKozlov
                  Nov 12 '18 at 12:26










                • @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                  – Holger
                  Nov 12 '18 at 12:27












                • @Holger 0^1 < 1^1 (should be greater)
                  – ZhekaKozlov
                  Nov 12 '18 at 12:29










                1




                1




                Why not just |x - 1|?
                – ZhekaKozlov
                Nov 12 '18 at 5:58




                Why not just |x - 1|?
                – ZhekaKozlov
                Nov 12 '18 at 5:58




                1




                1




                @ZhekaKozlov v^1
                – Holger
                Nov 12 '18 at 10:34




                @ZhekaKozlov v^1
                – Holger
                Nov 12 '18 at 10:34












                @Holger Excuse me?
                – ZhekaKozlov
                Nov 12 '18 at 12:26




                @Holger Excuse me?
                – ZhekaKozlov
                Nov 12 '18 at 12:26












                @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                – Holger
                Nov 12 '18 at 12:27






                @ZhekaKozlov value ^ 1 is even simpler than Math.abs(value - 1) (the difference is not as big as the difference of both to Math.pow, though).
                – Holger
                Nov 12 '18 at 12:27














                @Holger 0^1 < 1^1 (should be greater)
                – ZhekaKozlov
                Nov 12 '18 at 12:29






                @Holger 0^1 < 1^1 (should be greater)
                – ZhekaKozlov
                Nov 12 '18 at 12:29













                0














                You can try to do this step by step



                //order define here
                List<Integer> statusOrder= Arrays.asList(1,0,5,2);

                //define sort by status
                Comparator<User> byStatus = (u1, u2) -> {
                return Integer.compare(statusOrder.indexOf(u1.getStatus()), statusOrder.indexOf(u2.getStatus()));
                };

                //define sort by name
                Comparator<User> byName = Comparator.comparing(User::getUsername);

                //actualy sort
                users.sort(byStatus.thenComparing(byName));





                share|improve this answer


























                  0














                  You can try to do this step by step



                  //order define here
                  List<Integer> statusOrder= Arrays.asList(1,0,5,2);

                  //define sort by status
                  Comparator<User> byStatus = (u1, u2) -> {
                  return Integer.compare(statusOrder.indexOf(u1.getStatus()), statusOrder.indexOf(u2.getStatus()));
                  };

                  //define sort by name
                  Comparator<User> byName = Comparator.comparing(User::getUsername);

                  //actualy sort
                  users.sort(byStatus.thenComparing(byName));





                  share|improve this answer
























                    0












                    0








                    0






                    You can try to do this step by step



                    //order define here
                    List<Integer> statusOrder= Arrays.asList(1,0,5,2);

                    //define sort by status
                    Comparator<User> byStatus = (u1, u2) -> {
                    return Integer.compare(statusOrder.indexOf(u1.getStatus()), statusOrder.indexOf(u2.getStatus()));
                    };

                    //define sort by name
                    Comparator<User> byName = Comparator.comparing(User::getUsername);

                    //actualy sort
                    users.sort(byStatus.thenComparing(byName));





                    share|improve this answer












                    You can try to do this step by step



                    //order define here
                    List<Integer> statusOrder= Arrays.asList(1,0,5,2);

                    //define sort by status
                    Comparator<User> byStatus = (u1, u2) -> {
                    return Integer.compare(statusOrder.indexOf(u1.getStatus()), statusOrder.indexOf(u2.getStatus()));
                    };

                    //define sort by name
                    Comparator<User> byName = Comparator.comparing(User::getUsername);

                    //actualy sort
                    users.sort(byStatus.thenComparing(byName));






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 12 '18 at 4:26









                    Dang Nguyen

                    599221




                    599221























                        0














                        As you have mentioned you need custom ordering and that means you need to define somewhere that ordering either in HashMap<Status,Rank>> or one simple way add one more attribute say Integer rank; and you can define the rank based on your ordering for status attribute like say users.add(new User("A", 1,0)); here status 1 is sortest in order and its rank=0. And then you can use Comparator on rank attribute.



                        For e.g :



                        public class User {
                        public String username;
                        public Integer status;
                        public Integer rank;

                        public User(String username, Integer status, Integer rank)
                        {
                        this.username = username;
                        this.status = status;
                        this.rank = rank;
                        }
                        }


                        Comparator class :



                        class SortByRank implements Comparator<User> 
                        {
                        // Used for sorting in ascending order of
                        // rank number
                        public int compare(User a, User b)
                        {
                        return a.rank - b.rank;
                        }
                        }


                        Main Class :



                        class Main 
                        {
                        public static void main (String args)
                        {
                        List<User> users = new ArrayList();
                        users.add(new User("A", 1, 0));
                        users.add(new User("B", 5, 2));
                        users.add(new User("C", 0, 1));
                        users.add(new User("D", 1, 0));
                        users.add(new User("E", 5, 2));
                        users.add(new User("F", 0, 1));

                        System.out.println("Unsorted");
                        for (int i=0; i<users.size(); i++)
                        System.out.print(users.get(i).username);

                        Collections.sort(users, new SortByRank());

                        System.out.println("nSorted by Rank");
                        for (int i=0; i<users.size(); i++)
                        System.out.print(users.get(i).username);
                        }
                        }





                        share|improve this answer




























                          0














                          As you have mentioned you need custom ordering and that means you need to define somewhere that ordering either in HashMap<Status,Rank>> or one simple way add one more attribute say Integer rank; and you can define the rank based on your ordering for status attribute like say users.add(new User("A", 1,0)); here status 1 is sortest in order and its rank=0. And then you can use Comparator on rank attribute.



