Python numpy 数组 - 关闭最小区域

Question

我有一个代表图像的二维布尔 numpy 数组，我在其上调用 skimage.measure.label 来标记每个分段区域，给我一个 int [0,500] 的二维数组；该数组中的每个值都代表该像素的区域标签。我现在想删除最小的区域。例如，如果我的输入数组是形状 (n, n)，我希望将所有

0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 7, 8, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 2, 2, 2, 1, 1
4, 4, 4, 4, 2, 2, 2, 2, 1, 1
4, 6, 6, 4, 2, 2, 2, 3, 3, 3
4, 6, 6, 4, 5, 5, 5, 3, 3, 5
4, 4, 4, 4, 5, 5, 5, 5, 5, 5
4, 4, 4, 4, 5, 5, 5, 5, 5, 5

然后输出是，

0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 0, 0, 1, 1, 1  # 7 and 8 are replaced by 0
0, 0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 2, 2, 2, 1, 1
4, 4, 4, 4, 2, 2, 2, 2, 1, 1
4, 4, 4, 4, 2, 2, 2, 3, 3, 3  # 6 is gone, but 3 remains
4, 4, 4, 4, 5, 5, 5, 3, 3, 5
4, 4, 4, 4, 5, 5, 5, 5, 5, 5
4, 4, 4, 4, 5, 5, 5, 5, 5, 5

我研究了 skimage 形态学操作，包括 binary closing，但似乎没有一个适合我的用例。有什么建议吗？

Answer 1

您可以通过对每个标签对应的布尔区域执行二进制膨胀来实现此目的。通过这样做，您将找到每个区域的邻居数。使用它，您可以根据需要替换值。

示例代码：

import numpy as np
import scipy.ndimage

m = 5

arr = [[0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
       [0, 0, 7, 8, 0, 0, 0, 1, 1, 1],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
       [0, 0, 0, 0, 0, 2, 2, 2, 1, 1],
       [4, 4, 4, 4, 2, 2, 2, 2, 1, 1],
       [4, 6, 6, 4, 2, 2, 2, 3, 3, 3],
       [4, 6, 6, 4, 5, 5, 5, 3, 3, 5],
       [4, 4, 4, 4, 5, 5, 5, 5, 5, 5],
       [4, 4, 4, 4, 5, 5, 5, 5, 5, 5]]
arr = np.array(arr)
nval = np.max(arr) + 1

# Compute number of occurances of each number
counts, _ = np.histogram(arr, bins=range(nval + 1))

# Compute the set of neighbours for each number via binary dilation
c = np.array([scipy.ndimage.morphology.binary_dilation(arr == i)
              for i in range(nval)])

# Loop over the set of arrays with bad count and update them to the most common
# neighbour
for i in filter(lambda i: counts[i] < m, range(nval)):
    arr[arr == i] = np.argmax(np.sum(c[:, arr == i], axis=1))

这给出了预期的结果：

>>> arr.tolist()
[[0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
 [0, 0, 0, 0, 0, 2, 2, 2, 1, 1],
 [4, 4, 4, 4, 2, 2, 2, 2, 1, 1],
 [4, 4, 4, 4, 2, 2, 2, 3, 3, 3],
 [4, 4, 4, 4, 5, 5, 5, 3, 3, 5],
 [4, 4, 4, 4, 5, 5, 5, 5, 5, 5],
 [4, 4, 4, 4, 5, 5, 5, 5, 5, 5]]