python 使用zip合并相邻的列表项

1》使用zip()函数和iter()函数，来合并相邻的列表项

>>> x

[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> zip(*[iter(x)]*2)
[(1, 2), (3, 4), (5, 6), (7, 8)]
>>> zip(*[iter(x)]*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
>>> zip(*[iter(x)]*4)

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

之所以会出现上述结果，是因为：

>>> [iter(x)]*3
[<listiterator object at 0x02F4D790>, <listiterator object at0x02F4D790>, <listiterator object at0x02F4D790>]

可以看到，列表中的3个迭代器实际上是同一个迭代器！！！

2》   在1》的基础上，封装成一个函数，如下：

>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))
>>> group_adjacent(x,3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
>>> group_adjacent(x,2)
[(1, 2), (3, 4), (5, 6), (7, 8)]
>>> group_adjacent(x,1)
[(1,), (2,), (3,), (4,), (5,), (6,), (7,), (8,), (9,)]

3》使用zip()函数和切片操作，来合并相邻的表项

>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> zip(x[::2],x[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8)]
>>> zip(x[0::2],x[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8)]
>>> zip(x[0::3],x[1::3],x[2::3])
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
>>> zip(x[::3],x[1::3],x[2::3])
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]

4》  在3》的基础上，封装成函数，如下：

>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> group_adjacent = lambda a, k: zip(*[a[i::k] for i in range(k)])
>>> group_adjacent(x,3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
>>> group_adjacent(x,2)
[(1, 2), (3, 4), (5, 6), (7, 8)]
>>> group_adjacent(x,1)
[(1,), (2,), (3,), (4,), (5,), (6,), (7,), (8,), (9,)]

友情链接：

zip()函数，参考：python zip( )函数

iter()函数，参考：python iter( )函数

lambda函数，参考：python lambda函数基础
切片操作，参考：python切片操作

（完）