Combining pure functions containing compound expressions
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
add a comment |
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
add a comment |
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
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
edited 2 hours ago
asked 4 hours ago
theorist
1,077420
1,077420
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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.
Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g.,e1 = orderlessForm[a + b];
e2 = orderlessForm[b + a];
e1 == e2
givesa + b == b + a
instead ofTrue
. I've added an edit at the end of my question to make this requirement explicit.
– theorist
2 hours ago
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
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.
Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g.,e1 = orderlessForm[a + b];
e2 = orderlessForm[b + a];
e1 == e2
givesa + b == b + a
instead ofTrue
. I've added an edit at the end of my question to make this requirement explicit.
– theorist
2 hours ago
add a comment |
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.
Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g.,e1 = orderlessForm[a + b];
e2 = orderlessForm[b + a];
e1 == e2
givesa + b == b + a
instead ofTrue
. I've added an edit at the end of my question to make this requirement explicit.
– theorist
2 hours ago
add a comment |
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.
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.
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
givesa + b == b + a
instead ofTrue
. I've added an edit at the end of my question to make this requirement explicit.
– theorist
2 hours ago
add a comment |
Thanks Carl. One problem I notice is that orderlessForm alters some functionality. E.g.,e1 = orderlessForm[a + b];
e2 = orderlessForm[b + a];
e1 == e2
givesa + b == b + a
instead ofTrue
. 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
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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