Question

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，如测试neigboring值，如果North，East，South，West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，循环在1：n和1：m，但它显然非常慢，我想知道是否有办法更快地完成它。我找到了几个关于滑动窗口来计算平均值的链接，但我不知道如何跟踪我的索引来创建新数组。 Using strides for an efficient moving average filter

提前感谢您的意见。

Answer 1

假设A为输入数组，这是使用slicing和boolean 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

示例运行 -

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]])

这些问题主要出现在信号处理/图像处理领域。因此，您也可以使用2D convolution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

示例运行 -

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，另一种获取掩码的替代方法是 -

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

根据相邻值替换2d数组中的值

1 个答案: