python - 如何删除 2D 矩阵中值为 0 的前 n 列/行？

Question

引用上一个问题:

Remove all-zero rows in a 2D matrix

import numpy as np

data = np.array([[4, 1, 1, 2, 0, 4],
                 [3, 4, 3, 1, 4, 4],
                 [1, 4, 3, 1, 0, 0],
                 [0, 4, 4, 0, 4, 3],
                 [0, 0, 0, 0, 0, 0]])

data = data[~(data==0).all(1)]
print(data)

输出:

    [[4 1 1 2 0 4]
     [3 4 3 1 4 4]
     [1 4 3 1 0 0]
     [0 4 4 0 4 3]]

到目前为止一切顺利，但是如果我添加空列怎么办？

 data = np.array([[0, 4, 1, 1, 2, 0, 4],
                  [0, 3, 4, 3, 1, 4, 4],
                  [0, 0, 1, 4, 3, 1, 0],
                  [0, 0, 4, 4, 0, 4, 3],
                  [0, 0, 0, 0, 0, 0, 0]])

输出为

          [[0 4 1 1 2 0 4]
           [0 3 4 3 1 4 4]
           [0 1 4 3 1 0 0]
           [0 0 4 4 0 4 3]]

这不是我想要的。

基本上，如果我的矩阵是:

            [[0, 0, 0, 0, 0, 0, 0, 0, 0],
             [0, 0, 4, 1, 1, 2, 0, 4, 0],
             [0, 0, 3, 4, 3, 1, 4, 4, 0],
             [0, 0, 1, 4, 3, 1, 0, 0, 0],
             [0, 0, 0, 4, 4, 0, 4, 3, 0],
             [0, 0, 0, 0, 0, 0, 0, 0, 0]]

我期望的输出是

        [[4 1 1 2 0 4]
         [3 4 3 1 4 4]
         [1 4 3 1 0 0]
         [0 4 4 0 4 3]]

Answer 1

这是一种方法 -

def reduced_box(a):
    # Store shape info
    M,N = a.shape

    # Mask of valid places in the array
    mask = a!=0

    # Get boolean array with at least a valid one per row
    m_col = mask.any(1)

    # Get the starting and ending valid rows with argmax.
    # More info : https://stackoverflow.com/a/47269413/
    r0,r1 = m_col.argmax(), M-m_col[::-1].argmax()

    # Repeat for cols
    m_row = mask.any(0)
    c0,c1 = m_row.argmax(), N-m_row[::-1].argmax()

    # Finally slice with the valid indices as the bounding box limits
    return a[r0:r1,c0:c1]

示例运行 -

In [210]: a
Out[210]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 4, 1, 0, 2, 0, 4, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 4, 0, 1, 0, 0, 0],
       [0, 0, 0, 4, 0, 0, 4, 3, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0]])

In [211]: reduced_box(a)
Out[211]: 
array([[4, 1, 0, 2, 0, 4],
       [0, 0, 0, 0, 0, 0],
       [1, 4, 0, 1, 0, 0],
       [0, 4, 0, 0, 4, 3]])