12
$\begingroup$

I often find myself writing code that looks a bit like this:

f[x_Integer] := 
  With[
   {
     range = Range[2] + x
   },
   With[
     {
       a = range[[1]],
       b = range[[2]],
       c = g[range]
     },
     h[a,b,c]
   ]
 ];

It would be nice if I could avoid Withs and just write

f[x_Integer] := 
  Let[
   range = Range[2] + x,
   {a,b} = range,
   c = g[range]
   ,
   h[a,b,c]
 ];

which would then automatically expand to the above at definition time.

What I'm asking is a bit similar to this question. There are additional requirements however. The new scoping construct (Let in the above) should:

  • Group sequential disjoint assignments into single Withs.
  • Thread over List assignments.

Of course, it should not evaluate the left-hand-sides and the right-hand-sides of the assignments while expanding to Withs.

Any proposals for such a scoping construct? (I'll post my version soon).

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6
  • $\begingroup$ Your example doesn't need any scoping: f[x_Integer] := h[Sequence@@#,g@#]&@(Range[2]+x) $\endgroup$
    – Bob Hanlon
    Commented Nov 5, 2014 at 14:24
  • $\begingroup$ Why do you want to rewrite using With rather than preserving a higher abstraction such as LetL? $\endgroup$
    – Mr.Wizard
    Commented Nov 5, 2014 at 14:56
  • $\begingroup$ Also: what is the reason to prefer With over Module? Assignments such as {a,b} = range are simpler with the latter. $\endgroup$
    – Mr.Wizard
    Commented Nov 5, 2014 at 15:00
  • $\begingroup$ @Mr.Wizard This should be a generalization of LetL, which doesn't do the two points I mentioned (Leonid's answer below does). Module doesn't allow you to do threaded assignments in the first argument, forcing you to write Module[{a,b},{a,b}=Range[2];...], which is duplication I don't like. Also, it'd like to inject into held expressions -- another reason not to go with Module. $\endgroup$ Commented Nov 5, 2014 at 15:07
  • 3
    $\begingroup$ If you are ok with undocumented features, you can supply multiple arguments to With in Mathematica 10.3 and above: e.g. With[{c = d}, {b = c}, {a = b}, a] (You'll have to also tolerate the red syntax coloring in the front end, or turn it off manually) [and I just realized that it doesn't satisfy your second requirement of threading over List assignments] $\endgroup$
    – QuantumDot
    Commented May 4, 2016 at 20:03

1 Answer 1

14
$\begingroup$

With this helper function:

SetAttributes[partThread, HoldAll];
partThread[l___, rhs_] :=
  Join @@ Replace[
    MapIndexed[Append[#, First@#2] &, Thread[Hold[{l}]]],
    Hold[s_, i_] :> Hold[s = rhs[[i]]],
    {1}];

The following modification of LetL seems to work according to your specs:

ClearAll[Let, let];
SetAttributes[{Let, let}, HoldAll];

Let /: Verbatim[SetDelayed][lhs_, rhs : HoldPattern[Let[__, _]]] := 
   Block[{With}, Attributes[With] = {HoldAll};
      lhs := Evaluate[rhs /. HoldPattern[With[{}, b_]] :> b]
   ];

Let[args___, body_] := let[{args}, body, {}, {}];

let[{}, body_, {}, _] := With[{}, body];

let[{Set[{s___}, rhs_], rest___}, body_, dec_, syms_] :=
   Module[{temp},
     partThread[s, temp] /. Hold[d___] :>
        let[{temp = rhs, d, rest}, body, dec, syms]
   ];

let[
   {Set[sym_, rhs_], rest___}, 
   body_, 
   {decs___}, 
   {syms___}
] /; FreeQ[Unevaluated[rhs], Alternatives[syms]] :=
     let[{rest}, body, {decs, sym = rhs}, {syms, HoldPattern[sym]}];

let[{args___}, body_, {decs__}, _] :=
   Block[{With},
     Attributes[With] = {HoldAll};
     With[{decs},Evaluate[let[{args}, body, {}, {}]]]
   ];

This works quite similarly to LetL. What it does in addition to LetL is that it collects previous declarations into auxiliary lists stored as extra arguments of let, so that it can group together disjoint declarations. It also threads over arguments, using the partThread helper function. In all other respects it is the same code as LetL.

Here is your example:

f[x_Integer] := 
   Let[range = Range[2] + x, {a, b} = range, c = g[range], h[a, b, c]];

we can check what was generated:

?f

Global`f
  f[x_Integer]:=
    With[{range=Range[2]+x},
       With[{a=range[[1]],b=range[[2]],c=g[range]},h[a,b,c]]]
$\endgroup$
15
  • $\begingroup$ @MichaelE2 That was an evaluation leak, thanks for reporting. I seem to have fixed it, although perhaps not very elegantly. $\endgroup$ Commented Nov 5, 2014 at 14:40
  • $\begingroup$ You're welcome. It seems fixed now. $\endgroup$
    – Michael E2
    Commented Nov 5, 2014 at 14:51
  • $\begingroup$ @MichaelE2 Thanks for double-checking. I've not done this sort of things for some while, got a bit rusty. $\endgroup$ Commented Nov 5, 2014 at 14:52
  • $\begingroup$ That's a bit more elegant than what I had :). One question though: why doesn't Alternatives leak evaluation here? $\endgroup$ Commented Nov 5, 2014 at 15:09
  • $\begingroup$ @TeakeNutma because I wrapped symbols in HoldPattern, and use Unevaluated for the rhs. HoldPattern is exactly the right tool for the job here, since it is invisible to the pattern-matcher (and so FreeQ works fine). $\endgroup$ Commented Nov 5, 2014 at 15:09

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