단일 목록의 쌍

단일 목록의 쌍

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종종 나는 쌍으로 목록을 처리해야 할 필요성을 발견했습니다. 나는 그것을 수행하는 비단뱀적이고 효율적인 방법이 무엇인지 궁금해했으며 Google에서 이것을 발견했습니다.

pairs = zip(t[::2], t[1::2])

나는 그것이 충분히 비단뱀 적이라고 생각했지만, 최근 관용어 대 효율성에 관한 토론을 한 후 몇 가지 테스트를하기로 결정했습니다.

import time
from itertools import islice, izip

def pairs_1(t):
    return zip(t[::2], t[1::2]) 

def pairs_2(t):
    return izip(t[::2], t[1::2]) 

def pairs_3(t):
    return izip(islice(t,None,None,2), islice(t,1,None,2))

A = range(10000)
B = xrange(len(A))

def pairs_4(t):
    # ignore value of t!
    t = B
    return izip(islice(t,None,None,2), islice(t,1,None,2))

for f in pairs_1, pairs_2, pairs_3, pairs_4:
    # time the pairing
    s = time.time()
    for i in range(1000):
        p = f(A)
    t1 = time.time() - s

    # time using the pairs
    s = time.time()
    for i in range(1000):
        p = f(A)
        for a, b in p:
            pass
    t2 = time.time() - s
    print t1, t2, t2-t1

내 컴퓨터의 결과는 다음과 같습니다.

1.48668909073 2.63187503815 1.14518594742
0.105381965637 1.35109519958 1.24571323395
0.00257992744446 1.46182489395 1.45924496651
0.00251388549805 1.70076990128 1.69825601578

올바르게 해석하고 있다면 파이썬에서 목록, 목록 인덱싱 및 목록 분할을 구현하는 것이 매우 효율적이라는 것을 의미합니다. 편안함과 예상치 못한 결과입니다.

쌍으로 목록을 순회하는 또 다른 "더 나은"방법이 있습니까?

목록에 홀수 개의 요소가있는 경우 마지막 요소는 쌍에 포함되지 않습니다.

모든 요소가 포함되도록하는 올바른 방법은 무엇입니까?

I added these two suggestions from the answers to the tests:

def pairwise(t):
    it = iter(t)
    return izip(it, it)

def chunkwise(t, size=2):
    it = iter(t)
    return izip(*[it]*size)

These are the results:

0.00159502029419 1.25745987892 1.25586485863
0.00222492218018 1.23795199394 1.23572707176

Results so far

Most pythonic and very efficient:

pairs = izip(t[::2], t[1::2])

Most efficient and very pythonic:

pairs = izip(*[iter(t)]*2)

It took me a moment to grok that the first answer uses two iterators while the second uses a single one.

To deal with sequences with an odd number of elements, the suggestion has been to augment the original sequence adding one element (None) that gets paired with the previous last element, something that can be achieved with itertools.izip_longest().

Finally

Note that, in Python 3.x, zip() behaves as itertools.izip(), and itertools.izip() is gone.

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My favorite way to do it:

from itertools import izip

def pairwise(t):
    it = iter(t)
    return izip(it,it)

# for "pairs" of any length
def chunkwise(t, size=2):
    it = iter(t)
    return izip(*[it]*size)

When you want to pair all elements you obviously might need a fillvalue:

from itertools import izip_longest
def blockwise(t, size=2, fillvalue=None):
    it = iter(t)
    return izip_longest(*[it]*size, fillvalue=fillvalue)

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I'd say that your initial solution pairs = zip(t[::2], t[1::2]) is the best one because it is easiest to read (and in Python 3, zip automatically returns an iterator instead of a list).

To ensure that all elements are included, you could simply extend the list by None.

Then, if the list has an odd number of elements, the last pair will be (item, None).

>>> t = [1,2,3,4,5]
>>> t.append(None)
>>> zip(t[::2], t[1::2])
[(1, 2), (3, 4), (5, None)]
>>> t = [1,2,3,4,5,6]
>>> t.append(None)
>>> zip(t[::2], t[1::2])
[(1, 2), (3, 4), (5, 6)]

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I start with small disclaimer - don't use the code below. It's not Pythonic at all, I wrote just for fun. It's similar to @THC4k pairwise function but it uses iter and lambda closures. It doesn't use itertools module and doesn't support fillvalue. I put it here because someone might find it interesting:

pairwise = lambda t: iter((lambda f: lambda: (f(), f()))(iter(t).next), None)

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As far as most pythonic goes, I'd say the recipes supplied in the python source docs (some of which look a lot like the answers that @JochenRitzel provided) is probably your best bet ;)

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

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Is there another, "better" way of traversing a list in pairs?

I can't say for sure but I doubt it: Any other traversal would include more Python code which has to be interpreted. The built-in functions like zip() are written in C which is much faster.

Which would be the right way to ensure that all elements are included?

Check the length of the list and if it's odd (len(list) & 1 == 1), copy the list and append an item.

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>>> my_list = [1,2,3,4,5,6,7,8,9,10]
>>> my_pairs = list()
>>> while(my_list):
...     a = my_list.pop(0); b = my_list.pop(0)
...     my_pairs.append((a,b))
... 
>>> print(my_pairs)
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

참고URL : https://stackoverflow.com/questions/4628290/pairs-from-single-list