`Itertools.cycle`：大多数pythonic方式采取多个步骤？

Question

I've got some list, and I'm looking to cycle through it. The trouble is that I'm not interested in cycling through elements one at a time, but I want to cycle through by n elements at a time.

For instance, if my list is l = ["a", "b", "c", "d"] , then I would want the following output:

>>> from itertools import cycle
>>> gen = cycle(l)
>>> next(gen, 4) # I know this doesn't work -- take as pseudocode
d
>>> next(gen, 3)
c

I know that I can accomplish this with something like:

def generator_step(g, n):
    for item in range(n):
        next(g)
     return item

Answer 1

You can use itertools.islice to advance the cycle object before calling next:

from itertools import cycle, islice

c = cycle(l)
n = 4
print(next(islice(c, n-1, n)))
# d

Answer 2

In the documentation for itertools, there's a handy recipe for precisely this problem .

def consume(iterator, n):
    "Advance the iterator n-steps ahead. If n is none, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)