Re: Re: RE: Re: Re: easiest way to sort a list?

• To: mathgroup at smc.vnet.net
• Subject: [mg18722] Re: [mg18556] Re: [mg18525] RE: [mg18428] Re: [mg18344] Re: [mg18308] easiest way to sort a list?
• From: "Michael J. Sharpe" <msharpe at ucsd.edu>
• Date: Sat, 17 Jul 1999 02:36:42 -0400
• Organization: Univ of Calif San Diego
• References: <7mg54a\$1bq\$6@dragonfly.wolfram.com>
• Sender: owner-wri-mathgroup at wolfram.com

```I'm using Mathematica 3.0 on a G3 266. If I run just OrderedUnion[4] on
Table[Random[Integer, {1, 5000}], {10000}], I get times comparable for those with
OrderedUnion[1]. However, when I run OrderedUnion[2] on this particular data, the
kernel invariably freezes the machine or crashes it with a "bus error". On
smaller data sets, if it doesn't crash the machine, it slows subsequent
operations dramatically. Perhaps the anomolous timing you observed with
OrderedUnion[4] in Mathematica 4.0 is caused by some similar effect.

I believe the problem can be laid at the feet of recursion depth in the
construction of l0 in OrderedUnion[2]. After running the following similar code,
my machine as soon as I try to access li, say by asking for li[[1]].

li={};
For[k=1,k<=5000,k++,li={li,k}];
Flatten[li];

Michael Sharpe

Andrzej Kozlowski wrote:

> Actually, OrderedUnion[4] is dreadfully slow on my Mac for this last
> list. Here is what I get:
>
> In[16]:=
> Table[OrderedUnion[i][l3]; // Timing, {i, 1, 4}]
> Out[16]=
> {{1.16667 Second, Null}, {2.15 Second, Null}, {38.5833 Second,
>     Null}, {396.433 Second, Null}}
>
> And yet Sort and Union are indeed in version 4. You can see this when I
> add Colin Rose's function to the test list:
>
> OrderedUnion[5][lis_] :=
>
>   Part[lis, Sort[Map[Position[lis, #][[1]] &, Union[lis]] // Flatten]]
>
> Running all five in a fresh Mathematica 4.0 session I get:
>
> In[6]:=
> l = Table[Random[Integer, {1, 5000}], {10000}];
>
> In[7]:=
> Table[OrderedUnion[i][l]; // Timing, {i, 1, 5}]
> Out[7]=
> {{1.16667 Second, Null}, {2.06667 Second, Null}, {36.3 Second,
>     Null}, {403.417 Second, Null}, {4.93333 Second, Null}}
>
> with OrderedUnion[4] doing slightly worse than before. For a list with
> only a few digits I get:
>
> In[8]:=
> l1 = Table[Random[Integer, {1, 50}], {10000}];
>
> In[9]:=
> Table[OrderedUnion[i][l1]; // Timing, {i, 1, 5}]
> Out[9]=
> {{0.166667 Second, Null}, {0.05 Second, Null}, {1.01667 Second,
>     Null}, {0.566667 Second, Null}, {1.16667 Second, Null}}
>
> So on my machine it is only for such types of lists that OrderedUnion[4]
> does relatively well. There may be something like the Split problem
> involved here, but I do not think it's wrth the time and effort it would
> take to investigate this. I am now inclined to think that cross-platform
> timings in Mathematica are almost meaningless.
>
> On Sat, Jul 10, 1999, Carl K.Woll <carlw at fermi.phys.washington.edu> wrote:
>
> >Andrzej,
> >
> >Your results are very different from those I get with Mathematica V3.0
> on a 100
> >MHz PC
> >with Windows 95. For your l2 list, I got
> >
> >{{1.93 Second, Null}, {6.04 Second, Null},
> > {43.12 Second, Null}, {5.65 Second, Null}}
> >
> >If I try using the following list
> >
> >Table[Random[Integer, {1, 5000}], {10000}];
> >
> >I get the following timings
> >
> >{{7.47 Second, Null}, {14.4 Second, Null},
> > {177.24 Second, Null}, {6.98 Second, Null}}
> >
> >So, for some reason OrderedUnion[4] is much slower on your machine than it is
> >on mine.
