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







output-formatting sorting output canonicalization







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







    output-formatting sorting output canonicalization







    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







      output-formatting sorting output canonicalization







      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




      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





          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%2f188576%2fcombining-pure-functions-containing-compound-expressions%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f188576%2fcombining-pure-functions-containing-compound-expressions%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

          サソリ

          Image
          この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
          Read more

          広島県道265号伴広島線

          Image
          一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
          Read more

          Accessing regular linux commands in Huawei's Dopra Linux

          Image
          4 Backstory: I have a Huawei HG8245 router in my house and I want to change the default username/password. I've tried following this guide but the config file always gets reset. My ISP isn't helping either so I'm looking for other ways to accomplish this. The router has telnet access so this seems like a way to change the credentials. The first shell I gain access to looks like this: WAP> Pressing "?" gives me a list of available commands which I've pasted here, the two interesting ones are "su" and "shell". I can't seem to get su access, even after logging in as root (challenge password prompt, and then nothing happens), but shell gives me a proper linux shell. WAP>shell BusyBox v1.18.4 (2017-12-26 17:06:34 CST) built-in shell (ash) Enter 'hel
          Read more