Perform localized, evaluation-leak free replacements
Consider the following expression:
expr = Hold[
{
f[
{1, {Print@1}}
],
g[
{{{Print@1}}}
]
}
];
I'm looking for a way to apply a replacement rule to the contents of all expressions with head f. To give a more concrete example, let's say I want to replace Print by Echo, but only inside f.
Restrictions: I have no knowledge about the exact structure around and inside f - there could be more or less nesting going on. This means the replacement rule can't capture the wrapper with Hold attribute, nor can it capture the individual parts of f[…]. (And the evaluation of Print should be prevented of course)
evaluation replacement hold
edited 10 hours ago
Kuba♦
103k12201516
asked yesterday
Lukas Lang
6,4351929
add a comment |
Consider the following expression:
expr = Hold[
{
f[
{1, {Print@1}}
],
g[
{{{Print@1}}}
]
}
];
I'm looking for a way to apply a replacement rule to the contents of all expressions with head f. To give a more concrete example, let's say I want to replace Print by Echo, but only inside f.
Restrictions: I have no knowledge about the exact structure around and inside f - there could be more or less nesting going on. This means the replacement rule can't capture the wrapper with Hold attribute, nor can it capture the individual parts of f[…]. (And the evaluation of Print should be prevented of course)
evaluation replacement hold
edited 10 hours ago
Kuba♦
103k12201516
asked yesterday
Lukas Lang
6,4351929
add a comment |
Consider the following expression:
expr = Hold[
{
f[
{1, {Print@1}}
],
g[
{{{Print@1}}}
]
}
];
I'm looking for a way to apply a replacement rule to the contents of all expressions with head f. To give a more concrete example, let's say I want to replace Print by Echo, but only inside f.
Restrictions: I have no knowledge about the exact structure around and inside f - there could be more or less nesting going on. This means the replacement rule can't capture the wrapper with Hold attribute, nor can it capture the individual parts of f[…]. (And the evaluation of Print should be prevented of course)
evaluation replacement hold
edited 10 hours ago
Kuba♦
103k12201516
asked yesterday
Lukas Lang
6,4351929
Consider the following expression:
expr = Hold[
{
f[
{1, {Print@1}}
],
g[
{{{Print@1}}}
]
}
];
I'm looking for a way to apply a replacement rule to the contents of all expressions with head f. To give a more concrete example, let's say I want to replace Print by Echo, but only inside f.
Restrictions: I have no knowledge about the exact structure around and inside f - there could be more or less nesting going on. This means the replacement rule can't capture the wrapper with Hold attribute, nor can it capture the individual parts of f[…]. (And the evaluation of Print should be prevented of course)
evaluation replacement hold
evaluation replacement hold
edited 10 hours ago
Kuba♦
103k12201516
asked yesterday
Lukas Lang
6,4351929
edited 10 hours ago
Kuba♦
103k12201516
asked yesterday
Lukas Lang
6,4351929
edited 10 hours ago
Kuba♦
103k12201516
edited 10 hours ago
Kuba♦
103k12201516
edited 10 hours ago
Kuba♦
103k12201516
103k12201516
asked yesterday
Lukas Lang
6,4351929
asked yesterday
Lukas Lang
6,4351929
asked yesterday
Lukas Lang
6,4351929
6,4351929
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Here's a way making use of Block to cause f to be inert:
Block[{f},
SetAttributes[f, HoldAllComplete];
expr /. f[args__] :>
RuleCondition[f[args] /. Print -> Echo]
]
Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}]
Note of course that this only works if you have a pattern of the form _Symbol[...]
answered yesterday
b3m2a1
26.9k257154
add a comment |
I can think of one (ugly) way to do it:
Attributes[myHold] = {HoldAll};
expr /.
f[args__] :> With[
{res = myHold[args] /. Print -> Echo},
f @@ res /; True
] /.
HoldPattern[f_ @@ myHold[args__]] :> f[args]
(* Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}] *)
The idea is to wrap the contents of f inside a function with HoldAll attribute (not Hold, to be able to identify it uniquely later on). In a first step, the expression is returned with myHold[…] still in place. In a second round of replacements, myHold is stripped out again.
answered yesterday
Lukas Lang
6,4351929
1
+1. If we use the vanishing macro defined in (43096), we can writevanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].
– WReach
yesterday
add a comment |
Alternatively:
expr /.
foo_f :> RuleCondition[Hold[foo] /. Print -> Echo] /.
Hold[foo_f] :> foo
answered 10 hours ago
Kuba♦
103k12201516
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a way making use of Block to cause f to be inert:
Block[{f},
SetAttributes[f, HoldAllComplete];
expr /. f[args__] :>
RuleCondition[f[args] /. Print -> Echo]
]
Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}]
Note of course that this only works if you have a pattern of the form _Symbol[...]
answered yesterday
b3m2a1
26.9k257154
add a comment |
Here's a way making use of Block to cause f to be inert:
Block[{f},
SetAttributes[f, HoldAllComplete];
expr /. f[args__] :>
RuleCondition[f[args] /. Print -> Echo]
]
Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}]
Note of course that this only works if you have a pattern of the form _Symbol[...]
