Is there an expression for an infinite iterator?

Question

Is there a straight-forward expression that can produce an infinite iterator?

This is a purely theoretical question. No need for a "practical" answer here 🙂

For example, it is easy to use a generator expression to make a finite iterator:

my_gen = (0 for i in xrange(42))

However, to make an infinite one I need to "pollute" my namespace with a bogus function:

def _my_gen():
    while True:
        yield 0
my_gen = _my_gen()

Doing things in a separate file and import-ing later doesn’t count.

I also know that itertools.repeat does exactly this. I’m curious if there is a one-liner solution without that.

Asked By: hugomg

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Answer 1

None that doesn’t internally use another infinite iterator defined as a class/function/generator (not -expression, a function with yield). A generator expression always draws from anoter iterable and does nothing but filtering and mapping its items. You can’t go from finite items to infinite ones with only map and filter, you need while (or a for that doesn’t terminate, which is exactly what we can’t have using only for and finite iterators).

Trivia: PEP 3142 is superficially similar, but upon closer inspection it seems that it still requires the for clause (so no (0 while True) for you), i.e. only provides a shortcut for itertools.takewhile.

Answered By: user395760

Answer 2

Maybe you could use decorators like this for example:

def generator(first):
    def wrap(func):
        def seq():
            x = first
            while True:
                yield x
                x = func(x)
        return seq
    return wrap

Usage (1):

@generator(0)
def blah(x):
    return x + 1

for i in blah():
    print i

Usage (2)

for i in generator(0)(lambda x: x + 1)():
    print i

I think it could be further improved to get rid of those ugly (). However it depends on the complexity of the sequence that you wish to be able to create. Generally speaking if your sequence can be expressed using functions, than all the complexity and syntactic sugar of generators can be hidden inside a decorator or a decorator-like function.

Answered By: julx

Answer 3

itertools provides three infinite iterators:

count(start=0, step=1): 0, 1, 2, 3, 4, …
cycle(p): p[0], p[1], …, p[-1], p[0], …
repeat(x, times=∞): x, x, x, x, …

I don’t know of any others in the standard library.

Since you asked for a one-liner:

__import__("itertools").count()

Answered By: Katriel

Answer 4

Your OS may provide something that can be used as an infinite generator. Eg on linux

for i in (0 for x in open('/dev/urandom')):
    print i

obviously this is not as efficient as

for i in __import__('itertools').repeat(0)
    print i

Answered By: John La Rooy

Answer 5

for x in iter(int, 1): pass

Two-argument iter = zero-argument callable + sentinel value
int() always returns 0

Therefore, iter(int, 1) is an infinite iterator. There are obviously a huge number of variations on this particular theme (especially once you add lambda into the mix). One variant of particular note is iter(f, object()), as using a freshly created object as the sentinel value almost guarantees an infinite iterator regardless of the callable used as the first argument.

Answered By: ncoghlan

Answer 6

you can iterate over a callable returning a constant always different than iter()’s sentinel

g1=iter(lambda:0, 1)

Answered By: user237419

Answer 7

Quite ugly and crazy (very funny however), but you can build your own iterator from an expression by using some tricks (without “polluting” your namespace as required):

{ print("Hello world") for _ in
    (lambda o: setattr(o, '__iter__', lambda x:x)
            or setattr(o, '__next__', lambda x:True)
            or o)
    (type("EvilIterator", (object,), {}))() }

Answered By: Thomas Baruchel

Is there an expression for an infinite iterator?

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