[英]Is pairwise from itertools recipes always giving the same results as zip(a, a[1:])?
在相當長的一段時間里,我使用了itertools配方中的成對函數。 ( https://docs.python.org/3.4/library/itertools.html#itertools-recipes )
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return zip(a, b)
但是,這難道不總是提供與zip(iterable,iterable [1:])相同的結果嗎?
zip(iterable, iterable[1:])
僅適用於可切片的東西。 pairwise
適用於任意可迭代對象。
為了證明這一點,這是斐波那契數列的典型功能無限列表(令人討厭的列表)的繁復版本:
from itertools import tee
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return zip(a, b)
def cons(x, xs):
yield x
yield from xs
def take(count, l):
it = iter(l)
for i in range(count):
yield next(it)
def lazy_yield_from(get_iterable):
yield from get_iterable()
infinite_fibonacci, _if = tee(
cons(1, cons(1, lazy_yield_from(lambda: (a + b for a, b in pairwise(_if))))))
print(list(take(5, infinite_fibonacci)))
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