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MathGroup Archive 2006

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Re: Fastest method for comparing overlapping times in random time series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64979] Re: [mg64935] Fastest method for comparing overlapping times in random time series
  • From: ggroup at sarj.ca
  • Date: Fri, 10 Mar 2006 05:15:10 -0500 (EST)
  • References: <200603080559.AAA03078@smc.vnet.net>
  • Reply-to: ggroup at sarj.ca
  • Sender: owner-wri-mathgroup at wolfram.com

Robert,

I couldn't get your code to paste in without errors. In any case, I've
listed below my attempt that has no explicit For/Do/While loops. It
seems to be reasonably fast on short lists (on the order of 1000 time
intervals per list).

========== Begin Function ===========
compare[list1_, list2_] :=
  Module[
  {pos, list2pos, filtlist2pos, f, newintervals, myHead},
  SetAttributes[pos, Listable]; 
  pos[x_] := Module[{tmp},
    tmp = Union[Flatten[list1], {x}];
    tmp = First[Flatten[Position[tmp, x]]];
    (tmp + 1)/2];

  list2pos = Transpose[{pos /@ list2, list2}];
  filtlist2pos = 
    Select[
      list2pos,
      #1[[1,2]] - #1[[1,1]] != 0 || #1[[1,1]] != IntegerPart[#1[[1,1]]] & 
    ];
  
  f[{{strt_Integer, end_}, l2int_}] := 
    Module[
      {l1int, lenl1int},
      l1int = list1[[Range[strt, end]]];
      lenl1int = Length[l1int];
    
      If[ IntegerQ[end], 
        l1int,
        Join[
          l1int[[Range[1, lenl1int - 1]]], 
          {{l1int[[-1,1]], l2int[[2]]}}
        ]
      ] /. {} :> Sequence[]
    ];
         
  f[{{strt_, end_}, l2int_}] := 
    Module[
      {l1int, lenl1int}, 
      l1int = list1[[Range[strt - 1/2, end - 1/2]]];
      lenl1int = Length[l1int];
    
      If[ strt == end, 
        {l2int}, 
      
        If[ !IntegerQ[end],
          {
            {l2int[[1]], l1int[[1,2]]},
            l1int[[Range[2, lenl1int - 1]]],
            {l1int[[-1,1]], l2int[[2]]}
          }, 
          {
            {l2int[[1]], l1int[[1,2]]},
            l1int[[Range[2, lenl1int]]]
          }
        ] /. {} :> Sequence[]
      ]
    ]; 

  newintervals = f /@ filtlist2pos;
  Flatten[
    newintervals /. {(a___)?NumericQ} :> myHead[a]
  ] /. myHead[a___] :> {a}
]
==========  End  Function ===========

Let's test it:

list1 = list2 = {};
n = 10^3; 
For[i = 0, i <= n, i += Random[], AppendTo[list1, {i, i += 2*Random[]}]]
For[i = 0, i <= n, i += Random[], AppendTo[list2, {i, i += 2*Random[]}]]

Timing[compare[list1,list2];]

Out[84]=
{2.864Second, {675,649}}

Hope that helps


On Wednesday, March 8, 2006 at 12:59 AM, Prince-Wright, Robert G Sepco
wrote: 

> I have two lists, list1{ {t1,t1+dt1}, {t2,t2+dt2},..{ti,ti+dti}}, and 
> list2, each representing 'time(i)' and corresponding 'time(i) + 
> deltatime(i)'. The time(i) values are determined by an exponential 
> inter-arrival time model, and the durations are a scaled uniform random 
> variable. Both lists are ordered on time(i). You can think of list 1 as 
> representing periods when System 1 is not working, and list 2 as the 
> periods when System 2 is not working. Example lists are given as Cell 
> Expressions below together with code to convert to a ticker-tape Plot 
> (you may need to stretch the graphic to see clearly). The challenge is 
> to develop a fast method for determining the periods when both Systems 
> are not working, i.e. to create a list corresponding to the start and 
> finish times of the overlaps.

>  Thus far I have only managed to use a Do loop which is very slow for long lists!

>  Bob

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>  Robert Prince-Wright
>  Risk Management Engineer, EP Americas
>  Shell Exploration & Production Company
>  Woodcreek, 200 North Dairy Ashford
>  Houston, TX 77079, USA

>  Tel: +1 281 544 3016
>  Fax: +1 281 544 2238
>  Shell MeetMe Tel. +1 713 423 0600, Participant Code 62709127
>  Email: robert.prince-wright at shell.com
>  Internet: http://www.shell.com




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