Re: Need a nice way to do this
- To: mathgroup at smc.vnet.net
- Subject: [mg40295] Re: [mg40279] Need a nice way to do this
- From: Selwyn Hollis <selwynh at earthlink.net>
- Date: Sun, 30 Mar 2003 04:09:14 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Oops, that timing was wrong. Should have been
s = Table[Random[Integer, {1, 9}], {5000}];
First@Timing[countdiffs[s];]
2.36 Second
-- SH
Begin forwarded message:
> Steve,
>
> Nice problem. The following seems to work pretty well (and on the more
> general problem):
>
> countdiffs[s_List] := Module[{members, totals, g},
> members = Union[s];
> totals = Count[s, #] & /@ members;
> g = {};
> Scan[ Module[{i}, i = First@First@Position[members, #];
> PrependTo[g, Plus @@ totals - totals[[i]]];
> totals[[i]]--]&,
> Reverse[s] ];
> g ]
>
> s = Table[Random[Integer, {1, 9}], {10}]
>
> {3, 4, 4, 9, 9, 4, 1, 6, 1, 3}
>
> countdiffs[s]
>
> {0, 1, 1, 3, 3, 3, 6, 7, 7, 8}
>
> Since this will probably become a speed contest :) ...
>
> s = Table[Random[Integer, {1, 9}], {5000}];
> First@Timing[countdiffs[s];]
>
> 1.53 Second
>
> (4.1.5, Mac OS X, 1GHz DP)
> -----
> Selwyn Hollis
> http://www.math.armstrong.edu/faculty/hollis
>
>
> On Saturday, March 29, 2003, at 05:19 AM, Steve Gray wrote:
>
>> Given a list consisting of only two distinct values, such as
>> s={a,b,b,a,a,a,b,a,b,a,a}, I want to derive a list of equal length
>> g={0,1,1,2,2,2,4,3,5,4,4}. The rule is: for each position
>> 1<=p<=Length[s], look at list s and set g[[p]] to the number of
>> elements in s to the left of p which are not equal to s[[p]].
>> In a more general version, which I do not need now, s would
>> not be restricted to only two distinct values.
>>
>> Thank you for any ideas, including other applications where
>> this particular calculation is used. The current application is an
>> unusual conjecture in geometry.
>>
>>
>>
>