python - 来自 itertools 食谱的成对总是给出与 zip(a, a[1 :])?

Question

很长一段时间我都使用 itertools 食谱中的成对函数。 ( 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)

但这不是总是给出与 zip(iterable, iterable[1:]) 相同的结果吗？

Answer 1

zip(iterable, iterable[1:]) 只适用于可以切片的东西。 pairwise 适用于任意可迭代对象。

为了演示，这是斐波那契数列的典型函数式无限列表(此处为生成器)的极其复杂的版本:

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)))