                          For e.g :



                          public class User {
                          public String username;
                          public Integer status;
                          public Integer rank;

                          public User(String username, Integer status, Integer rank)
                          {
                          this.username = username;
                          this.status = status;
                          this.rank = rank;
                          }
                          }


                          Comparator class :



                          class SortByRank implements Comparator<User> 
                          {
                          // Used for sorting in ascending order of
                          // rank number
                          public int compare(User a, User b)
                          {
                          return a.rank - b.rank;
                          }
                          }


                          Main Class :



                          class Main 
                          {
                          public static void main (String args)
                          {
                          List<User> users = new ArrayList();
                          users.add(new User("A", 1, 0));
                          users.add(new User("B", 5, 2));
                          users.add(new User("C", 0, 1));
                          users.add(new User("D", 1, 0));
                          users.add(new User("E", 5, 2));
                          users.add(new User("F", 0, 1));

                          System.out.println("Unsorted");
                          for (int i=0; i<users.size(); i++)
                          System.out.print(users.get(i).username);

                          Collections.sort(users, new SortByRank());

                          System.out.println("nSorted by Rank");
                          for (int i=0; i<users.size(); i++)
                          System.out.print(users.get(i).username);
                          }
                          }





                          share|improve this answer


























                            0












                            0








                            0






                            As you have mentioned you need custom ordering and that means you need to define somewhere that ordering either in HashMap<Status,Rank>> or one simple way add one more attribute say Integer rank; and you can define the rank based on your ordering for status attribute like say users.add(new User("A", 1,0)); here status 1 is sortest in order and its rank=0. And then you can use Comparator on rank attribute.



                            For e.g :



                            public class User {
                            public String username;
                            public Integer status;
                            public Integer rank;

                            public User(String username, Integer status, Integer rank)
                            {
                            this.username = username;
                            this.status = status;
                            this.rank = rank;
                            }
                            }


                            Comparator class :



                            class SortByRank implements Comparator<User> 
                            {
                            // Used for sorting in ascending order of
                            // rank number
                            public int compare(User a, User b)
                            {
                            return a.rank - b.rank;
                            }
                            }


                            Main Class :



                            class Main 
                            {
                            public static void main (String args)
                            {
                            List<User> users = new ArrayList();
                            users.add(new User("A", 1, 0));
                            users.add(new User("B", 5, 2));
                            users.add(new User("C", 0, 1));
                            users.add(new User("D", 1, 0));
                            users.add(new User("E", 5, 2));
                            users.add(new User("F", 0, 1));

                            System.out.println("Unsorted");
                            for (int i=0; i<users.size(); i++)
                            System.out.print(users.get(i).username);

                            Collections.sort(users, new SortByRank());

                            System.out.println("nSorted by Rank");
                            for (int i=0; i<users.size(); i++)
                            System.out.print(users.get(i).username);
                            }
                            }





                            share|improve this answer














                            As you have mentioned you need custom ordering and that means you need to define somewhere that ordering either in HashMap<Status,Rank>> or one simple way add one more attribute say Integer rank; and you can define the rank based on your ordering for status attribute like say users.add(new User("A", 1,0)); here status 1 is sortest in order and its rank=0. And then you can use Comparator on rank attribute.



                            For e.g :



                            public class User {
                            public String username;
                            public Integer status;
                            public Integer rank;

                            public User(String username, Integer status, Integer rank)
                            {
                            this.username = username;
                            this.status = status;
                            this.rank = rank;
                            }
                            }


                            Comparator class :



                            class SortByRank implements Comparator<User> 
                            {
                            // Used for sorting in ascending order of
                            // rank number
                            public int compare(User a, User b)
                            {
                            return a.rank - b.rank;
                            }
                            }


                            Main Class :



                            class Main 
                            {
                            public static void main (String args)
                            {
                            List<User> users = new ArrayList();
                            users.add(new User("A", 1, 0));
                            users.add(new User("B", 5, 2));
                            users.add(new User("C", 0, 1));
                            users.add(new User("D", 1, 0));
                            users.add(new User("E", 5, 2));
                            users.add(new User("F", 0, 1));

                            System.out.println("Unsorted");
                            for (int i=0; i<users.size(); i++)
                            System.out.print(users.get(i).username);

                            Collections.sort(users, new SortByRank());

                            System.out.println("nSorted by Rank");
                            for (int i=0; i<users.size(); i++)
                            System.out.print(users.get(i).username);
                            }
                            }






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 12 '18 at 4:36

























                            answered Nov 12 '18 at 4:28









                            apandey846

                            8441019




                            8441019






























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