Question

可能重复：
  How do you split a list into evenly sized chunks in Python?
  python: convert “5,4,2,4,1,0” into [[5, 4], [2, 4], [1, 0]]

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

- ＆GT;

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

有没有简单的方法可以做到这一点，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)]

How does zip(*[iter(s)]*n) work in Python?

Answer 2

如果你真的希望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zip截断的左侧元素，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

您也可以在此使用numpy.reshape：

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)]

Answer 5

这是使用递归执行此操作的一种“聪明”方式：

from itertools import chain

def groupsof(n, xs):
    if len(xs) < n:
        return [xs]
    else:
        return chain([xs[0:n]], groupsof(n, xs[n:]))

print list(groupsof(3, [1,2,3,4,5,6,7,8,9,10,11,12,13]))

如何将python列表拆分成相同大小的块？

5 个答案: