任何一种优雅的方式都可以使列表包含一些整数，从而使列表包含某些元组

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

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

pairs = zip(*[iter(a)]*2)

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

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)]) 

从activestate，一个n元组的配方，而不仅仅是2元组

b = []
for i in range(0,len(a),2):
    b.append((a[i],a[i+1]))
a = b

>>> 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)]