Re(2): Re: RE: Re: Re: easiest way to sort a list?

*To*: mathgroup at smc.vnet.net*Subject*: [mg18598] Re(2): [mg18556] Re: [mg18525] RE: [mg18428] Re: [mg18344] Re: [mg18308] easiest way to sort a list?*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Tue, 13 Jul 1999 01:01:29 -0400*Sender*: owner-wri-mathgroup at wolfram.com

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: [mg18598] [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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