Gerard Flanagan wrote: > [EMAIL PROTECTED] wrote: > > hello, > > > > 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]] > > > > (6, not following 3, so 3 => [3:4] ; 7, 8 following 6 so 6, 7, 8 => > > [6:9], and so on) > > > > i was able to to it without generators/yield but i think it could be > > better with them, may be do you an idea? > > > > best regards, > > > > a list comprehension/itertools version (this won't work with an empty > list): > > from itertools import groupby > > a = [3, 6, 7, 8, 12, 13, 15] > > result = [[3, 4], [6,9], [12, 14], [15, 16]] > > b = [ list(g)[0] for k,g in groupby(range(a[0],a[-1]+2), lambda x: > x in a)] > > c = [ b[i:i+2] for i in range(0,len(a),2) ] > > assert c == result > > > Gerard
change len(a) to len(b) in case a has duplicates like [3,3,3,6,7,8,12,13,15] Gerard -- http://mail.python.org/mailman/listinfo/python-list
