Numpy 在三维阵列的每个二维切片上的随机位置填充块

例如，我有3D numpy阵列，如下所示：

array([[[ 0,  1,  2,  3],
        [ 4,  5,  6,  7],
        [ 8,  9, 10, 11],
        [12, 13, 14, 15]],

       [[16, 17, 18, 19],
        [20, 21, 22, 23],
        [24, 25, 26, 27],
        [28, 29, 30, 31]]])

是否有一种方法可以对其进行索引，例如，我选择第一个平面中2x2元素的右上角，以及第二个平面中的中心2x2元素子阵列？这样我就可以把元素2,3,6,7,21,22,25,26归零：

array([[[ 0,  1,  0,  0],
        [ 4,  5,  0,  0],
        [ 8,  9, 10, 11],
        [12, 13, 14, 15]],

       [[16, 17, 18, 19],
        [20,  0,  0, 23],
        [24,  0,  0, 27],
        [28, 29, 30, 31]]])

我有一批图像，我需要调零出一个固定大小的小窗口，但在批中每个图像的不同随机位置。第一个维度是图像的数量

大概是这样的： a[：，x:x+2，y:y+2]=0

其中x和y是向量，对于a的每个第一维度具有不同的值

方法1：这里有一种方法主要基于线性索引-

方法2：对于使用-

方法3：使用旧的trusty循环-

def random_block_fill_loopy(a, N, fillval=0):
    # a is input array
    # N is blocksize

    # Store shape info
    m,n,r = a.shape

    # Generate random start indices for second and third axes keeping proper
    # distance from the boundaries for the block to be accomodated within.
    idx0 = np.random.randint(0,n-N+1,m)
    idx1 = np.random.randint(0,r-N+1,m)

    # Iterate through first and use slicing to assign fillval.
    for i in range(m):
        a[i, idx0[i]:idx0[i]+N, idx1[i]:idx1[i]+N] = fillval        
    return a

样本运行-

In [357]: a = np.arange(2*4*7).reshape(2,4,7)

In [358]: a
Out[358]: 
array([[[ 0,  1,  2,  3,  4,  5,  6],
        [ 7,  8,  9, 10, 11, 12, 13],
        [14, 15, 16, 17, 18, 19, 20],
        [21, 22, 23, 24, 25, 26, 27]],

       [[28, 29, 30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39, 40, 41],
        [42, 43, 44, 45, 46, 47, 48],
        [49, 50, 51, 52, 53, 54, 55]]])

In [359]: random_block_fill_adv(a, N=3, fillval=0)
Out[359]: 
array([[[ 0,  0,  0,  0,  4,  5,  6],
        [ 7,  0,  0,  0, 11, 12, 13],
        [14,  0,  0,  0, 18, 19, 20],
        [21, 22, 23, 24, 25, 26, 27]],

       [[28, 29, 30, 31, 32, 33, 34],
        [35, 36, 37, 38,  0,  0,  0],
        [42, 43, 44, 45,  0,  0,  0],
        [49, 50, 51, 52,  0,  0,  0]]])

有趣的事情：在原地填充，如果我们继续运行random_block_fill_adva，N=3，fillval=0，我们最终会得到全零a。因此，还需要验证代码

运行时测试

---

因此，选择哪一个取决于第一个轴的长度和块大小。

@Abdou如前所述-基本上我需要将一块固定大小的块归零，但数组中每个平面的随机位置不同。所以，不会有一个特定的输出，但是是的，OP可以列出一个可能的输出。我更新了问题，希望它更清楚我要做什么。@Divakar，提取和归零是两件不同的事情。这就是我要求获得预期输出的原因。

def random_block_fill_loopy(a, N, fillval=0):
    # a is input array
    # N is blocksize

    # Store shape info
    m,n,r = a.shape

    # Generate random start indices for second and third axes keeping proper
    # distance from the boundaries for the block to be accomodated within.
    idx0 = np.random.randint(0,n-N+1,m)
    idx1 = np.random.randint(0,r-N+1,m)

    # Iterate through first and use slicing to assign fillval.
    for i in range(m):
        a[i, idx0[i]:idx0[i]+N, idx1[i]:idx1[i]+N] = fillval        
    return a

In [357]: a = np.arange(2*4*7).reshape(2,4,7)

In [358]: a
Out[358]: 
array([[[ 0,  1,  2,  3,  4,  5,  6],
        [ 7,  8,  9, 10, 11, 12, 13],
        [14, 15, 16, 17, 18, 19, 20],
        [21, 22, 23, 24, 25, 26, 27]],

       [[28, 29, 30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39, 40, 41],
        [42, 43, 44, 45, 46, 47, 48],
        [49, 50, 51, 52, 53, 54, 55]]])

In [359]: random_block_fill_adv(a, N=3, fillval=0)
Out[359]: 
array([[[ 0,  0,  0,  0,  4,  5,  6],
        [ 7,  0,  0,  0, 11, 12, 13],
        [14,  0,  0,  0, 18, 19, 20],
        [21, 22, 23, 24, 25, 26, 27]],

       [[28, 29, 30, 31, 32, 33, 34],
        [35, 36, 37, 38,  0,  0,  0],
        [42, 43, 44, 45,  0,  0,  0],
        [49, 50, 51, 52,  0,  0,  0]]])

In [579]: a = np.random.randint(0,9,(10000,4,4))

In [580]: %timeit random_block_fill_lidx(a, N=2, fillval=0)
     ...: %timeit random_block_fill_adv(a, N=2, fillval=0)
     ...: %timeit random_block_fill_loopy(a, N=2, fillval=0)
     ...: 
1000 loops, best of 3: 545 µs per loop
1000 loops, best of 3: 891 µs per loop
100 loops, best of 3: 10.6 ms per loop

In [581]: a = np.random.randint(0,9,(1000,40,40))

In [582]: %timeit random_block_fill_lidx(a, N=10, fillval=0)
     ...: %timeit random_block_fill_adv(a, N=10, fillval=0)
     ...: %timeit random_block_fill_loopy(a, N=10, fillval=0)
     ...: 
1000 loops, best of 3: 739 µs per loop
1000 loops, best of 3: 671 µs per loop
1000 loops, best of 3: 1.27 ms per loop