On Mon, 25 Aug 2008 17:04:07 +0000, Grzegorz Staniak wrote: > On 25.08.2008, Terry Reedy <[EMAIL PROTECTED]> wroted: > >> The newish sorted() and reversed() built-ins were meant to complement >> list.sort and list.reverse, not replace them. > > BTW, is there a reason why sorted() on a list returns a list, while > reversed() on the same list returns an iterator?
Until the day that somebody discovers how to sort a list without seeing all the items first, there's no point in sorted() returning an iterator. It has to generate a copy of the entire list to sort it, and so might as well just return the list -- there's no advantage to turning it into an iterator after you've already built the list. On the other hand, reversed() can supply it's items lazily. Although it does need access to the entire source, it doesn't need to return an entire list. It can just return the items one at a time, starting from the last one. That however does mean there's one gotcha: if you mutate a list after calling sorted() on it, the result of the sorted() doesn't change. But the same doesn't hold for reversed(): >>> L = range(5) >>> it = reversed(L) # expecting [4, 3, 2, 1, 0] as an iterator >>> it.next() 4 >>> L[3] = 'mutate' >>> it.next() 'mutate' The solution to that is simple: call list(reversed(L)). Or don't mutate the original. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
