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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













6












$begingroup$


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.










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
    Mar 19 at 21:31















6












$begingroup$


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.










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
    Mar 19 at 21:31













6












6








6


3



$begingroup$


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.










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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 19 at 21:09









leastactionleastaction

244210




244210







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    Mar 19 at 21:31












  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    Mar 19 at 21:31







1




1




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31










4 Answers
4






active

oldest

votes


















5












$begingroup$

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
    Mar 19 at 21:38


















5












$begingroup$

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
    Mar 19 at 21:31










  • $begingroup$
    @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
    $endgroup$
    – Kuba
    Mar 19 at 21:32










  • $begingroup$
    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
    Mar 19 at 21:36






  • 1




    $begingroup$
    @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
    $endgroup$
    – Kuba
    Mar 19 at 21:38



















4












$begingroup$

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









$endgroup$




















    4












    $begingroup$

    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









      5












      $begingroup$

      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
        Mar 19 at 21:38















      5












      $begingroup$

      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
        Mar 19 at 21:38













      5












      5








      5





      $begingroup$

      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 Mar 19 at 21:35









      Carl WollCarl Woll

      71.4k394186




      71.4k394186











      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        Mar 19 at 21:38
















      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        Mar 19 at 21:38















      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      Mar 19 at 21:38




      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      Mar 19 at 21:38











      5












      $begingroup$

      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
        Mar 19 at 21:31










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        Mar 19 at 21:32










      • $begingroup$
        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
        Mar 19 at 21:36






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba
        Mar 19 at 21:38
















      5












      $begingroup$

      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
        Mar 19 at 21:31










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        Mar 19 at 21:32










      • $begingroup$
        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
        Mar 19 at 21:36






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba
        Mar 19 at 21:38














      5












      5








      5





      $begingroup$

      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 Mar 19 at 21:28









      KubaKuba

      107k12210530




      107k12210530











      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        Mar 19 at 21:31










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        Mar 19 at 21:32










      • $begingroup$
        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
        Mar 19 at 21:36






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba
        Mar 19 at 21:38

















      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        Mar 19 at 21:31










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        Mar 19 at 21:32










      • $begingroup$
        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
        Mar 19 at 21:36






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba
        Mar 19 at 21:38
















      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      Mar 19 at 21:31




      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      Mar 19 at 21:31












      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba
      Mar 19 at 21:32




      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba
      Mar 19 at 21:32












      $begingroup$
      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
      Mar 19 at 21:36




      $begingroup$
      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
      Mar 19 at 21:36




      1




      1




      $begingroup$
      @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
      $endgroup$
      – Kuba
      Mar 19 at 21:38





      $begingroup$
      @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
      $endgroup$
      – Kuba
      Mar 19 at 21:38












      4












      $begingroup$

      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









      $endgroup$

















        4












        $begingroup$

        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









        $endgroup$















          4












          4








          4





          $begingroup$

          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 Mar 19 at 21:35









          MarcoBMarcoB

          37.9k556114




          37.9k556114





















              4












              $begingroup$

              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











              $endgroup$

















                4












                $begingroup$

                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











                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  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 Mar 19 at 21:43

























                  answered Mar 19 at 21:36









                  mikadomikado

                  6,7171929




                  6,7171929



























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