Convert defined variables to rule list












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







share|improve this question











$endgroup$








  • 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
    9 hours ago










  • $begingroup$
    Is OwnValues /@ Unevaluated@{x, y, z} OK?
    $endgroup$
    – xzczd
    2 hours ago
















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







share|improve this question











$endgroup$








  • 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
    9 hours ago










  • $begingroup$
    Is OwnValues /@ Unevaluated@{x, y, z} OK?
    $endgroup$
    – xzczd
    2 hours ago














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







share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago









march

17.2k22769




17.2k22769










asked 9 hours ago









Chris KChris K

6,58421841




6,58421841








  • 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
    9 hours ago










  • $begingroup$
    Is OwnValues /@ Unevaluated@{x, y, z} OK?
    $endgroup$
    – xzczd
    2 hours ago














  • 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
    9 hours ago










  • $begingroup$
    Is OwnValues /@ Unevaluated@{x, y, z} OK?
    $endgroup$
    – xzczd
    2 hours ago








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
9 hours ago




$begingroup$
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
9 hours ago












$begingroup$
Is OwnValues /@ Unevaluated@{x, y, z} OK?
$endgroup$
– xzczd
2 hours ago




$begingroup$
Is OwnValues /@ Unevaluated@{x, y, z} OK?
$endgroup$
– xzczd
2 hours ago










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} *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    @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
    8 hours ago












  • $begingroup$
    I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    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
    8 hours ago





















3












$begingroup$

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} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago











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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} *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    @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
    8 hours ago












  • $begingroup$
    I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    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
    8 hours ago


















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} *)





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    @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
    8 hours ago












  • $begingroup$
    I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    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
    8 hours ago
















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} *)





share|improve this answer











$endgroup$



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} *)






share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 9 hours ago









marchmarch

17.2k22769




17.2k22769












  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    @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
    8 hours ago












  • $begingroup$
    I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    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
    8 hours ago




















  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    @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
    8 hours ago












  • $begingroup$
    I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
    $endgroup$
    – Chris K
    8 hours ago










  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    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
    8 hours ago


















$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
8 hours ago




$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
8 hours ago












$begingroup$
@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
8 hours ago






$begingroup$
@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
8 hours ago














$begingroup$
I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
$endgroup$
– Chris K
8 hours ago




$begingroup$
I think the trick from your link works: variableToRule[var_] := SymbolName[Unevaluated@var] -> var seems OK
$endgroup$
– Chris K
8 hours ago












$begingroup$
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
8 hours ago




$begingroup$
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
8 hours ago












$begingroup$
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
8 hours ago






$begingroup$
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
8 hours ago













3












$begingroup$

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} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago
















3












$begingroup$

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} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago














3












3








3





$begingroup$

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} *)





share|improve this answer









$endgroup$



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} *)






share|improve this answer












share|improve this answer



share|improve this answer










answered 8 hours ago









ShredderroyShredderroy

1,5701115




1,5701115












  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago


















  • $begingroup$
    Thanks - any way to keep the variables' values afterwards?
    $endgroup$
    – Chris K
    8 hours ago
















$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
8 hours ago




$begingroup$
Thanks - any way to keep the variables' values afterwards?
$endgroup$
– Chris K
8 hours ago


















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