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Re: Complement replacement
*To*: mathgroup at smc.vnet.net
*Subject*: [mg58633] Re: Complement replacement
*From*: konstantpi at mail15.com
*Date*: Sun, 10 Jul 2005 16:52:01 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
thanks very much for the answers,
my previous question was submitted, because at first i do not know
about the Complement function, i know that a function like this must
be in the Union, Intersection family, but somehow i have not noticed
the Complement in the documentation so i wrote my old way program:
pp = Table[Random[Integer, {1, 1000}], {i, 1000}];
r = Table[i,{i,1000}];
(lst = {}; Do[If[MemberQ[pp, r[[i]]] == False,
lst = Join[lst, {r[[i]]}];
]
, {i, 1, Length[pp]}];
lst = Union[lst, lst];) // Timing
{0.22 Second, Null}
on P4 2 GHz 128k cache 256M ram
after i have written my letter i have discovered the Complement
function, and noticed its speed, so i keep my original message to
compare different strategies and methods which may presented by this
group marvelous contributors
Thanks for all
From: Andrzej Kozlowski <akozlowski at gmail.com> ( Add to address
To: mathgroup at smc.vnet.net
book)
Subject: [mg58633] Re: [mg58608] Complement replacement
On 10 Jul 2005, at 18:12, konstantpi at mail15.com wrote:
> hi
> in the list:
> pp=Table[Random[Integer, {1, 1000}], {i, 1000}];
> how could i know which numbers from 1 to 1000 does not exist in the
> pp List.
> but without using:
> Complement[Table[i,{i,1000}],pp]
> regards
>
>
Perhaps you should explain why you do not want to use Complement as
it will be very hard if not impossible to match its performance.
Indeed, on ny machine I get:
In[1]:=
pp=Table[Random[Integer, {1, 1000}], {i, 1000}];
In[2]:=
a=Complement[Range[1000],pp];//Timing
Out[2]=
{0.001657 Second,Null}
while using the most obvious alternative:
In[3]:=
b=Select[Range[1000],Not[MemberQ[pp,#]]&];//Timing
Out[3]=
{0.408538 Second,Null}
In[4]:=
Union[b]==a
Out[4]=
True
The difference in performance is huge. There are a number of ways
that give a better performance but I can't find any that beets
Complement. Here s one that is at least better than the most obvious
way above. First define a function f:
SetAttributes[f, Listable]
Scan[(f[#]=Sequence[])&,pp];//Timing
{0.020849 Second,Null}
f[x_] := x
(note that evaluating this defintion itself takes more time than
running Complement)
now:
In[8]:=
c=f[Range[1000]];//Timing
Out[8]=
{0.008274 Second,Null}
In[9]:=
Union[c]==a
Out[9]=
True
Andrzej Kozlowski
Chiba, Japan
> Peter Pein <petsie at dordos.net>:
> Please tell us why you want to do so.
>
> Cleaning up and generating testing data (I took a million values
instead
> of thousand for distinguishable timings):
>
> In[1]:= Clear[n, pp, missing];
> n = 10^6;
> pp = Table[Random[Integer, {1, n}], {n}];
>
> just as reference timing:
>
> First[Timing[missing[1] = Complement[Range[n], pp]; ]]
> Out[4] 0.844*Second
>
> Sort the random list joined with the whole range, sort and split
this
> list and get the unique elements:
>
> In[5]: First[Timing[missing[2] Cases[Split[Sort[Join[pp, Range
[n]]]],
> {x_Integer} :> x];
> ]]
> Out[5] 1.875*Second
>
> Method 'mind the gap' ;-) :
> Build ranges out of the gaps in the random list:
>
> In[6]:First[Timing[missing[3] Inner[Range, Drop[#1, -1] + 1, Drop
[#1, 1]
> - 1, Join]&
> [Union[pp, {0, n + 1}]];
> ]]
> Out[6] 1.609*Second
>
> Verification:
>
> In[7]: SameQ @@ missing /@ {1, 2, 3}
> Out[7] True
>
>
>
> --
> Peter Pein
> Berlin
> http://people.freenet.de/Peter_Berlin/
>
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