是否有任何优雅的方式使列表包含一些整数成为包含一些元组的列表？

Question

我有一个清单

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

是否有任何优雅的方式使它们成对工作？ 我的预期是

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

Answer 1

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

是一种常见的习语

Answer 2

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

这当然假设len（a）是偶数

Answer 3

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)]) 

来自activestate，n元组的配方，而不仅仅是2元组

Answer 4

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

Answer 5

尝试使用切片和拉链。

>>> 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)]