python/numpy组合子数组一次4行

Question

我有一个由每行分开的numpy数组:

splitArray:


[[0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0]]

我希望每4行合并一次splitArray,最后一个子阵列不一定必须是4,而只剩下剩下的剩余部分.

下面是我希望拥有的数组:

joinedArray:


[[0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0],
 [0,   0,   0,   0,   0,   0,  0,
  0,   0,   0,   0,   0,   0,  0]]

Answer 1

使用list-comp:

[a[i:i+4] for i in range(0, len(a), 4)]
#[array([[0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0]]),
# array([[0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0]]),
# array([[0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0]]),
# array([[0, 0, 0, 0, 0, 0, 0],
#        [0, 0, 0, 0, 0, 0, 0]])]

Answer 2

作为纯粹的 Numpythonic 方法，您可以通过创建从分块数到行数的范围（并将分块数step作为range. 然后使用np.split()来分割你的数组：

In [24]: def chunk_array(arr, ch):
    ...:     x = arr.shape[0]
    ...:     return np.split(a, np.arange(ch, x, ch))
    ...: 
    ...: 

演示：

In [25]: chunk_array(a, 4)
Out[25]: 
[array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]])]

In [26]: chunk_array(a, 3)
Out[26]: 
[array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]]), array([[0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0]])]

如果您希望连接分块数组，您可以在列表理解中使用 @jpp 的答案与np.concatenate()和或稍有不同。map

In [75]: def chunk_array(arr, ch):
    ...:     x = arr.shape[0]
    ...:     return [np.concatenate(subs) for subs in np.split(arr, np.arange(ch, x, ch))]