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MathGroup Archive 2005

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Re: letrec/named let

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56777] Re: [mg56707] letrec/named let
  • From: DrBob <drbob at bigfoot.com>
  • Date: Thu, 5 May 2005 06:04:37 -0400 (EDT)
  • References: <200505040433.AAA06220@smc.vnet.net> <opsp9mdot7iz9bcq@monster.ma.dl.cox.net> <1115235138.31794.120.camel@30-5-214.wireless.csail.mit.edu> <opsp9n4xswiz9bcq@monster.ma.dl.cox.net> <1115245251.31821.123.camel@30-5-214.wireless.csail.mit.edu>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Call the previous solution "noDispatch"; the "dispatch" solution will be much faster for large lists:

Clear[noDispatch, dispatch]
noDispatch[s_List] := Module[{u = Union@s},
     s /. Thread[u -> Range@Length@u]]
dispatch[s_List] := Module[{u = Union@s},
     s /. Dispatch@Thread[u -> Range@Length@u]]

data=Table[Random[],{1000}];
Timing[one=dispatch@data;]
Timing[two=noDispatch@data;]
one==two

{0.016 Second,Null}

{0.468 Second,Null}

True

That's almost 30 times faster, but double the size of the list, and now it's 122 times faster:

data=Table[Random[],{2000}];
Timing[one=dispatch@data;]
Timing[two=noDispatch@data;]
one==two

{0.016 Second,Null}

{1.953 Second,Null}

True


With 3000 elements, the speed ratio is 300 to 1, so Dispatch is well worth the trouble.

I have a feeling there might be a faster method using a variant of UnsortedUnion -- see the examples in Help for Union -- but it's just a feeling, so far.

Bobby

On Wed, 04 May 2005 18:20:51 -0400, Daniel Roy <droy at MIT.EDU> wrote:

> Ahah.. This works.  Very good solution!  (much faster than my solution
> whose performance is dependent on the absolute size of the values !
> ugly)
>
> dan
>
> On Wed, 2005-05-04 at 14:51 -0500, DrBob wrote:
>> Clear@CompressNumericalSequence
>> CompressNumericalSequence[s_List] := Module[{u = Union@s},
>>      s /. Thread[u -> Range@Length@u]
>>      ]
>> CompressNumericalSequence@{5, 1, 1, 2, 3, 4}
>>
>> {5,1,1,2,3,4}
>>
>> Bobby
>>
>> On Wed, 04 May 2005 15:32:18 -0400, Daniel Roy <droy at MIT.EDU> wrote:
>>
>> > I just figured out why the suggestions I've been given fail on my test
>> > cases.  Its because the sets that I am passing can have repeated digits.
>> >
>> > CompressNumericalSequence@{10,2,2,4,7,8}
>> > {5,1,1,2,3,4}
>> >
>> > Otherwise, your solution of Ordering@Ordering is MUCH faster.
>> > Unfortuantely, it gives the wrong answer on the above test.
>> >
>> > -dan
>> >
>> >
>> > On Wed, 2005-05-04 at 14:14 -0500, DrBob wrote:
>> >> I have no idea what "named let" does -- not a hell of a lot, I suspect -- but here's a simple replacement for the other routine:
>> >>
>> >> CompressNumericalSequence[s_List]:=Ordering@Ordering@s
>> >> CompressNumericalSequence@{10,2,4,7,8}
>> >>
>> >> {5,1,2,3,4}
>> >>
>> >> Bobby
>> >>
>> >> On Wed, 4 May 2005 00:33:53 -0400 (EDT), Daniel Roy <droy at mit.edu> wrote:
>> >>
>> >> > hi.  i'm a lisper/schemer and i'm working with mathematica.  i
>> >> > appreciate the lisp-like nature of mathematica but i can't seem to
>> >> > easily replicate some of the functionality i like which is forcing me to
>> >> > write ugly side-effect code.
>> >> >
>> >> > for instance, how do you do the equivalent of a "named let" in
>> >> > mathematica (NOTE! I know i can take the max of a list, this is just a
>> >> > simple example of a named let)
>> >> >
>> >> > (define (max-of-list lst)
>> >> >   (let loop ((lst (cdr lst))
>> >> >              (best (car lst)))
>> >> >     (if (null? lst)
>> >> >         best
>> >> >         (loop (cdr lst)
>> >> >               (if (> (car lst) best)
>> >> >                   (car lst)
>> >> >                   best)))))
>> >> >(max-of-list '(1 2 3 4 5 2))
>> >> >> 5
>> >> >
>> >> > Here is a mathematica function to compress a sequence numerically.
>> >> > here is one attempt using functions where i pass the function to
>> >> > itself... there has to be a better way
>> >> >
>> >> > CompressNumericalSequence[S_] := Module[
>> >> >       {C = Function[{C, R, i},
>> >> >             If[i < Max[R],
>> >> >               If[Length[Position[R, i]] == 0,
>> >> >                 C[C, (If[# > i, # - 1, #]) & /@ R, i],
>> >> >                 C[C, R, i + 1]],
>> >> >               R]]},
>> >> >       C[C, S, 1]];
>> >> >
>> >> > CompressNumericalSequence[{10, 2, 4, 7, 8}]
>> >> > {5, 1, 2, 3, 4}
>> >> >
>> >> > Also, is it possible to do letrec in mathematica?  (essentially, i know
>> >> > i can do recursive function declarations at the top level... my question
>> >> > is whether i can do them at lower levels?)...
>> >> >
>> >> > thanks, dan
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >>
>> >>
>> >>
>> >
>> >
>> >
>> >
>>
>>
>>
>
>
>
>



-- 
DrBob at bigfoot.com


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