Python 是否有一种优雅的方法可以使一个列表包含一些整数，从而使一个列表包含一些元组？

我有一张单子

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

有什么优雅的方法可以让它们成对工作吗？ 我的预期结果是

[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

这是一个常见的成语

[(a[2*i], a[2*i+1] ) for i in range(len(a)/2)]

这当然是假设len（a）是偶数

def group(lst, n):
    """group([0,3,4,10,2,3], 2) => [(0,3), (4,10), (2,3)]

    Group a list into consecutive n-tuples. Incomplete tuples are
    discarded e.g.

    >>> group(range(10), 3)
    [(0, 1, 2), (3, 4, 5), (6, 7, 8)]
    """
    return zip(*[lst[i::n] for i in range(n)]) 

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从activestate，n元组的配方，而不仅仅是2元组

尝试使用切片和zip

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> zip(a[::2],a[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

---

“有没有优雅的方式……”在Python中，答案总是肯定的；）你能解释一下吗？

def group(lst, n):
    """group([0,3,4,10,2,3], 2) => [(0,3), (4,10), (2,3)]

    Group a list into consecutive n-tuples. Incomplete tuples are
    discarded e.g.

    >>> group(range(10), 3)
    [(0, 1, 2), (3, 4, 5), (6, 7, 8)]
    """
    return zip(*[lst[i::n] for i in range(n)]) 

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> zip(a[::2],a[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]