python - 如果收藏中有任何东西可以进行这种细分吗？| Python

Question

(a,b), (c,d), (e,f), (g,h) => (b, c), (d, e), (f, g)

所以我们省略第一个和最后一个并处理反向段

我第三次打这个，总是我写自己的代码......

Answer 1

collections在或中都没有这样的东西itertools。

可以chain.from_iterable成对，用 删除第一个和最后一个元素islice，然后使用grouper配方将它们变成对，但您仍然需要自己定义grouper函数。

我会为此定义我自己的优化实用程序函数。

def shift(pairs):
    it = iter(pairs)
    (_, p) = next(it)
    for (n, p2) in it:
        yield (p, n)
        p = p2

Answer 2

You can do this:

>>> tups=[('a','b'), ('c','d'), ('e','f'), ('g','h')]
>>> zip(*[iter([e for t in tups for e in t][1:-1])]*2)
[('b', 'c'), ('d', 'e'), ('f', 'g')]

Edit

The code I posted is most assuredly NOT 'an order of magnitude slower' than using itertools or KennyTM's solution:

from __future__ import print_function
import itertools
from collections import defaultdict

li=[('a','b'), ('c','d'), ('e','f'), ('g','h')]

def t1(tups):
    return zip(*[iter([e for t in tups for e in t][1:-1])]*2)

def t2(tups):
    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 itertools.izip_longest(fillvalue=fillvalue, *args)        
    flat=itertools.chain.from_iterable(tups)
    sl=itertools.islice(flat,1,len(tups)*2-1)        
    return list(grouper(sl,2))

def t3(pairs):
    it = iter(pairs)
    (_, p) = next(it)
    for (n, p2) in it:
        yield (p, n)
        p = p2                    

if __name__ == '__main__':   
    import timeit 
    n=1000000
    print(t1(li),t2(li), list(t3(li)))
    assert t1(li)==t2(li)==list(t3(li))
    print(timeit.timeit("t1(li)", setup="from __main__ import t1, li",number=n))
    print(timeit.timeit("t2(li)", setup="from __main__ import t2, li, itertools",number=n))
    print(timeit.timeit("list(t3(li))", setup="from __main__ import t3, li",number=n))

Prints:

3.02761888504
4.29241204262
2.12067294121

Which shows they are all roughly the same speed. KennyTM's is fastes, but not by an order of magnitude.