将元素插入numy数组并获得所有滚动排列

你在这里要求的是 Toeplitz Matrix ,即:

一种矩阵,其中从左到右的每个降序对角线都是常数

scipy 具有易于使用的实现:

from scipy.linalg import toeplitz

def magic_toeplitz(arr, to_add):
    return toeplitz(np.hstack([to_add, arr[::-1]]), np.hstack([to_add, arr]))

a = [6,7,8,9]
add = [1]
magic_toeplitz(a, add)

array([[1, 6, 7, 8, 9],
       [9, 1, 6, 7, 8],
       [8, 9, 1, 6, 7],
       [7, 8, 9, 1, 6],
       [6, 7, 8, 9, 1]])

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

out = toeplitz(np.hstack([[np.nan], B[::-1]]), np.hstack([np.nan, B]))
out = np.tile(out, (len(A), 1, 1))
m = np.ma.array(out, mask=np.isnan(out))
vals = np.repeat(A, (B.shape[0] + 1)**2).reshape(out.shape)
print(m.filled(vals))

array([[[1, 6, 7, 8, 9],
        [9, 1, 6, 7, 8],
        [8, 9, 1, 6, 7],
        [7, 8, 9, 1, 6],
        [6, 7, 8, 9, 1]],

       [[2, 6, 7, 8, 9],
        [9, 2, 6, 7, 8],
        [8, 9, 2, 6, 7],
        [7, 8, 9, 2, 6],
        [6, 7, 8, 9, 2]],

       [[3, 6, 7, 8, 9],
        [9, 3, 6, 7, 8],
        [8, 9, 3, 6, 7],
        [7, 8, 9, 3, 6],
        [6, 7, 8, 9, 3]],

       [[4, 6, 7, 8, 9],
        [9, 4, 6, 7, 8],
        [8, 9, 4, 6, 7],
        [7, 8, 9, 4, 6],
        [6, 7, 8, 9, 4]],

       [[5, 6, 7, 8, 9],
        [9, 5, 6, 7, 8],
        [8, 9, 5, 6, 7],
        [7, 8, 9, 5, 6],
        [6, 7, 8, 9, 5]]])

2D

我们可以利用 np.lib.stride_tricks.as_strided 基于 scikit-image's view_as_windows 去拿推拉窗。 More info on use of as_strided based view_as_windows .

1D 数组以及要添加的新值( 1 在这种情况下)然后 仔细观察。作为一个视图,它将是非常有效的,看起来像这样-

from skimage.util.shape import view_as_windows

def rolling_add(a,val=1):
    a_ext = np.r_[a,val,a]
    return view_as_windows(a_ext,len(a)+1,1)[::-1]

np.lib.u技巧。当你大步走的时候 要避免翻转部分: [::-1] ,但读者可能很难理解这种设置。

样品运行-

In [254]: a = np.array([6, 7, 8, 9])

In [255]: rolling_add(a)
Out[255]: 
array([[1, 6, 7, 8, 9],
       [9, 1, 6, 7, 8],
       [8, 9, 1, 6, 7],
       [7, 8, 9, 1, 6],
       [6, 7, 8, 9, 1]])

In [263]: a = np.random.randint(0,10,10000)

In [264]: %timeit rolling_add(a)
10000 loops, best of 3: 58 µs per loop

3D 案例

三维

def rolling_add3D(a,add_ar):
    a_ext = np.r_[a,0,a]
    a_ext2 = np.repeat(a_ext[None],len(add_ar),0)
    a_ext2[:,len(a)] = add_ar
    return view_as_windows(a_ext2,(1,len(a)+1))[...,0,:][:,::-1]

样品运行-

In [292]: a
Out[292]: array([6, 7, 8, 9])

In [293]: rolling_add3D(a,[1,2,3])
Out[293]: 
array([[[1, 6, 7, 8, 9],
        [9, 1, 6, 7, 8],
        [8, 9, 1, 6, 7],
        [7, 8, 9, 1, 6],
        [6, 7, 8, 9, 1]],

       [[2, 6, 7, 8, 9],
        [9, 2, 6, 7, 8],
        [8, 9, 2, 6, 7],
        [7, 8, 9, 2, 6],
        [6, 7, 8, 9, 2]],

       [[3, 6, 7, 8, 9],
        [9, 3, 6, 7, 8],
        [8, 9, 3, 6, 7],
        [7, 8, 9, 3, 6],
        [6, 7, 8, 9, 3]]])

非常大的数组上的计时-

In [294]: a = np.random.randint(0,10,10000)

In [295]: %timeit rolling_add3D(a,[1,2,3])
10000 loops, best of 3: 83.7 µs per loop

性能将与要添加的阵列长度成正比。所以,添加一个 1000 元素排列成 10000 输入数组的长度为-

In [301]: a = np.random.randint(0,10,10000)

In [302]: add_array = np.random.randint(0,10,1000)

In [303]: %timeit rolling_add3D(a,add_array)
100 loops, best of 3: 16.9 ms per loop

for i in A:
   new_list = B[:]
   new_list.append(i)
   print(new_list)
   for q in B:
        new_list = [new_list[-1]]+new_list[:-1]
        print(new_list)
   print("-"*15)