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

*To*: mathgroup at smc.vnet.net*Subject*: [mg64954] Re: Fastest method for comparing overlapping times in random time series*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 10 Mar 2006 05:14:45 -0500 (EST)*Organization*: The University of Western Australia*References*: <dulspn$3an$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <dulspn$3an$1 at smc.vnet.net>, "Prince-Wright, Robert G SEPCO" <robert.prince-wright at shell.com> 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. > ... > 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. How about using Interval arithmetic? ints = Outer[IntervalIntersection, Interval /@ list1, Interval /@ list2] List @@ IntervalUnion @@ Flatten[ints] For your data I get {{0, 0}, {7.9785370986659725, 7.9795152152343825}, {23.964317142793323, 24.053482621096574}, {31.09821651735643, 31.141635693642247}, {32.51349069353666, 32.5469642626066}} Perhaps this will be sufficiently fast? Note that using Outer is inefficient because IntervalIntersection is commutative, so that each intersection is computed twice. This can be overcome by using Map instead of Outer. > 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). This Cell Expression was corrupted by some spurious ">" characters (possibly due to the code being "quoted"). For example, > RowBox[> {"{", > RowBox[{ > "12.599440581844581", ",", "12.69944058184458"}], "}"}], Cheers, Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul