如何有效證明 NumPy 數組的非零元素合理？

將非零元素對齊到特定一側（通常在搜尋中有用）是一種常見操作，可以直接對 NumPy 陣列完成。以下是如何實現的 -

import numpy as np

def justify(a, invalid_val=0, axis=1, side='left'):    
    """
    Justifies a 2D array

    Parameters
    ----------
    A : ndarray
        Input array to be justified
    axis : int
        Axis along which justification is to be made
    side : str
        Direction of justification. It could be 'left', 'right', 'up', 'down'
        It should be 'left' or 'right' for axis=1 and 'up' or 'down' for axis=0.

    """

    if invalid_val is np.nan:
        mask = ~np.isnan(a)
    else:
        mask = a!=invalid_val
    justified_mask = np.sort(mask,axis=axis)
    if (side=='up') | (side=='left'):
        justified_mask = np.flip(justified_mask,axis=axis)
    out = np.full(a.shape, invalid_val) 
    if axis==1:
        out[justified_mask] = a[mask]
    else:
        out.T[justified_mask.T] = a.T[mask.T]
    return out

上面的程式碼片段可以沿著選定的軸在四個可能方向中的任何一個方向對齊 2D 數組 -

# sample input array
a = np.array([[1, 0, 2, 0], 
              [3, 0, 4, 0], 
              [5, 0, 6, 0], 
              [0, 7, 0, 8]])

# shift to left
print(justify(a, axis=0, side='up'))
# shift to down
print(justify(a, axis=0, side='down'))
# shift to left
print(justify(a, axis=1, side='left'))
# shift to right
print(justify(a, axis=1, side='right'))

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