Question

鉴于N 2d numpy数组，有没有一种简洁的方法可以在对角线上“堆叠”或“闩锁”它们，用0填充任何新的插槽？例如。给出：

arr1 = np.array([[1, 2],
                 [3, 4]])

arr2 = np.array([[9, 8, 7],
                 [6, 5, 4],
                 [3, 2, 1]])

我想创建：

arr = np.array([[1, 2, 0, 0, 0],
                [3, 4, 0, 0, 0], 
                [0, 0, 9, 8, 7],
                [0, 0, 6, 5, 4],
                [0, 0, 3, 2, 1]])

Answer 1

There's a function for that.

scipy.linalg.block_diag(arr1, arr2)

需要任意多个参数：

scipy.linalg.block_diag(*list_of_arrays)

Answer 2

尝试：

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并且，为了验证它是否有效，我们可以显示>>> arr = np.zeros((5, 5)) >>> arr[:2, :2] = arr1 >>> arr[2:, 2:] = arr2：

arr

Answer 3

我刚刚发现这似乎完全符合我的需要：

https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.linalg.block_diag.html

>> from scipy.linalg import block_diag
>>> A = [[1, 0],
...      [0, 1]]
>>> B = [[3, 4, 5],
...      [6, 7, 8]]
>>> C = [[7]]
>>> block_diag(A, B, C)
[[1 0 0 0 0 0]
 [0 1 0 0 0 0]
 [0 0 3 4 5 0]
 [0 0 6 7 8 0]
 [0 0 0 0 0 7]]

Answer 4

np有一个相对较新的block函数，用于连接嵌套的数组列表。但它需要指定零填充块：

In [15]: np.block([[arr1, np.zeros((2,3),int)], [np.zeros((3,2),int), arr2]])
Out[15]: 
array([[1, 2, 0, 0, 0],
       [3, 4, 0, 0, 0],
       [0, 0, 9, 8, 7],
       [0, 0, 6, 5, 4],
       [0, 0, 3, 2, 1]])

将numpy数组堆叠到对角

4 个答案: