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Sort with assumptions


variable sized lists and using lists as variablesRetaining and reusing a one-to-one mapping from a sortSorting a matrix alphanumericallyRearranging a ListHow can I check if one expression implies another?Generating an Array of VectorsImporting, sorting and exporting listsDeleting Lonely Numbers From a ListApplying multiple functions to a single column in a tableFind positions in which list elements are equal













4












$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.






list-manipulation symbolic array sorting







share|improve this question









$endgroup$







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    16 mins ago















4












$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.






list-manipulation symbolic array sorting







share|improve this question









$endgroup$







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    16 mins ago













4












4








4


1



$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.






list-manipulation symbolic array sorting







share|improve this question









$endgroup$




I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.






list-manipulation symbolic array sorting




list-manipulation symbolic array sorting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 38 mins ago









leastactionleastaction

22729




22729







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    16 mins ago












  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    16 mins ago







1




1




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
16 mins ago




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
16 mins ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Here is a possibility:



sortWithAssumptions[list_, assum_] := Module[order,
 order[a_, b_] := Simplify[a < b, assum];
 Sort[list, order]
]


For your example:



sortWithAssumptions[
 0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
 Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm



$left-x_3-x_9,-x_9,0,x_7right$




Another example:



sortWithAssumptions[
 0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
 Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm



$left-x_3-x_9,-x_9,0,x_7,x_9right$







share|improve this answer









$endgroup$












  • $begingroup$
    Thank you, Carl!
    $endgroup$
    – leastaction
    9 mins ago


















3












$begingroup$

How about:



list[[Ordering[list /. _Subscript -> 1]]]



-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




So basically we sort it the way it would be sorted with all subscripts == 1.






share|improve this answer









$endgroup$












  • $begingroup$
    Thanks! Seems to work like a charm, but can you shed some light on why?
    $endgroup$
    – leastaction
    16 mins ago










  • $begingroup$
    @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
    $endgroup$
    – Kuba♦
    15 mins ago










  • $begingroup$
    I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
    $endgroup$
    – leastaction
    12 mins ago






  • 1




    $begingroup$
    @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
    $endgroup$
    – Kuba♦
    9 mins ago


















2












$begingroup$

Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





share|improve this answer









$endgroup$




















    2












    $begingroup$

    In this case, we can use RankedMin and FullSimplify to get the answer you seek



    Assuming[
     Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0, 
     FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
    (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


    This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






    share|improve this answer











    $endgroup$












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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
       order[a_, b_] := Simplify[a < b, assum];
       Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$












      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        9 mins ago















      2












      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
       order[a_, b_] := Simplify[a < b, assum];
       Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$












      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        9 mins ago













      2












      2








      2





      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
       order[a_, b_] := Simplify[a < b, assum];
       Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$



      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
       order[a_, b_] := Simplify[a < b, assum];
       Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
       0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
       Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$








      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 12 mins ago









      Carl WollCarl Woll

      70.8k394184




      70.8k394184











      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        9 mins ago
















      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        9 mins ago















      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      9 mins ago




      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      9 mins ago











      3












      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$












      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        16 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba♦
        15 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        12 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba♦
        9 mins ago















      3












      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$












      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        16 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba♦
        15 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        12 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba♦
        9 mins ago













      3












      3








      3





      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$



      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 20 mins ago









      Kuba♦Kuba

      106k12209530




      106k12209530











      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        16 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba♦
        15 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        12 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba♦
        9 mins ago
















      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        16 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba♦
        15 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        12 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba♦
        9 mins ago















      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      16 mins ago




      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      16 mins ago












      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba♦
      15 mins ago




      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba♦
      15 mins ago












      $begingroup$
      I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
      $endgroup$
      – leastaction
      12 mins ago




      $begingroup$
      I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
      $endgroup$
      – leastaction
      12 mins ago




      1




      1




      $begingroup$
      @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
      $endgroup$
      – Kuba♦
      9 mins ago




      $begingroup$
      @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
      $endgroup$
      – Kuba♦
      9 mins ago











      2












      $begingroup$

      Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

      (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





      share|improve this answer









      $endgroup$

















        2












        $begingroup$

        Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

        (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





        share|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

          (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





          share|improve this answer









          $endgroup$



          Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

          (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 12 mins ago









          MarcoBMarcoB

          37.6k556113




          37.6k556113





















              2












              $begingroup$

              In this case, we can use RankedMin and FullSimplify to get the answer you seek



              Assuming[
               Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0, 
               FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
              (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


              This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






              share|improve this answer











              $endgroup$

















                2












                $begingroup$

                In this case, we can use RankedMin and FullSimplify to get the answer you seek



                Assuming[
                 Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0, 
                 FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






                share|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  In this case, we can use RankedMin and FullSimplify to get the answer you seek



                  Assuming[
                   Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0, 
                   FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                  (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                  This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






                  share|improve this answer











                  $endgroup$



                  In this case, we can use RankedMin and FullSimplify to get the answer you seek



                  Assuming[
                   Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0, 
                   FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                  (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                  This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 4 mins ago

























                  answered 11 mins ago









                  mikadomikado

                  6,6971929




                  6,6971929



























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