[EMAIL PROTECTED] wrote: > i'm looking for a way to have a list of number grouped by consecutive > interval, after a search, for example : > > [3, 6, 7, 8, 12, 13, 15] > > => > > [[3, 4], [6,9], [12, 14], [15, 16]]
Know your itertools. From the examples section[1]: """ # Find runs of consecutive numbers using groupby. The key to the # solution is differencing with a range so that consecutive numbers all # appear in same group. >>> data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28] >>> for k, g in groupby(enumerate(data), lambda (i,x):i-x): ... print map(operator.itemgetter(1), g) ... [1] [4, 5, 6] [10] [15, 16, 17, 18] [22] [25, 26, 27, 28] """ So I think something like this should work: >>> import itertools >>> def intervals(numbers): ... def key((i, n)): ... return i - n ... for key, group in itertools.groupby(enumerate(numbers), key): ... group = list(group) ... _, first_n = group[0] ... _, last_n = group[-1] ... yield first_n, last_n + 1 ... >>> list(intervals([3, 6, 7, 8, 12, 13, 15])) [(3, 4), (6, 9), (12, 14), (15, 16)] If you really need lists instead of tuples, just put brackets around the terms in the yield statement. [1] http://docs.python.org/lib/itertools-example.html STeVe -- http://mail.python.org/mailman/listinfo/python-list
