本文实例讲述了Python使用zip合并相邻列表项的方法。分享给大家供大家参考，具体如下：
1》使用zip()函数和iter()函数，来合并相邻的列表项
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>>> 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)]
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之所以会出现上述结果，是因为：
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>>> [iter(x)]*3
[<listiterator object at 0x02F4D790>, <listiterator object at0x02F4D790>, <listiterator object at 0x02F4D790>]
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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：
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>>> 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,)]
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3》使用zip()函数和切片操作，来合并相邻的表项
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>>> 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)]
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4》 在3》的基础上，封装成函数，如下：
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>>> 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,)]
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参考文章：
python zip()函数http://www.zzvips.com/article/134351.html
python iter()函数http://www.zzvips.com/article/134350.html
python lambda函数基础http://www.zzvips.com/article/134346.html
python切片操作http://www.zzvips.com/article/134345.html
希望本文所述对大家Python程序设计有所帮助。
原文链接：http://blog.csdn.net/sxingming/article/details/51476869