[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, > > J.
I did a proceedural version, then re-read your post and did a generator based version ;-) === interv1 === >>> inlist = [3, 6, 7, 8, 12, 13, 15] >>> tmp = [] >>> for i,val in enumerate(inlist): ... if i==0: ... tmp.append(val) ... elif val != valinc: ... tmp += [valinc, val] ... valinc = val+1 ... >>> tmp.append(valinc) >>> tmp [3, 4, 6, 9, 12, 14, 15, 16] >>> tmp[0::2] [3, 6, 12, 15] >>> tmp[1::2] [4, 9, 14, 16] >>> zip(tmp[0::2], tmp[1::2]) [(3, 4), (6, 9), (12, 14), (15, 16)] >>> === END interv1 === === interv2 === >>> def interv2(inlist): ... for i,val in enumerate(inlist): ... if i==0: ... tmp = val ... elif val != valinc: ... yield [tmp, valinc] ... tmp = val ... valinc = val+1 ... yield [tmp, valinc] ... >>> list(interv2(inlist)) [[3, 4], [6, 9], [12, 14], [15, 16]] === END interv2 === - Paddy. -- http://mail.python.org/mailman/listinfo/python-list
