python现在,接下来,n迭代

Question

编写一个通用函数,可以迭代任何可迭代的返回现在,下一对.

def now_nxt(iterable):
    iterator = iter(iterable)
    nxt = iterator.__next__()
    for x in iterator:
        now = nxt
        nxt = x
        yield (now,nxt) 

for i in now_nxt("hello world"):
    print(i)

('h', 'e')
('e', 'l')
('l', 'l')
('l', 'o')
('o', ' ')
(' ', 'w')
('w', 'o')
('o', 'r')
('r', 'l')
('l', 'd')

我一直在考虑编写函数的最佳方法,其中可以设置每个元组中的项目数.

例如,如果是的话

func("hello",n=3)

结果将是:

('h','e','l')
('e','l','l')
('l','l','o')

我是新手使用timeit,所以请指出我在这里做错了什么:

import timeit

def n1(iterable, n=1):
    #now_nxt_deque
    from collections import deque
    deq = deque(maxlen=n)
    for i in iterable:
        deq.append(i)
        if len(deq) == n:
            yield tuple(deq)

def n2(sequence, n=2):
    # now_next
    from itertools import tee
    iterators = tee(iter(sequence), n)
    for i, iterator in enumerate(iterators):
        for j in range(i):
            iterator.__next__()
    return zip(*iterators)

def n3(gen, n=2):
    from itertools import tee, islice
    gens = tee(gen, n)
    gens = list(gens)
    for i, gen in enumerate(gens):
        gens[i] = islice(gens[i], i, None) 
    return zip(*gens)


def prin(func):
    for x in func:
        yield x

string = "Lorem ipsum tellivizzle for sure ghetto, consectetuer adipiscing elit."

print("func 1: %f" %timeit.Timer("prin(n1(string, 5))", "from __main__ import n1, string, prin").timeit(100000))
print("func 2: %f" %timeit.Timer("prin(n2(string, 5))", "from __main__ import n2, string, prin").timeit(100000))
print("func 3: %f" %timeit.Timer("prin(n3(string, 5))", "from __main__ import n3, string, prin").timeit(100000))

结果:

$  py time_this_function.py 
func 1: 0.163129
func 2: 2.383288
func 3: 1.908363

Answer 1

我的建议是,

from collections import deque

def now_nxt_deque(iterable, n=1):
    deq = deque(maxlen=n)
    for i in iterable:
        deq.append(i)
        if len(deq) == n:
            yield tuple(deq)

for i in now_nxt_deque("hello world", 3):
    print(i)

('h', 'e', 'l')
('e', 'l', 'l')
('l', 'l', 'o')
('l', 'o', ' ')
('o', ' ', 'w')
(' ', 'w', 'o')
('w', 'o', 'r')
('o', 'r', 'l')
('r', 'l', 'd')

Answer 2

这是一个非常简单的方法:

• n使用using 克隆迭代器时间itertools.tee
• 推进i迭代器i时间
• izip 他们都在一起

import itertools

def now_next(sequence, n=2):
    iterators = itertools.tee(iter(sequence), n)
    for i, iterator in enumerate(iterators):
        for j in range(i):
            iterator.next()
    return itertools.izip(*iterators)