From: 3Jane on 8 May 2010 23:33 You could interpret [[1,2,3,4],[5,6,7,8]] as a tree and your task as traversal of its leaves. All solutions before would not work with trees with bigger height. Here is how to traverse such trees recursively: def eventualPrint(x): for v in x: if isinstance(v, list): eventualPrint(x) else: print(v) Then eventualPrint(a) does the job. This would only cope with lists as proper nodes. More general tests than isinstance() could be tried, hasattr(x, 'getitem') would match all sequence types for example. Also "for v in x:" should perhaps tested for exceptions. Optimal directives for both alternatives depend on the scope of the code's purpose. As often the price for generality is performance here. Good luck, Joost From: Tim Chase on 9 May 2010 07:42 On 05/08/2010 10:33 PM, 3Jane wrote:> You could interpret [[1,2,3,4],[5,6,7,8]] as a tree and > your task as traversal of its leaves. All solutions before > would not work with trees with bigger height. > > Here is how to traverse such trees recursively: > > def eventualPrint(x): > for v in x: > if isinstance(v, list): eventualPrint(x) > else: print(v) > > Then eventualPrint(a) does the job. Caveat: ...but wanders off into the bushes for recursive lists :) x = [1,2,3] x.append (x) eventualPrint(x) Not a particular concern in this case since I'm now two levels of hypothetical removed from the OP's question (from flatten one layer to recursively flattening arbitrary nestings to recursively flattening arbitrary & self-referential nestings)... For the record, I'm in the itertools camp of proposed solutions to the OP's question. -tkc