回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>编写一个通用函数,它可以迭代任何返回now,next pairs的iterable。在</p>
<pre><code>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')
</code></pre>
<p>我一直在考虑编写一个函数的最佳方法,在这个函数中可以设置每个元组中的项数。在</p>
<p>例如如果它是</p>
^{pr2}$
<p>结果是:</p>
<pre><code>('h','e','l')
('e','l','l')
('l','l','o')
</code></pre>
<p>我刚开始使用timeit,因此如果我在这里做了什么错事,请指出:</p>
<pre><code>import timeit
def n1(iterable, n=1):
#now_nxt_deque
from collections import deque
deq = deque(maxlen=n)
for i in iterable:
deq.<a href="https://www.cnpython.com/list/append" class="inner-link">append</a>(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))
</code></pre>
<p>结果:</p>
<pre><code>$ py time_this_function.py
func 1: 0.163129
func 2: 2.383288
func 3: 1.908363
</code></pre>