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/