The problem

Often, using an iterator lazily is better than generating a sequence (like the one you get from a list comprehension). For example, compare these two scripts:

The first one has to read the entire file into memory, which can be a problem for huge files, but usually you don't have files that big.

A more serious problem is that it has to read the entire file before it can do any work. Even with moderately-sized files, the corresponding delay at startup can make debugging more painful (each run takes seconds instead of milliseconds before you can see whether the first results are correct). And it can even have a significant performance cost (by preventing the file reads from interleaving with the network reads, which takes away caching opportunities from the OS/filesystem/drive).

The problem with the second one is that you can only iterate a file once. If you try to iterate it a second time, it'll be empty (because you've already iterated the whole thing). So, the following code gives you 2 instead of 4:

Of course little of this is specific to files. If you have an generator that requires a lot of CPU work to run, running it to completion before you get started causes the same startup delay, and can prevent pipelining of work, which can have a huge cost in CPU cache misses. But just leaving it as a generator means you can't iterate through it repeatedly, or you'll get nothing each time but the first.

The solution

So, is there a way to get an iterable that's lazy the first time you run through it like an iterator, but restartable like a sequence?

Well, you could build a complete "lazy sequence" class, but that's a lot of work, with some fiddly edge cases to deal with if in order to handle the full interface properly (including things like indexing and slicing with negative values).

Fortunately, you don't need the full interface. You need __next__ to store the values as you create them, and __iter__ to give you a new iterator that shares the same storage.

The easy way

As it turns out, that's exactly what itertools.tee does:
The problem is that it's tedious and error-prone to have to call tee explicitly each time you want to iterate. But you can easily wrap this up:
Now Reiterable is as easy to use as list, and gives you the benefit you care about (being able to iterate the values repeatedly) without the cost (iterating the entire thing up front).

Performance

The documentation shows you how tee is implemented. And if you don't understand it, it's probably worth copying the pure-Python implementation from the docs and stepping through what it does.

But the basic idea is this: Each time you call tee, it creates two new deques and two new generators, both tied to the original iterator. Whenever either generator needs a value that wasn't yet produced, it's taken from the original iterator and added to both deques, and then of course immediately popped from one.

So, the first time through Reiterable, it iterates the values on demand, and copies each value to the spare generator's deque. Each subsequent time, it's doing the same, but from an iterator over the spare deque instead of from the original iterator. So the values get moved from one deque to the next, with no wasted space and very little wasted time, right?

Well, not quite. This is hard to see with the C implementation of tee, or even the generator-based implementation given in the docs, but it you build a class-based implementation, you can see what's going on. Unfortunately, the class implementation seems to break the online interactive visualizer, so you'll need to copy the code below and run it locally:

    def tee(iterable, n=2):
        class gen(object):
            it = iter(iterable)
            deques = [collections.deque() for i in range(n)]
            def __init__(self, d):
                self.d = d
            def __iter__(self):
                return self
            def __next__(self):
                if not self.d:              # when the local deque is empty
                    newval = next(gen.it)      # fetch a new value and
                    for d in gen.deques:        # load it to all the deques
                        d.append(newval)
                return self.d.popleft()
        return tuple(gen(d) for d in gen.deques)

    class Reiterable(object):
        def __init__(self, iterable):
            self.iterable = iterable
        def __iter__(self):
            self.iterable, t = itertools.tee(self.iterable)
            return t

    f = io.StringIO('abc\ndef\n')
    f = Reiterable(f)
    for i in range(3):
        list(f)
    print(f.iterable.it.it.it)

Algorithmic analysis

Reiterable is building up a chain of tee objects. It is moving the values from one deque to the next, so all but the highest are empty, but there is still a deque and a tee wrapper object. Each value iterated is just moved from the highest deque on the chain to the new deque, so the wasted time per iteration step is minimal, but when you run out of values, it has to run through the whole chain to discover that they're all empty before the new iterator can be declared empty.

So, to iterate N items M times, instead of wasting N space to hold a copy of the iterable, you're wasting N+M space to hold a copy of the iterable and a chain of M empty tees and deques. And instead of NM time for the iteration, it's NM+M/2 time for the iteration plus the extra empty checks (which is still O(NM), of course)>

So, there's no algorithmic cost, except in edge cases when M >> N, which is a very strange use case. (If N is tiny, you really should just use a list; if M is gigantic, that almost always means you're doing a nested iteration that you can just flip over.)

Real-life performance

The real cost is the added overhead of having to go through the tee's generator for each value instead of just going through a list iterator. Which you can time pretty easily, so let's try it:

    In [66]: def func():
        ...      f = (i for i in range(1000000))
        ...      sum(f)
    In [67]: %timeit func()]
    10 loops, best of 3: 73.2 ms per loop

    In [68]: def func():
        ...      f = (i for i in range(1000000))
        ...      sum(list(f))
    In [68]: %timeit func()]
    10 loops, best of 3: 101 ms per loop

    In [69]: def func():
        ...      f = (i for i in range(1000000))
        ...      sum(Reiterable(f))
    In [70]: %timeit func()]
    10 loops, best of 3: 108 ms per loop

So, there is an additional performance cost to building a tee out of an iterator vs. building a list… but it's only about 25% higher.