Force evaluation of user-defined function
$begingroup$
Consider a function that sorts its arguments based on a function called order
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2]
expr2 = f[2, 1]
(* f[1, 2] *)
(* f[1, 2] *)
However, let's imagine that we change the ordering function
order[1] = 2;
order[2] = 1;
Surprisingly, expr1 and expr2 still evaluate to f[1,2]
expr1
expr2
(* f[1, 2] *)
(* f[1, 2] *)
However, if we introduce f[1, 2] we obtain the correct result
f[1, 2]
(* f[2, 1] *)
Is there a way to force the reevaluation of expr to obtain f[2, 1]? I am looking for a function along the lines of
ForceEvaluation@expr1
ForceEvaluation@expr2
(* f[2, 1] *)
(* f[2, 1] *)
functions function-construction evaluation
asked Dec 19 '18 at 17:38
AGimAGim
465
$endgroup$
add a comment |
$begingroup$
Consider a function that sorts its arguments based on a function called order
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2]
expr2 = f[2, 1]
(* f[1, 2] *)
(* f[1, 2] *)
However, let's imagine that we change the ordering function
order[1] = 2;
order[2] = 1;
Surprisingly, expr1 and expr2 still evaluate to f[1,2]
expr1
expr2
(* f[1, 2] *)
(* f[1, 2] *)
However, if we introduce f[1, 2] we obtain the correct result
f[1, 2]
(* f[2, 1] *)
Is there a way to force the reevaluation of expr to obtain f[2, 1]? I am looking for a function along the lines of
ForceEvaluation@expr1
ForceEvaluation@expr2
(* f[2, 1] *)
(* f[2, 1] *)
functions function-construction evaluation
asked Dec 19 '18 at 17:38
AGimAGim
465
$endgroup$
1
$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
1
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26
add a comment |
$begingroup$
Consider a function that sorts its arguments based on a function called order
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2]
expr2 = f[2, 1]
(* f[1, 2] *)
(* f[1, 2] *)
However, let's imagine that we change the ordering function
order[1] = 2;
order[2] = 1;
Surprisingly, expr1 and expr2 still evaluate to f[1,2]
expr1
expr2
(* f[1, 2] *)
(* f[1, 2] *)
However, if we introduce f[1, 2] we obtain the correct result
f[1, 2]
(* f[2, 1] *)
Is there a way to force the reevaluation of expr to obtain f[2, 1]? I am looking for a function along the lines of
ForceEvaluation@expr1
ForceEvaluation@expr2
(* f[2, 1] *)
(* f[2, 1] *)
functions function-construction evaluation
asked Dec 19 '18 at 17:38
AGimAGim
465
$endgroup$
Consider a function that sorts its arguments based on a function called order
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2]
expr2 = f[2, 1]
(* f[1, 2] *)
(* f[1, 2] *)
However, let's imagine that we change the ordering function
order[1] = 2;
order[2] = 1;
Surprisingly, expr1 and expr2 still evaluate to f[1,2]
expr1
expr2
(* f[1, 2] *)
(* f[1, 2] *)
However, if we introduce f[1, 2] we obtain the correct result
f[1, 2]
(* f[2, 1] *)
Is there a way to force the reevaluation of expr to obtain f[2, 1]? I am looking for a function along the lines of
ForceEvaluation@expr1
ForceEvaluation@expr2
(* f[2, 1] *)
(* f[2, 1] *)
functions function-construction evaluation
functions function-construction evaluation
asked Dec 19 '18 at 17:38
AGimAGim
465
asked Dec 19 '18 at 17:38
AGimAGim
465
edited Dec 19 '18 at 17:46
AGim
asked Dec 19 '18 at 17:38
AGimAGim
465
asked Dec 19 '18 at 17:38
AGimAGim
465
asked Dec 19 '18 at 17:38
AGimAGim
465
465
1
$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
1
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26
add a comment |
1
$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
1
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26
1
1
$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
1
1
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Stumbled across the solution to this in tutorial/ControllingInfiniteEvaluation in the documentation. The last paragraph says:
Some of the trickiest cases occur when you have rules that depend on complicated
/;conditions (see "Putting Constraints on Patterns"). One particularly awkward case is when the condition involves a "global variable". The Wolfram Language may think that the evaluation is finished because the expression did not change. However, a side effect of some other operation could change the value of the global variable, and so should lead to a new result in the evaluation. The best way to avoid this kind of difficulty is not to use global variables in/;conditions. If all else fails, you can typeUpdate[s]to tell the Wolfram Language to update all expressions involvings.Updatetells the Wolfram Language to update absolutely all expressions.
The symbol to updated in this case is f, rather than order, but:
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2];
expr2 = f[2, 1];
{expr1, expr2}
{f[1, 2], f[1, 2]}
order[1] = 2;
order[2] = 1;
Update[f];
{expr1, expr2}
{f[2, 1], f[2, 1]}
Clunky, but it works.
