如何将python列表拆分成相同大小的块？

Question

可能重复:
如何在Python中将列表拆分为大小均匀的块？
python:将"5,4,2,4,1,0"转换为[[5,4],[2,4],[1,0]]

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

- >

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

是否有简单的方法来做到这一点,没有明确的'for'？

Answer 1

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

如何zip(*[iter(s)]*n)在Python 中工作？

Answer 2

如果你真的想要子元素是列表与元组:

In [9]: [list(t) for t in zip(*[iter(range(1,10))]*3)]
Out[9]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

或者,如果要包含将被截断的左侧元素zip,请使用切片语法:

In [16]: l=range(14)

In [17]: [l[i:i+3] for i in range(0,len(l),3)]
Out[17]: [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13]]

Answer 3

你也可以numpy.reshape在这里使用:

import numpy as np

x = np.array([1,2,3,4,5,6,7,8,9])

new_x = np.reshape(x, (3,3))

结果:

>>> new_x
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

Answer 4

>>> map(None,*[iter(s)]*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]