Combining pure functions containing compound expressions












1














Background



For display purposes, I sometimes find it desirable to defeat Mathematica's canonical ordering of variables, and instead have it output terms in the order in which I type them. E.g., instead of this:



b-a
v=v0+a t
a8+a9+a10



–a + b



a t + v0



a10 + a8 + a9




...I'd prefer this:




b – a



v0 + a t



a8 + a9 + a10




This can be easily accomplished by removing the Orderless attributes from Plus and Times, entering the expression, and then immediately restoring those attributes (it is important to do the latter, since their absence alters basic functionality; e.g., without Orderless, Plus doesn't recognize that a + b is mathematically identical to b + a).



ClearAttributes[Plus, Orderless]
ClearAttributes[Times, Orderless]
c (b-a)
SetAttributes[Plus, Orderless]
SetAttributes[Times, Orderless]



c (b – a)




Question



How do I implement the above as a function? I.e., I'd like to define a function such that:



func[c (b-a)]
c (b-a) (*confirm that Orderless is restored to both Plus and Times*)



c (b – a)



(–a + b) c




Thus far I've only managed the following, which achieves the desired output, but requires I enter a function, then the expression, then another function:



func1 := (ClearAttributes[Plus, Orderless]; ClearAttributes[Times, Orderless]) &;
func2 := (SetAttributes[Plus, Orderless]; SetAttributes[Times, Orderless]) &;
func1[_]; c (b-a)
func2[_]
c (b-a)



c (b – a)



(–a + b) c




How do I combine these into a single function that takes the expression as its argument?



EDIT: Note that I want to alter only the display format, not any other functionality. E.g., MMA should still recognize that e1 and e2 are mathematically equivalent:



func1[_]
e1 = c (b - a)
e2 = (-a + b) c
func2[_]
e1 == e2



c (b - a)



(-a + b) c



True











share|improve this question





























    1














    Background



    For display purposes, I sometimes find it desirable to defeat Mathematica's canonical ordering of variables, and instead have it output terms in the order in which I type them. E.g., instead of this:



    b-a
    v=v0+a t
    a8+a9+a10



    –a + b



    a t + v0



    a10 + a8 + a9




    ...I'd prefer this:




    b – a



    v0 + a t



    a8 + a9 + a10




    This can be easily accomplished by removing the Orderless attributes from Plus and Times, entering the expression, and then immediately restoring those attributes (it is important to do the latter, since their absence alters basic functionality; e.g., without Orderless, Plus doesn't recognize that a + b is mathematically identical to b + a).



    ClearAttributes[Plus, Orderless]
    ClearAttributes[Times, Orderless]
    c (b-a)
    SetAttributes[Plus, Orderless]
    SetAttributes[Times, Orderless]



    c (b – a)




    Question



    How do I implement the above as a function? I.e., I'd like to define a function such that:



    func[c (b-a)]
    c (b-a) (*confirm that Orderless is restored to both Plus and Times*)



    c (b – a)



    (–a + b) c




    Thus far I've only managed the following, which achieves the desired output, but requires I enter a function, then the expression, then another function:



    func1 := (ClearAttributes[Plus, Orderless]; ClearAttributes[Times, Orderless]) &;
    func2 := (SetAttributes[Plus, Orderless]; SetAttributes[Times, Orderless]) &;
    func1[_]; c (b-a)
    func2[_]
    c (b-a)



    c (b – a)



    (–a + b) c




    How do I combine these into a single function that takes the expression as its argument?



    EDIT: Note that I want to alter only the display format, not any other functionality. E.g., MMA should still recognize that e1 and e2 are mathematically equivalent:



    func1[_]
    e1 = c (b - a)
    e2 = (-a + b) c
    func2[_]
    e1 == e2



    c (b - a)



    (-a + b) c



    True











    share|improve this question



























      1












      1








      1







      Background



      For display purposes, I sometimes find it desirable to defeat Mathematica's canonical ordering of variables, and instead have it output terms in the order in which I type them. E.g., instead of this:



      b-a
      v=v0+a t
      a8+a9+a10



      –a + b



      a t + v0



      a10 + a8 + a9




      ...I'd prefer this:




      b – a



      v0 + a t



      a8 + a9 + a10




      This can be easily accomplished by removing the Orderless attributes from Plus and Times, entering the expression, and then immediately restoring those attributes (it is important to do the latter, since their absence alters basic functionality; e.g., without Orderless, Plus doesn't recognize that a + b is mathematically identical to b + a).



      ClearAttributes[Plus, Orderless]
      ClearAttributes[Times, Orderless]
      c (b-a)
      SetAttributes[Plus, Orderless]
      SetAttributes[Times, Orderless]



      c (b – a)




      Question



      How do I implement the above as a function? I.e., I'd like to define a function such that:



      func[c (b-a)]
      c (b-a) (*confirm that Orderless is restored to both Plus and Times*)



      c (b – a)



      (–a + b) c




      Thus far I've only managed the following, which achieves the desired output, but requires I enter a function, then the expression, then another function:



      func1 := (ClearAttributes[Plus, Orderless]; ClearAttributes[Times, Orderless]) &;
      func2 := (SetAttributes[Plus, Orderless]; SetAttributes[Times, Orderless]) &;
      func1[_]; c (b-a)
      func2[_]
      c (b-a)



      c (b – a)



      (–a + b) c




      How do I combine these into a single function that takes the expression as its argument?



      EDIT: Note that I want to alter only the display format, not any other functionality. E.g., MMA should still recognize that e1 and e2 are mathematically equivalent:



      func1[_]
      e1 = c (b - a)
      e2 = (-a + b) c
      func2[_]
      e1 == e2



      c (b - a)



      (-a + b) c



      True











      share|improve this question















      Background



      For display purposes, I sometimes find it desirable to defeat Mathematica's canonical ordering of variables, and instead have it output terms in the order in which I type them. E.g., instead of this:



      b-a
      v=v0+a t
      a8+a9+a10



      –a + b



      a t + v0



      a10 + a8 + a9




      ...I'd prefer this:




      b – a



      v0 + a t



      a8 + a9 + a10




      This can be easily accomplished by removing the Orderless attributes from Plus and Times, entering the expression, and then immediately restoring those attributes (it is important to do the latter, since their absence alters basic functionality; e.g., without Orderless, Plus doesn't recognize that a + b is mathematically identical to b + a).



      ClearAttributes[Plus, Orderless]
      ClearAttributes[Times, Orderless]
      c (b-a)
      SetAttributes[Plus, Orderless]
      SetAttributes[Times, Orderless]



      c (b – a)




      Question



      How do I implement the above as a function? I.e., I'd like to define a function such that:



      func[c (b-a)]
      c (b-a) (*confirm that Orderless is restored to both Plus and Times*)



      c (b – a)



      (–a + b) c




      Thus far I've only managed the following, which achieves the desired output, but requires I enter a function, then the expression, then another function:



      func1 := (ClearAttributes[Plus, Orderless]; ClearAttributes[Times, Orderless]) &;
      func2 := (SetAttributes[Plus, Orderless]; SetAttributes[Times, Orderless]) &;
      func1[_]; c (b-a)
      func2[_]
      c (b-a)



      c (b – a)



      (–a + b) c




      How do I combine these into a single function that takes the expression as its argument?



      EDIT: Note that I want to alter only the display format, not any other functionality. E.g., MMA should still recognize that e1 and e2 are mathematically equivalent:



      func1[_]
      e1 = c (b - a)
      e2 = (-a + b) c
      func2[_]
      e1 == e2



      c (b - a)



      (-a + b) c



      True








      output-formatting sorting output canonicalization






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago

























      asked 4 hours ago









      theorist

      1,077420




      1,077420






















          1 Answer
          1






          active

          oldest

          votes


















          4














          You could define a format that does this for you:



          SetAttributes[orderlessForm, HoldFirst]

          MakeBoxes[orderlessForm[expr_], form_] ^:= Internal`InheritedBlock[{Times, Plus},
          ClearAttributes[{Times, Plus}, Orderless];
          MakeBoxes[expr, form]
          ]


          Then:



          orderlessForm[c (b - a)]



          c (b - a)




          And, the usual output when not using the wrapper:



          c (b - a)



          (-a + b) c




          Note that the HoldFirst attribute does most of the work.






          share|improve this answer





















          • Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
            – theorist
            2 hours ago











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4














          You could define a format that does this for you:



          SetAttributes[orderlessForm, HoldFirst]

          MakeBoxes[orderlessForm[expr_], form_] ^:= Internal`InheritedBlock[{Times, Plus},
          ClearAttributes[{Times, Plus}, Orderless];
          MakeBoxes[expr, form]
          ]


          Then:



          orderlessForm[c (b - a)]



          c (b - a)




          And, the usual output when not using the wrapper:



          c (b - a)



          (-a + b) c




          Note that the HoldFirst attribute does most of the work.






          share|improve this answer





















          • Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
            – theorist
            2 hours ago
















          4














          You could define a format that does this for you:



          SetAttributes[orderlessForm, HoldFirst]

          MakeBoxes[orderlessForm[expr_], form_] ^:= Internal`InheritedBlock[{Times, Plus},
          ClearAttributes[{Times, Plus}, Orderless];
          MakeBoxes[expr, form]
          ]


          Then:



          orderlessForm[c (b - a)]



          c (b - a)




          And, the usual output when not using the wrapper:



          c (b - a)



          (-a + b) c




          Note that the HoldFirst attribute does most of the work.






          share|improve this answer





















          • Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
            – theorist
            2 hours ago














          4












          4








          4






          You could define a format that does this for you:



          SetAttributes[orderlessForm, HoldFirst]

          MakeBoxes[orderlessForm[expr_], form_] ^:= Internal`InheritedBlock[{Times, Plus},
          ClearAttributes[{Times, Plus}, Orderless];
          MakeBoxes[expr, form]
          ]


          Then:



          orderlessForm[c (b - a)]



          c (b - a)




          And, the usual output when not using the wrapper:



          c (b - a)



          (-a + b) c




          Note that the HoldFirst attribute does most of the work.






          share|improve this answer












          You could define a format that does this for you:



          SetAttributes[orderlessForm, HoldFirst]

          MakeBoxes[orderlessForm[expr_], form_] ^:= Internal`InheritedBlock[{Times, Plus},
          ClearAttributes[{Times, Plus}, Orderless];
          MakeBoxes[expr, form]
          ]


          Then:



          orderlessForm[c (b - a)]



          c (b - a)




          And, the usual output when not using the wrapper:



          c (b - a)



          (-a + b) c




          Note that the HoldFirst attribute does most of the work.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          Carl Woll

          67k387175




          67k387175












          • Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
            – theorist
            2 hours ago


















          • Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
            – theorist
            2 hours ago
















          Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
          – theorist
          2 hours ago




          Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g., e1 = orderlessForm[a + b]; e2 = orderlessForm[b + a]; e1 == e2 gives a + b == b + a instead of True. I've added an edit at the end of my question to make this requirement explicit.
          – theorist
          2 hours ago


















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