我有一些清单,我正在寻找它。麻烦的是我对一次循环一个元素不感兴趣,但我希望一次循环n
个元素。
例如,如果我的列表是l = ["a", "b", "c", "d"]
,那么我想要以下输出:
>>> from itertools import cycle
>>> gen = cycle(l)
>>> next(gen, 4) # I know this doesn't work -- take as pseudocode
d
>>> next(gen, 3)
c
我知道我可以通过以下方式实现这一目标:
def generator_step(g, n):
for item in range(n):
next(g)
return item
答案 0 :(得分:2)
您可以使用itertools.islice
推进循环对象,然后再调用下一个:
from itertools import cycle, islice
c = cycle(l)
n = 4
print(next(islice(c, n-1, n)))
# d
答案 1 :(得分:1)
在itertools的文档中,有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)