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
$endgroup$
add a comment |
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1 Answer
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active
oldest
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oldest
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$begingroup$
Stumbled across the solution to this in tutorial/ControllingInfiniteEvaluation in the documentation. The last paragraph says:
Some of the trickiest cases occur when you have rules that depend on complicated
/;conditions (see "Putting Constraints on Patterns"). One particularly awkward case is when the condition involves a "global variable". The Wolfram Language may think that the evaluation is finished because the expression did not change. However, a side effect of some other operation could change the value of the global variable, and so should lead to a new result in the evaluation. The best way to avoid this kind of difficulty is not to use global variables in/;conditions. If all else fails, you can typeUpdate[s]to tell the Wolfram Language to update all expressions involvings.Updatetells the Wolfram Language to update absolutely all expressions.
The symbol to updated in this case is f, rather than order, but:
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2];
expr2 = f[2, 1];
{expr1, expr2}
{f[1, 2], f[1, 2]}
order[1] = 2;
order[2] = 1;
Update[f];
{expr1, expr2}
{f[2, 1], f[2, 1]}
Clunky, but it works.
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
$endgroup$
add a comment |
$begingroup$
Stumbled across the solution to this in tutorial/ControllingInfiniteEvaluation in the documentation. The last paragraph says:
Some of the trickiest cases occur when you have rules that depend on complicated
/;conditions (see "Putting Constraints on Patterns"). One particularly awkward case is when the condition involves a "global variable". The Wolfram Language may think that the evaluation is finished because the expression did not change. However, a side effect of some other operation could change the value of the global variable, and so should lead to a new result in the evaluation. The best way to avoid this kind of difficulty is not to use global variables in/;conditions. If all else fails, you can typeUpdate[s]to tell the Wolfram Language to update all expressions involvings.Updatetells the Wolfram Language to update absolutely all expressions.
The symbol to updated in this case is f, rather than order, but:
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2];
expr2 = f[2, 1];
{expr1, expr2}
{f[1, 2], f[1, 2]}
order[1] = 2;
order[2] = 1;
Update[f];
{expr1, expr2}
{f[2, 1], f[2, 1]}
Clunky, but it works.
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
$endgroup$
add a comment |
$begingroup$
Stumbled across the solution to this in tutorial/ControllingInfiniteEvaluation in the documentation. The last paragraph says:
Some of the trickiest cases occur when you have rules that depend on complicated
/;conditions (see "Putting Constraints on Patterns"). One particularly awkward case is when the condition involves a "global variable". The Wolfram Language may think that the evaluation is finished because the expression did not change. However, a side effect of some other operation could change the value of the global variable, and so should lead to a new result in the evaluation. The best way to avoid this kind of difficulty is not to use global variables in/;conditions. If all else fails, you can typeUpdate[s]to tell the Wolfram Language to update all expressions involvings.Updatetells the Wolfram Language to update absolutely all expressions.
The symbol to updated in this case is f, rather than order, but:
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2];
expr2 = f[2, 1];
{expr1, expr2}
{f[1, 2], f[1, 2]}
order[1] = 2;
order[2] = 1;
Update[f];
{expr1, expr2}
{f[2, 1], f[2, 1]}
Clunky, but it works.
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
$endgroup$
Stumbled across the solution to this in tutorial/ControllingInfiniteEvaluation in the documentation. The last paragraph says:
Some of the trickiest cases occur when you have rules that depend on complicated
/;conditions (see "Putting Constraints on Patterns"). One particularly awkward case is when the condition involves a "global variable". The Wolfram Language may think that the evaluation is finished because the expression did not change. However, a side effect of some other operation could change the value of the global variable, and so should lead to a new result in the evaluation. The best way to avoid this kind of difficulty is not to use global variables in/;conditions. If all else fails, you can typeUpdate[s]to tell the Wolfram Language to update all expressions involvings.Updatetells the Wolfram Language to update absolutely all expressions.
The symbol to updated in this case is f, rather than order, but:
Clear[f, order]
f[x_, y_] /; order[x] > order[y] := f[y, x]
order[1] = 1;
order[2] = 2;
expr1 = f[1, 2];
expr2 = f[2, 1];
{expr1, expr2}
{f[1, 2], f[1, 2]}
order[1] = 2;
order[2] = 1;
Update[f];
{expr1, expr2}
{f[2, 1], f[2, 1]}
Clunky, but it works.
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
answered Dec 19 '18 at 18:05
eyorbleeyorble
5,1031826
5,1031826
add a comment |
add a comment |
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$begingroup$
While I appreciate the accept, I'd recommend holding off for at least a few more hours. There are plenty of users who know more about Mathematica than I who might be able to provide a better answer, but once an answer is accepted many people will start skipping the question entirely.
$endgroup$
– eyorble
Dec 19 '18 at 18:12
$begingroup$
Thanks for the comment, since I am new I did not consider this. It is certainly an obscure behaviour of Mathematica, and it would be great to hear more about it!
$endgroup$
– AGim
Dec 19 '18 at 18:17
1
$begingroup$
Related, but not as minimal an example: (139461)
$endgroup$
– Mr.Wizard♦
Dec 19 '18 at 21:26