5
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I'm trying to modify Plus but am running into trouble with it being Listable:

ClearAll[f, g, h]
Attributes[f] = {Listable};
h /: f[x_h, l_List] := 0
h /: g[x_h, l_List] := 0
f[h[1], {1, 1}]   (* {f[h[1], 1], f[h[1], 1]}, not OK I want 0 *)
g[h[1], {1,1}]    (* 0 as expected *)

How can I make the UpValue (or equivalent) have higher priority than the listability?

EDIT: I ended up wanting to do this again and figured I'd fix the Plus properly. Here it is working using Sashas answer and Mr.Wizards $Pre method:

    $Pre =.
ClearAll[myPlus]; Attributes[myPlus] = {Orderless};
Unprotect[InterpolatingFunction]; UpValues[InterpolatingFunction] = {};
InterpolatingFunction /: 
 myPlus[y_InterpolatingFunction[t_Symbol], l_List] := 
 Interpolation[
   MapThread[List, {y["Grid"], l + # & /@ y["ValuesOnGrid"]}], 
   InterpolationOrder -> First[y["InterpolationOrder"]]][t]
myPlus[other__] := +other
$Pre = Function[x, Unevaluated@x /. Plus -> myPlus, HoldAllComplete];
    Protect[InterpolatingFunction];

    y = Interpolation[Table[{i, RandomReal[{0, 1}, 2]}, {i, 1, 10}]];
    ParametricPlot[{y[t], y[t] - {1, 1}}, {t, 1, 10}]
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5
  • $\begingroup$ You can't do that, or I'll be very surprised. Change your design somehow. It's hard to help in that without knowing a little bit about the bigger picture. Perhaps for your case adding the "listability" as e:f[_, l_List]:=Thread@Unevaluated@e or something similar, but that depends on the case at hand $\endgroup$
    – Rojo
    Commented Feb 5, 2013 at 13:55
  • $\begingroup$ Oh, f is Plus, my bad. $\endgroup$
    – Rojo
    Commented Feb 5, 2013 at 13:57
  • $\begingroup$ Yes, I agree. Not possible (I even tried with $Pre and $Post to fool the evaluator, but you cannot do that). Just program Listable like f[x_, l_List] := Thread[f, {x, l}]; and all is good. $\endgroup$ Commented Feb 5, 2013 at 14:01
  • $\begingroup$ @Rojo Updated with bigger picture $\endgroup$
    – ssch
    Commented Feb 5, 2013 at 15:37
  • 1
    $\begingroup$ I think this is one of these things which are really hard to make different generally and consistently, since one would have to explicitly go against the standard evaluation sequence. I would reconsider the design of whatever you try to achieve with this. While I will be the first to suggest workarounds which change system's behavior in many cases, I also think that admitting and accepting certain limitations of the system can sometimes be more productive. $\endgroup$ Commented Feb 5, 2013 at 20:48

2 Answers 2

3
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This seems disgusting, but here it goes

h /: Plus[x_h, l_List] :=
 withPlusListability[True][

  blabla; 0
  ]

withPlusListability[bool_: True | False] := Function[code,
   Internal`InheritedBlock[{Plus},
    Unprotect[Plus];
    If[bool, SetAttributes, ClearAttributes][Plus, Listable];
    code
    ], HoldFirst];

withH = withPlusListability[False];

So

withH[
  Print[h[3] + {4, 5}];
  Print[h[3] + 7];
 ];

prints

(*
0
7+h[3]
*)

While you evaluate code iniside withPlusListability[True|False], it takes care that Plus has|doesn't have the Listable attribute, without changing it globally. h's definition will only have a chance of matching with an unlistable Plus.

Plus is one of those symbols that are so special you really try not to mess with. As @Mr.Wizard warned, this will likely break for packed arrays, because it probably has been optimized to cut some corners.

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3
  • $\begingroup$ @MrWizard, surprisingly, h[3] + RandomReal[{-1, 1}, 10]; works. Could you try it in v7? $\endgroup$
    – Rojo
    Commented Feb 5, 2013 at 14:31
  • $\begingroup$ Cool, didn't know about Internal`InheritedBlock $\endgroup$
    – ssch
    Commented Feb 5, 2013 at 15:56
  • $\begingroup$ Rojo: yes, it works. I need to take a closer look at this. $\endgroup$
    – Mr.Wizard
    Commented Feb 6, 2013 at 1:28
3
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Contrary to Rolf Mertig's comment I believe $Pre does work.

First define your function:

h /: myPlus[_h, _List] := 0
myPlus[other__] := +other

Then set $Pre:

$Pre = Function[x, Unevaluated@x /. Plus -> myPlus, HoldAllComplete];

Test:

h[1] + {4, 5, 6}
0
z[1] + {4, 5, 6}
{4 + z[1], 5 + z[1], 6 + z[1]}
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9
  • $\begingroup$ @RolfMertig is pessimistic (comments) these days $\endgroup$
    – Rojo
    Commented Feb 5, 2013 at 14:36
  • 2
    $\begingroup$ +other!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! HAAAAAAAAAAA +1 $\endgroup$
    – Rojo
    Commented Feb 5, 2013 at 14:39
  • $\begingroup$ You make it look so easy, thanks! $\endgroup$
    – ssch
    Commented Feb 5, 2013 at 15:59
  • $\begingroup$ @Rojo I'm afraid I don't understand your outburst. Unintentional humor? Does "plus other" mean something in contemporary parlance? $\endgroup$
    – Mr.Wizard
    Commented Feb 5, 2013 at 16:36
  • $\begingroup$ @ssch Glad I could help. :-) $\endgroup$
    – Mr.Wizard
    Commented Feb 5, 2013 at 16:37

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