提取索引

Question

我有一个NumPy的阵列仅由0和1的元素如下：

import numpy as np 
data = np.array([[1, 1 , 0 , 0 , 0 , 0 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , **1** , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , 1 , 1 , 1 , 1], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 0 , 0 , 0 , 0 , 0 , 0]]) 

我必须找出被1中2所包围的元件1的索引在每个方向上2个像素。

预期答案的位置以粗体显示。

我在寻找更简单快捷的方法。

Answer 1

这是很容易与basic morphological operation：

import numpy as np 
from scipy.ndimage.morphology import binary_erosion 


data = np.array([[1, 1 , 0 , 0 , 0 , 0 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , 1 , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , 1 , 1 , 1 , 1], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 0 , 0 , 0 , 0 , 0 , 0]]) 

expected = np.array([[0, 0 , 0 , 0 , 0 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 0 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 0 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 1 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 0 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 0 , 0 , 0 , 0], 
        [0, 0 , 0 , 0 , 0 , 0 , 0 , 0]]) 

# otherwise known as np.ones((5, 5)) 
structuring_element = np.array([[1, 1, 1, 1, 1], 
           [1, 1, 1, 1, 1], 
           [1, 1, 1, 1, 1], 
           [1, 1, 1, 1, 1], 
           [1, 1, 1, 1, 1]]) 

# will be of dtype np.bool but you can convert with .astype(np.int) 
# if you really need 
result = binary_erosion(data, structuring_element) 

print(result) 

print(np.allclose(result, expected)) 

Answer 2

你可以使用一些signal processing -

import numpy as np 
from scipy import signal 

# Input 
data = np.array([[1, 1 , 0 , 0 , 0 , 0 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , 1 , 1 , 1 , 0], 
       [0, 0 , 1 , 1 , 1 , 1 , 1 , 1], 
       [1, 1 , 1 , 1 , 1 , 1 , 1 , 0], 
       [1, 1 , 0 , 0 , 0 , 0 , 0 , 0]]) 

# Extent of the search   
extent = 2; 

# Kernel to be used with 2D convolution     
kernel = np.ones((2*extent+1,2*extent+1)) 

# Perform 2D convolution with input data and kernel 
filt_out = signal.convolve2d(data, kernel, boundary='symm', mode='same') 

# Find where the convolution resulted in a perfect score, 
# i.e is equal to the number of elements in kernel 
R,C = np.where(filt_out == kernel.size) 

输出 -

In [66]: print(R,C) 
[3] [4] 

---

列在此节是一种替代方法与ndimage执行相同的卷积与之前的做法，保持其余步骤相同。下面的代码，以获得卷积输出filt_out -

import scipy.ndimage 
filt_out = scipy.ndimage.convolve(data,kernel)