On Fri, 12 Oct 2012 00:21:57 +0200, Hans Mulder wrote: > On 9/10/12 04:39:28, rusi wrote: >> On Oct 9, 7:34 am, rusi <rustompm...@gmail.com> wrote: >>> How about a 2-paren version? >>> >>>>>> x = [1,2,3] >>>>>> reduce(operator.add, [['insert', a] for a in x]) >>> >>> ['insert', 1, 'insert', 2, 'insert', 3] >> >> Or if one prefers the different parens on the other side: >> >>>>> reduce(operator.add, (['insert', a] for a in x)) >> ['insert', 1, 'insert', 2, 'insert', 3] > > Or, if you don't want to import the operator module: > > sum((['insert', a] for a in x), [])
Which is also O(N**2) like the reduce solution above. That means that it will seem perfectly fine when you test it using a a hundred or so items, then some day you'll pass it a list with ten million items and it will take 36 hours to complete. I'm serious by the way. By my tests, increasing the number of items in the list by a factor of ten increases the time taken by between 30 and 300 times. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