answered yesterday
b3m2a1
26.9k257154
add a comment |
Here's a way making use of Block to cause f to be inert:
Block[{f},
SetAttributes[f, HoldAllComplete];
expr /. f[args__] :>
RuleCondition[f[args] /. Print -> Echo]
]
Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}]
Note of course that this only works if you have a pattern of the form _Symbol[...]
answered yesterday
b3m2a1
26.9k257154
Here's a way making use of Block to cause f to be inert:
Block[{f},
SetAttributes[f, HoldAllComplete];
expr /. f[args__] :>
RuleCondition[f[args] /. Print -> Echo]
]
Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}]
Note of course that this only works if you have a pattern of the form _Symbol[...]
answered yesterday
b3m2a1
26.9k257154
answered yesterday
b3m2a1
26.9k257154
answered yesterday
b3m2a1
26.9k257154
answered yesterday
b3m2a1
26.9k257154
26.9k257154
add a comment |
add a comment |
I can think of one (ugly) way to do it:
Attributes[myHold] = {HoldAll};
expr /.
f[args__] :> With[
{res = myHold[args] /. Print -> Echo},
f @@ res /; True
] /.
HoldPattern[f_ @@ myHold[args__]] :> f[args]
(* Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}] *)
The idea is to wrap the contents of f inside a function with HoldAll attribute (not Hold, to be able to identify it uniquely later on). In a first step, the expression is returned with myHold[…] still in place. In a second round of replacements, myHold is stripped out again.
answered yesterday
Lukas Lang
6,4351929
1
+1. If we use the vanishing macro defined in (43096), we can writevanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].
– WReach
yesterday
add a comment |
I can think of one (ugly) way to do it:
Attributes[myHold] = {HoldAll};
expr /.
f[args__] :> With[
{res = myHold[args] /. Print -> Echo},
f @@ res /; True
] /.
HoldPattern[f_ @@ myHold[args__]] :> f[args]
(* Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}] *)
The idea is to wrap the contents of f inside a function with HoldAll attribute (not Hold, to be able to identify it uniquely later on). In a first step, the expression is returned with myHold[…] still in place. In a second round of replacements, myHold is stripped out again.
answered yesterday
Lukas Lang
6,4351929
1
+1. If we use the vanishing macro defined in (43096), we can writevanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].
– WReach
yesterday
add a comment |
I can think of one (ugly) way to do it:
Attributes[myHold] = {HoldAll};
expr /.
f[args__] :> With[
{res = myHold[args] /. Print -> Echo},
f @@ res /; True
] /.
HoldPattern[f_ @@ myHold[args__]] :> f[args]
(* Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}] *)
The idea is to wrap the contents of f inside a function with HoldAll attribute (not Hold, to be able to identify it uniquely later on). In a first step, the expression is returned with myHold[…] still in place. In a second round of replacements, myHold is stripped out again.
answered yesterday
Lukas Lang
6,4351929
I can think of one (ugly) way to do it:
Attributes[myHold] = {HoldAll};
expr /.
f[args__] :> With[
{res = myHold[args] /. Print -> Echo},
f @@ res /; True
] /.
HoldPattern[f_ @@ myHold[args__]] :> f[args]
(* Hold[{f[{1, {Echo[1]}}], g[{{{Print[1]}}}]}] *)
The idea is to wrap the contents of f inside a function with HoldAll attribute (not Hold, to be able to identify it uniquely later on). In a first step, the expression is returned with myHold[…] still in place. In a second round of replacements, myHold is stripped out again.
answered yesterday
Lukas Lang
6,4351929
answered yesterday
Lukas Lang
6,4351929
answered yesterday
Lukas Lang
6,4351929
answered yesterday
Lukas Lang
6,4351929
6,4351929
1
+1. If we use the vanishing macro defined in (43096), we can writevanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].
– WReach
yesterday
add a comment |
1
+1. If we use the vanishing macro defined in (43096), we can writevanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].
– WReach
yesterday
1
1
+1. If we use the vanishing macro defined in (43096), we can write
vanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].– WReach
yesterday
+1. If we use the vanishing macro defined in (43096), we can write
vanishing[{h}, expr /. x_f :> RuleCondition[h[x] /. Print -> Echo]].– WReach
yesterday
add a comment |
Alternatively:
expr /.
foo_f :> RuleCondition[Hold[foo] /. Print -> Echo] /.
Hold[foo_f] :> foo
answered 10 hours ago
Kuba♦
103k12201516
add a comment |
Alternatively:
expr /.
foo_f :> RuleCondition[Hold[foo] /. Print -> Echo] /.
Hold[foo_f] :> foo
answered 10 hours ago
Kuba♦
103k12201516
add a comment |
Alternatively:
expr /.
foo_f :> RuleCondition[Hold[foo] /. Print -> Echo] /.
Hold[foo_f] :> foo
answered 10 hours ago
Kuba♦
103k12201516
Alternatively:
expr /.
foo_f :> RuleCondition[Hold[foo] /. Print -> Echo] /.
Hold[foo_f] :> foo
answered 10 hours ago
Kuba♦
103k12201516
answered 10 hours ago
Kuba♦
103k12201516
answered 10 hours ago
Kuba♦
103k12201516
answered 10 hours ago
Kuba♦
103k12201516
103k12201516
add a comment |
add a comment |
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