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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 at 0x02F4D790>]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:
iter()函数,参考:
lambda函数,参考:
切片操作,参考:(完)
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