我有一个数组(n,m):

 [255 100 255]
 [100 255 100]
 [255 100 255]


我需要创建一个新数组,例如在测试相邻值的情况下,如果北,东,南,西的ALL都等于100,则我的值设置为100:

 [255 100 255]
 [100 100 100]
 [255 100 255]


我有一个简单的解决方案,可以在1:n和1:m上循环,但是显然很慢,我想知道是否有办法更快地做到这一点。
我发现了几个有关滑动窗口以计算平均值的链接,但是我看不到如何跟踪索引来创建新数组。
Using strides for an efficient moving average filter

预先感谢您的输入。

最佳答案

假设A作为输入数组,这是使用slicingboolean indexing的一种方法-

# Get west, north, east & south elements for [1:-1,1:-1] region of input array
W = A[1:-1,:-2]
N  = A[:-2,1:-1]
E = A[1:-1,2:]
S  = A[2:,1:-1]

# Check if all four arrays have 100 for that same element in that region
mask = (W == 100) & (N == 100) & (E == 100) & (S == 100)

# Use the mask to set corresponding elements in a copy version as 100s
out = A.copy()
out[1:-1,1:-1][mask] = 100


样品运行-

In [90]: A
Out[90]:
array([[220,  93, 205,  82,  23, 210,  22],
       [133, 228, 100,  27, 210, 186, 246],
       [196, 100,  73, 100,  86, 100,  53],
       [195, 131, 100, 142, 100, 214, 100],
       [247,  73, 117, 116,  24, 100,  50]])

In [91]: W = A[1:-1,:-2]
    ...: N  = A[:-2,1:-1]
    ...: E = A[1:-1,2:]
    ...: S  = A[2:,1:-1]
    ...: mask = (W == 100) & (N == 100) & (E == 100) & (S == 100)
    ...:
    ...: out = A.copy()
    ...: out[1:-1,1:-1][mask] = 100
    ...:

In [92]: out
Out[92]:
array([[220,  93, 205,  82,  23, 210,  22],
       [133, 228, 100,  27, 210, 186, 246],
       [196, 100, 100, 100,  86, 100,  53],
       [195, 131, 100, 142, 100, 100, 100],
       [247,  73, 117, 116,  24, 100,  50]])




这种问题主要在信号处理/图像处理领域中看到。因此,您也可以将2D convolution用作替代解决方案,例如-

from scipy import signal
from scipy import ndimage

# Use a structuring elements with north, west, east and south elements as 1s
strel = ndimage.generate_binary_structure(2, 1)

# 2D Convolve to get 4s at places that are surrounded by 1s
mask = signal.convolve2d((A==100).astype(int),strel,'same')==4

# Use the mask to set corresponding elements in a copy version as 100
out = A.copy()
out[mask] = 100


样品运行-

In [119]: A
Out[119]:
array([[108, 184,   0, 176, 131,  86, 201],
       [ 22,  47, 100,  78, 151, 196, 221],
       [185, 100, 142, 100, 121, 100,  24],
       [201, 101, 100, 138, 100,  20, 100],
       [127, 227, 217,  19, 206, 100,  43]])

In [120]: strel = ndimage.generate_binary_structure(2, 1)
     ...: mask = signal.convolve2d((A==100).astype(int),strel,'same')==4
     ...:
     ...: out = A.copy()
     ...: out[mask] = 100
     ...:

In [121]: out
Out[121]:
array([[108, 184,   0, 176, 131,  86, 201],
       [ 22,  47, 100,  78, 151, 196, 221],
       [185, 100, 100, 100, 121, 100,  24],
       [201, 101, 100, 138, 100, 100, 100],
       [127, 227, 217,  19, 206, 100,  43]])


更简单的方法是使用ndimage.binary_closing,这正是closing的预期操作。因此,获取面罩的另一种方法是-

strel = ndimage.generate_binary_structure(2, 1)
mask = ndimage.binary_closing(A==100, structure=strel)

09-10 04:42
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