On Tue, 10 May 2005 15:14:47 +0200, Max M <[EMAIL PROTECTED]> wrote: >I am writing a "find-free-time" function for a calendar. There are a lot >of time spans with start end times, some overlapping, some not. > >To find the free time spans, I first need to convert the events into a >list of non overlapping time spans "meta-spans". > >This nice ascii graph should show what I mean. > >1) --- >2) --- >3) --- >4) ----- >5) ----- > > >> ------- -------- # meta spans > >I can then iterate through the meta-spans and find non-busy times. > >I have written the class below, but it is rather O^2, so I wondered if >anybody has an idea for a better approach? > >
>>> spans = [(0,3), (4,7), (2,5), (9,14), (12,17)] Non-recommended one-liner: >>> reduce(lambda L,se: (L and se[0]<=L[-1][1] and (L.append((L.pop()[0], >>> se[1])) or L)) or L.append(se) or L ,sorted(spans), []) [(0, 7), (9, 17)] Regards, Bengt Richter -- http://mail.python.org/mailman/listinfo/python-list
