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performance tuning - Comparing LetL and Module efficiency



I was recently introduced to the LetL macro thanks to Leonid's answer to one of my prior questions. I was, needless to say, impressed by the simplicity of its recursive definition. However, I noticed that it may not necessarily be optimized. As it is defined, if my LetL statement contains a definition which does not need to be nested, then it will call With unnecessarily:


testLetL := LetL[{x = 1, y = 2, z = 2 x y}, {x, y, z}]
?testLetL
(* testLetL:=With[{x=1},With[{y=2},With[{z=2 x y},{x,y,z}]]] *)

So I compared it to Module:


testModule := Module[{x = 1, y = 2, z}, z = 2 x y; {x, y, z}]
(Do[#, {i, 5000000}] // AbsoluteTiming) & /@ {testLetL, testModule}
(* {{0.9390537, Null}, {0.9270530, Null}} *)


As you can see, there doesn't seem to be much speed gained in using LetL - essentially nested Withs - instead of Module. I thought perhaps that it was the extra With being called that was slowing things down. So I tried another test:


testLetL2 := LetL[{x = 1, y = 2 x }, {x, y}]
testModule2 := Module[{x = 1, y}, y = 2 x ; {x, y}]
(Do[#, {i, 5000000}] // AbsoluteTiming) & /@ {testLetL2, testModule2}
(* {{0.9270531, Null}, {0.9120521, Null}} *)

This again showed that they were pretty much the same, if not Module being a bit faster.



My question is, then:


Is LetL simply used for convenience or are my tests missing something?




Answer



The main point of LetL is just replacement of nested With, not necessarily the speed gain. Now, why would one want to use nested With in place of Module:



  • Immutable code (same advantages as With - no side effects in the body)

  • Use variables defined earlier in definitions of variables defined later.


In fact, if you want the second property, you will either have to have nested Module-s as well, or make side effects in the body.


That said, LetL should be pretty fast. You should not normally see large timing difference between LetL and equivalent nested With. Moreover, for functions defined via SetDelayed, LetL expands into equivalent nested With at definition-time, so there is no run-time performance penalty at all.


And yes, your tests missed the point, since pure function is #-& notation evaluates its argument, so you were actually testing already evaluated expressions. Try this:


Do[testLetL2,{i,50000}]//AbsoluteTiming

Do[testModule2,{i,50000}]//AbsoluteTiming

(*
   {0.190430,Null}
   {0.276367,Null}
*)

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