> >That is, for the list l2, OrderedUnion[4] is 66 times slower than
> >OrderedUnion[1] on your
> >machine, but it is only 3 times slower on my machine. This is odd, since
> >OrderedUnion[4]
> >uses Sort and Union, and I thought Colin Rose suggested that Sort and
> Union are
> >much
> >faster in Version 4.0 than in Version 3.0. Hence, I would have thought that
> >OrderedUnion[4] would have worked even better on your machine.
> >
> >Carl Woll
> >Dept of Physics
> >U of Washington
> >
> >Andrzej Kozlowski wrote:
> >
> >> Here are some more results on my PowerBook G3 (233 mghz, Mathematica 4.0):
> >>
> >> In[1]:=
> >> OrderedUnion[1][li_] :=
> >>   Block[{i, Sequence},
> >>                   i[n_] := (i[n] = Sequence[]; n);
> >>                   i /@ li]
> >>
> >> OrderedUnion[2][li_] :=
> >>   Block[{i, counter = 0, l0 = {}, m = Length[Union[li]], Sequence},
> >>     i[n_] := (i[n] = Sequence[]; counter = counter + 1; n);
> >>    Scan[If[counter < m, l0 = {l0, i[#]}, Return[Flatten[l0]]] &, li];
> >>     Flatten[l0]]
> >>
> >> OrderedUnion[3][h_[args__]] :=
> >>   Fold[ If[ FreeQ[#1, #2, 1], Join[#1, h[#2]], #1] & , h[], {args}]
> >>
> >> OrderedUnion[4][li_] :=
> >>   Module[{tmp = Transpose[{li, Range[Length[li]]}]},
> >>     tmp = Union[tmp, SameTest -> (#1[[1]] === #2[[1]] &)];
> >>     Last /@ Sort[RotateLeft /@ tmp]]
> >>
> >> First a long list with a small number of distinnct elements:
> >>
> >> In[5]:=
> >> l1 = Table[Random[Integer, {1, 10}], {10000}];
> >>
> >> In[6]:=
> >> Table[OrderedUnion[i][l1]; // Timing, {i, 1, 4}]
> >> Out[6]=
> >> {{0.183333 Second, Null}, {0.0333333 Second, Null},
> >>
> >>   {0.75 Second, Null}, {0.533333 Second, Null}}
> >>
> >> Next a long list with quite many distinct elements:
> >>
> >> l2 = Table[Random[Integer, {1, 1000}], {10000}];
> >>
> >> In[8]:=
> >> Table[OrderedUnion[i][l2]; // Timing, {i, 1, 4}]
> >> Out[8]=
> >> {{0.3 Second, Null}, {1.05 Second, Null},
> >>
> >>   {8.48333 Second, Null}, {22.0333 Second, Null}}
> >>
> >> --
> >> Andrzej Kozlowski
> >> Toyama International University
> >> JAPAN
> >> http://sigma.tuins.ac.jp
> >> http://eri2.tuins.ac.jp
> >>
> >> ----------
> >> >From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
To: mathgroup at smc.vnet.net
> >> To: mathgroup at smc.vnet.net
> >> >To: mathgroup at smc.vnet.net
> >> >Subject: [mg18722] [mg18556] [mg18525] RE: [mg18428] Re: [mg18344] Re: [mg18308]
> >easiest way to
> >> sort a list?
> >> >Date: Fri, Jul 9, 1999, 11:32 AM
> >> >
> >>
> >> > I have a new approach to add to this long thread.
> >> >
> >> > In[1]:=
> >> > OrderedUnion5[h_[args__]]:=
> >> >   Fold[ If[ FreeQ[#1,#2,1], Join[#1,h[#2]], #1]& ,h[],{args}]
> >> >
> >> > In[2]:=
> >> > OrderedUnion5[{5,3,5,3,1,2,2,3,4,5,3}]
> >> >
> >> > Out[2]=
> >> > {5,3,1,2,4}
> >> >
> >> >
> >> > I would like to see how the timings of different methods compare (version
> >> > above included).  I would do it, but I am working under Windows 98, and I
> >> > understand Microsoft made it impossible for Timing to give accurate
> >results.
