Sort with assumptions












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.






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
    yesterday
















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.






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
    yesterday














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.






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 yesterday









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
    yesterday














  • 1




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








1




1




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




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










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
    yesterday



















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
    yesterday










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










  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
    $endgroup$
    – Kuba♦
    yesterday





















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











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "387"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193600%2fsort-with-assumptions%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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
        yesterday
















      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
        yesterday














      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 yesterday









      Carl WollCarl Woll

      70.9k394185




      70.9k394185












      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        yesterday


















      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        yesterday
















      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      yesterday




      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      yesterday











      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
        yesterday










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










      • $begingroup$
        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
        yesterday






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba♦
        yesterday


















      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
        yesterday










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










      • $begingroup$
        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
        yesterday






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba♦
        yesterday
















      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 yesterday









      Kuba♦Kuba

      107k12210530




      107k12210530












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










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










      • $begingroup$
        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
        yesterday






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba♦
        yesterday




















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










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










      • $begingroup$
        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
        yesterday






      • 1




        $begingroup$
        @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
        $endgroup$
        – Kuba♦
        yesterday


















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




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












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




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












      $begingroup$
      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
      yesterday




      $begingroup$
      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
      yesterday




      1




      1




      $begingroup$
      @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
      $endgroup$
      – Kuba♦
      yesterday






      $begingroup$
      @leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
      $endgroup$
      – Kuba♦
      yesterday













      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 yesterday









          MarcoBMarcoB

          37.7k556113




          37.7k556113























              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 yesterday

























                  answered yesterday









                  mikadomikado

                  6,7171929




                  6,7171929






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematica Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193600%2fsort-with-assumptions%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      數位音樂下載

                      Image
                      數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                      Read more

                      When can things happen in Etherscan, such as the picture below?

                      Image
                      2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
                      Read more

                      格利澤436b

                      Image
                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                      Read more