以顺时针方式用1D数组填充2D数组-NumPy/Python

Question

假设有两个数组：

inner_array = np.array([[3, 3, 3],
[6, 6, 6],
[9, 9, 9]])
outer_array = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16])

创建这样的数组最干净的方法是什么：

[[ 1,  2,  3,  4, 5],
[16,  3,  3,  3, 6],
[15,  6,  6,  6, 7],
[14,  9,  9,  9, 8],
[13, 12, 11, 10, 9]]

如果可以控制外阵列的起始位置，那就更好了

Answer 1

以下是选择起始位置作为附加参数的一种方法-

def fill_around(inner_array, outer_array, origin=0):
outer_array = np.roll(outer_array, origin)
m,n = inner_array.shape
out = np.pad(inner_array,(1,1))

s = np.split(outer_array, np.cumsum([n+2,m,n+2]))
out[[0,-1]] = [s[0],s[2][::-1]]
out[1:-1,::n+1] = np.dstack([s[3][::-1],s[1]])
return out

给定数据的样本运行-

In [181]: fill_around(inner_array, outer_array)
Out[181]: 
array([[ 1,  2,  3,  4,  5],
[16,  3,  3,  3,  6],
[15,  6,  6,  6,  7],
[14,  9,  9,  9,  8],
[13, 12, 11, 10,  9]])

设置起始位置-

In [8]: fill_around(inner_array, outer_array, origin=2)
Out[8]: 
array([[15, 16,  1,  2,  3],
[14,  3,  3,  3,  4],
[13,  6,  6,  6,  5],
[12,  9,  9,  9,  6],
[11, 10,  9,  8,  7]])