Convert defined variables to rule list
Clash Royale CLAN TAG#URR8PPP
3
$begingroup$
I'd like to make a function that takes a list of variables and returns a corresponding rule list with the current values of the variables. E.g.
x = 1;
y = 2;
VariablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> z *)
Is this even possible?
function-construction evaluation replacement
edited Feb 6 at 20:52
march
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
$endgroup$
add a comment |
3
$begingroup$
I'd like to make a function that takes a list of variables and returns a corresponding rule list with the current values of the variables. E.g.
x = 1;
y = 2;
VariablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> z *)
Is this even possible?
function-construction evaluation replacement
edited Feb 6 at 20:52
march
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
$endgroup$
1
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order toClearthem, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.
$endgroup$
– march
Feb 6 at 20:34
1
$begingroup$
IsOwnValues /@ Unevaluated@x, y, zOK?
$endgroup$
– xzczd
Feb 7 at 3:28
add a comment |
3
3
3
$begingroup$
I'd like to make a function that takes a list of variables and returns a corresponding rule list with the current values of the variables. E.g.
x = 1;
y = 2;
VariablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> z *)
Is this even possible?
function-construction evaluation replacement
edited Feb 6 at 20:52
march
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
$endgroup$
I'd like to make a function that takes a list of variables and returns a corresponding rule list with the current values of the variables. E.g.
x = 1;
y = 2;
VariablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> z *)
Is this even possible?
function-construction evaluation replacement
function-construction evaluation replacement
edited Feb 6 at 20:52
march
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
edited Feb 6 at 20:52
march
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
edited Feb 6 at 20:52
march
17.3k22769
edited Feb 6 at 20:52
march
17.3k22769
edited Feb 6 at 20:52
march
17.3k22769
17.3k22769
asked Feb 6 at 20:15
Chris KChris K
6,75421842
asked Feb 6 at 20:15
Chris KChris K
6,75421842
asked Feb 6 at 20:15
Chris KChris K
6,75421842
6,75421842
1
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order toClearthem, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.
$endgroup$
– march
Feb 6 at 20:34
1
$begingroup$
IsOwnValues /@ Unevaluated@x, y, zOK?
$endgroup$
– xzczd
Feb 7 at 3:28
add a comment |
1
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order toClearthem, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.
$endgroup$
– march
Feb 6 at 20:34
1
$begingroup$
IsOwnValues /@ Unevaluated@x, y, zOK?
$endgroup$
– xzczd
Feb 7 at 3:28
1
1
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order to
Clear them, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.$endgroup$
– march
Feb 6 at 20:34
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order to
Clear them, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.$endgroup$
– march
Feb 6 at 20:34
1
1
$begingroup$
Is
OwnValues /@ Unevaluated@x, y, z OK?$endgroup$
– xzczd
Feb 7 at 3:28
$begingroup$
Is
OwnValues /@ Unevaluated@x, y, z OK?$endgroup$
– xzczd
Feb 7 at 3:28
add a comment |
2 Answers
2
active
oldest
votes
4
$begingroup$
Update 2
Based on a suggestion by Somos, the following version is nicer. According to what the OP wants:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := SymbolName@Unevaluated@var -> var
and according to my original interpretation of the problem:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := Module[val = var, Clear@var; var -> val]
Update 1
After some comments from the OP, it seems they want instead something like
variableToRule[var_] := SymbolName@Unevaluated@var -> var
instead.
Original Post
Here's a first iteration. First define the helper function,
ClearAll@variableToRule
SetAttributes[variableToRule, HoldAll]
variableToRule[var_] := Module[val = var
, Clear@var
; var -> val
]
Then, the function is
ClearAll@variablesToRules
SetAttributes[variablesToRules, HoldAll]
variablesToRules[vars_List] := variableToRule /@ Unevaluated@vars
This uses the trick from this answer.
Then,
x = 1; y = 2; z = 3;
variablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:38
marchmarch
17.3k22769
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values ofxandy, etc. in expressions that contain those variables? In that place, you don't wantxandyset before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get1 -> 1, 2 -> 2.
