How do I reverse an itertools.chain object?

Question

My function creates a chain of generators:

def bar(num):
    import itertools
    some_sequence = (x*1.5 for x in range(num))
    some_other_sequence = (x*2.6 for x in range(num))
    chained = itertools.chain(some_sequence, some_other_sequence)
    return chained

My function sometimes needs to return chained in reversed order. Conceptually, the following is what I would like to be able to do:

if num < 0:
    return reversed(chained)
return chained

Unfortunately:

>>> reversed(chained)
TypeError: argument to reversed() must be a sequence

What are my options?

This is in some realtime graphic rendering code so I don't want to make it too complicated/slow.

EDIT: When I first posed this question I hadn't thought about the reversibility of generators. As many have pointed out, generators can't be reversed.

I do in fact want to reverse the flattened contents of the chain; not just the order of the generators.

Based on the responses, there is no single call I can use to reverse an itertools.chain, so I think the only solution here is to use a list, at least for the reverse case, and perhaps for both.

Answer 1

reversed only works on objects that support len and indexing. You have to first generate all results of a generator before wrapping reversed around them.

However, you could easily do this:

def bar(num):
    import itertools
    some_sequence = (x*1.5 for x in range(num, -1, -1))
    some_other_sequence = (x*2.6 for x in range(num, -1, -1))
    chained = itertools.chain(some_other_sequence, some_sequence)
    return chained

Answer 2

def reversed2(iter):
    return reversed(list(iter))

Answer 3

In theory you can't because chained objects may even contain infinite sequences such as itertools.count(...).

You should try to reverse your generators/sequences or use reversed(iterable) for each sequence if applicable and then chain them together last-to-first. Of course this highly depends on your use case.

Answer 4

if num < 0:
    lst = list(chained)
    lst.reverse()
    return lst
else:
    return chained

reversed() needs an actual sequence, because it iterates it backwards by index, and that wouldn't work for a generator (which only has the notion of "next" item).

Since you will need to unroll the whole generator anyway for reversing, the most efficient way is to read it to a list and reverse the list in-place with the .reverse() method.

Answer 5

itertools.chain would need to implement __reversed__() (this would be best) or __len__() and __getitem__()

Since it doesn't, and there's not even a way to access the internal sequences you'll need to expand the entire sequence to be able to reverse it.

reversed(list(CHAIN_INSTANCE))

It would be nice if chain would make __reversed__() available when all the sequences are reversable, but currently it does not do that. Perhaps you can write your own version of chain that does

Answer 6

You cannot reverse generators by definition. The interface of a generator is the iterator, which is a container that supports only forward iteration. When you want to reverse a iterator, you have to collect all it's items first and reverse them after that.

Use lists instead or generate the sequences backwards from the start.