> >> >
> >> > Regards,
> >> > Ted Ersek
> >> >
> >> > --------------------
> >> > Hi Hartmut (and newsgroup),
> >> >
> >> > Thanks for unraveling the linear nature of the Block[{i... function. Nice
> >> > explanation!
> >> >
> >> > Since this topic shows no sign of slowing down, I thought I would present
> >a
> >> > couple refined versions of my two functions. First, an improvement of the
> >> > "linear" solution:
> >> >
> >> > OrderedUnion[li_]:=Block[{i,Sequence},
> >> >     i[n_]:=(i[n]=Sequence[];n);
> >> >     i/@li]
> >> >
> >> > By including Sequence in the Block variables, Mathematica doesn't waste
> >time
> >> > eliminating Sequences until after the Block is exited. This seems to be a
> >> > 5-10% or so improvement in speed.
> >> >
> >> > Secondly, an improvement of my O(n log n) solution:
> >> >
> >> > OrderedUnion[3][li_]:=Module[{tmp=Transpose[{li,Range[Length[li]]}]},
> >> >   tmp=Union[tmp, SameTest->(#1[[1]]===#2[[1]]&)];
> >> >   Last/@Sort[RotateLeft/@tmp]]
> >> >
> >> > The key idea here is that Sort with the default ordering function is much
> >> > faster than Sort with a user supplied ordering function. So, RotateLeft
> >> > switches the order of the index and the element to allow Sort to be used
> >> > with its default ordering function. For lists with a large number of
> >> > distinct elements, this seems to improve the speed by a factor of 3.
> >> >
> >> > Carl Woll
> >> > Dept of Physics
> >> > U of Washington
> >> >
> >> > Wolf, Hartmut wrote:
> >> >
> >> >> Hello everone,
> >> >>
> >> >> I just sent a letter to Carl K.Woll, part of which I would like to
> >> >> publish to the community:
> >> >>
> >> >> Hello Carl,
> >> >>
> >> >> congratulations for your brilliant solution of "easiest way to sort a
> >> >> list?" At the first glance, ... I was already struck.
> >> >>
> >> >> According to my measurements, you proposed not only the fastest
> >> >> solution, but also the second fastest, the fastest O[n log n] version.
> >> >> ....
> >> >>
> >> >> The question that concerned me was, is your solution really O[n]? My
> >> >> preoccupation before was that O[n log n] could be the best achievable.
> >> >> Clearly your solution is O[n] for 'smaller' samples, but the question
> >> >> was, is it still O[n] for very large samples?  So I did some
> >> >> measurements, which I'd like to show you. They are in the two attached
> >> >> notebooks (...not for you, unless you request it from
> >> >> mailto:hwolf at debis.com )
> >> >>
> >> >> What is the interpretation of the graphs? In fact your solution is as
> >> >> far as I could calculate seemingly linear overall. There are interesting
> >> >> wiggles in it. The algorithm seems to become increasingly worse, when it
> >> >> makes up it's mind and "starts anew". Well I think this are the traces
> >> >> of Mathematica's Extendible Hashing algorithm for the symbol store. When
> >> >> hashing becomes more and more inefficient (due to collisions) then the
> >> >> store ist extended. (This conforms to my observation of memory usage
> >> >> with Windows NT Task Manager during the calculation, where I say periods
> >> >> of rapid increase of used memory and times of slower grow.) So it's this
> >> >> extension, which keeps your algorithm linear.
> >> >>
> >> >> ....
> >> >>
> >> >> With kind regards, your
> >> >>         Hartmut Wolf
> >> >
> >
> >
> >
> >
> >
>
> Andrzej Kozlowski
> Toyama International University
> JAPAN
> http://sigma.tuins.ac.jp/
> http://eri2.tuins.ac.jp/

```

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