$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:variableToRule[var_] := SymbolName[Unevaluated@var] -> varseems OK
$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
|
show 6 more comments
4
$begingroup$
Pass in the names of the symbols as strings, and the rest is quite easy:
ClearAll[varsToRules];
varsToRules[s_] := With[t = Map[Symbol, s],
s // Apply[ClearAll];
MapThread[Rule, Symbol /@ s, t]
];
ClearAll[x, y];
x, y, z = 1, 2, 3;
varsToRules["x", "y", "z"]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
Update 2
Based on a suggestion by Somos, the following version is nicer. According to what the OP wants:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := SymbolName@Unevaluated@var -> var
and according to my original interpretation of the problem:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := Module[val = var, Clear@var; var -> val]
Update 1
After some comments from the OP, it seems they want instead something like
variableToRule[var_] := SymbolName@Unevaluated@var -> var
instead.
Original Post
Here's a first iteration. First define the helper function,
ClearAll@variableToRule
SetAttributes[variableToRule, HoldAll]
variableToRule[var_] := Module[val = var
, Clear@var
; var -> val
]
Then, the function is
ClearAll@variablesToRules
SetAttributes[variablesToRules, HoldAll]
variablesToRules[vars_List] := variableToRule /@ Unevaluated@vars
This uses the trick from this answer.
Then,
x = 1; y = 2; z = 3;
variablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:38
marchmarch
17.3k22769
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values ofxandy, etc. in expressions that contain those variables? In that place, you don't wantxandyset before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get1 -> 1, 2 -> 2.
$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:variableToRule[var_] := SymbolName[Unevaluated@var] -> varseems OK
$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
|
show 6 more comments
4
$begingroup$
Update 2
Based on a suggestion by Somos, the following version is nicer. According to what the OP wants:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := SymbolName@Unevaluated@var -> var
and according to my original interpretation of the problem:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := Module[val = var, Clear@var; var -> val]
Update 1
After some comments from the OP, it seems they want instead something like
variableToRule[var_] := SymbolName@Unevaluated@var -> var
instead.
Original Post
Here's a first iteration. First define the helper function,
ClearAll@variableToRule
SetAttributes[variableToRule, HoldAll]
variableToRule[var_] := Module[val = var
, Clear@var
; var -> val
]
Then, the function is
ClearAll@variablesToRules
SetAttributes[variablesToRules, HoldAll]
variablesToRules[vars_List] := variableToRule /@ Unevaluated@vars
This uses the trick from this answer.
Then,
x = 1; y = 2; z = 3;
variablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:38
marchmarch
17.3k22769
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values ofxandy, etc. in expressions that contain those variables? In that place, you don't wantxandyset before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get1 -> 1, 2 -> 2.
$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:variableToRule[var_] := SymbolName[Unevaluated@var] -> varseems OK
$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
|
show 6 more comments
4
4
4
$begingroup$
Update 2
Based on a suggestion by Somos, the following version is nicer. According to what the OP wants:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := SymbolName@Unevaluated@var -> var
and according to my original interpretation of the problem:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := Module[val = var, Clear@var; var -> val]
Update 1
After some comments from the OP, it seems they want instead something like
variableToRule[var_] := SymbolName@Unevaluated@var -> var
instead.
Original Post
Here's a first iteration. First define the helper function,
ClearAll@variableToRule
SetAttributes[variableToRule, HoldAll]
variableToRule[var_] := Module[val = var
, Clear@var
; var -> val
]
Then, the function is
ClearAll@variablesToRules
SetAttributes[variablesToRules, HoldAll]
variablesToRules[vars_List] := variableToRule /@ Unevaluated@vars
This uses the trick from this answer.
Then,
x = 1; y = 2; z = 3;
variablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:38
marchmarch
17.3k22769
$endgroup$
Update 2
Based on a suggestion by Somos, the following version is nicer. According to what the OP wants:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := SymbolName@Unevaluated@var -> var
and according to my original interpretation of the problem:
SetAttributes[variableToRule, HoldAll, Listable]
variableToRule[var_] := Module[val = var, Clear@var; var -> val]
Update 1
After some comments from the OP, it seems they want instead something like
variableToRule[var_] := SymbolName@Unevaluated@var -> var
instead.
Original Post
Here's a first iteration. First define the helper function,
ClearAll@variableToRule
SetAttributes[variableToRule, HoldAll]
variableToRule[var_] := Module[val = var
, Clear@var
; var -> val
]
Then, the function is
ClearAll@variablesToRules
SetAttributes[variablesToRules, HoldAll]
variablesToRules[vars_List] := variableToRule /@ Unevaluated@vars
This uses the trick from this answer.
Then,
x = 1; y = 2; z = 3;
variablesToRules[x, y, z]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:38
marchmarch
17.3k22769
edited Feb 6 at 23:16
answered Feb 6 at 20:38
marchmarch
17.3k22769
answered Feb 6 at 20:38
marchmarch
17.3k22769
answered Feb 6 at 20:38
marchmarch
17.3k22769
17.3k22769
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values ofxandy, etc. in expressions that contain those variables? In that place, you don't wantxandyset before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get1 -> 1, 2 -> 2.
$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:variableToRule[var_] := SymbolName[Unevaluated@var] -> varseems OK
$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
|
show 6 more comments
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values ofxandy, etc. in expressions that contain those variables? In that place, you don't wantxandyset before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get1 -> 1, 2 -> 2.
$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:variableToRule[var_] := SymbolName[Unevaluated@var] -> varseems OK
$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values of
x and y, etc. in expressions that contain those variables? In that place, you don't want x and y set before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get 1 -> 1, 2 -> 2.$endgroup$
– march
Feb 6 at 21:02
$begingroup$
@ChrisK. Your question confuses me. If you're making replacement rules, isn't the whole point that you will use them to replace the values of
x and y, etc. in expressions that contain those variables? In that place, you don't want x and y set before-hand. What is it that you are trying to do here? Is this just for display purposes or something? If you keep the variables set, then you will get 1 -> 1, 2 -> 2.$endgroup$
– march
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:
variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
I think the trick from your link works:
variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK$endgroup$
– Chris K
Feb 6 at 21:02
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
But then, what is the point of the replacement rule? Because that won't actually work as a replacement rule, because you have the symbol name (which is a string), instead of the symbol itself.
$endgroup$
– march
Feb 6 at 21:04
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
$begingroup$
It's a bit hard to explain, but I've got an inner function that takes a list of rules as an argument, which needs to be defined in an outer function where the variables are already defined. Anyhow I think I'm sorted now. Thanks!
$endgroup$
– Chris K
Feb 6 at 21:07
|
show 6 more comments
4
$begingroup$
Pass in the names of the symbols as strings, and the rest is quite easy:
ClearAll[varsToRules];
varsToRules[s_] := With[t = Map[Symbol, s],
s // Apply[ClearAll];
MapThread[Rule, Symbol /@ s, t]
];
ClearAll[x, y];
x, y, z = 1, 2, 3;
varsToRules["x", "y", "z"]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
add a comment |
4
$begingroup$
Pass in the names of the symbols as strings, and the rest is quite easy:
ClearAll[varsToRules];
varsToRules[s_] := With[t = Map[Symbol, s],
s // Apply[ClearAll];
MapThread[Rule, Symbol /@ s, t]
];
ClearAll[x, y];
x, y, z = 1, 2, 3;
varsToRules["x", "y", "z"]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
$endgroup$
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
add a comment |
4
4
4
$begingroup$
Pass in the names of the symbols as strings, and the rest is quite easy:
ClearAll[varsToRules];
varsToRules[s_] := With[t = Map[Symbol, s],
s // Apply[ClearAll];
MapThread[Rule, Symbol /@ s, t]
];
ClearAll[x, y];
x, y, z = 1, 2, 3;
varsToRules["x", "y", "z"]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
$endgroup$
Pass in the names of the symbols as strings, and the rest is quite easy:
ClearAll[varsToRules];
varsToRules[s_] := With[t = Map[Symbol, s],
s // Apply[ClearAll];
MapThread[Rule, Symbol /@ s, t]
];
ClearAll[x, y];
x, y, z = 1, 2, 3;
varsToRules["x", "y", "z"]
(* x -> 1, y -> 2, z -> 3 *)
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
answered Feb 6 at 20:45
ShredderroyShredderroy
1,5901116
1,5901116
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
add a comment |
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
Feb 6 at 20:57
add a comment |
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1
$begingroup$
This is a super interesting question. This post is related, because you will need to get the symbol names in order to
Clearthem, but it's not clear how to short-circuit the evaluation when you will be feeding a list of variable names to the function rather than just the variable name.$endgroup$
– march
Feb 6 at 20:34
1
$begingroup$
Is
OwnValues /@ Unevaluated@x, y, zOK?$endgroup$
– xzczd
Feb 7 at 3:28