numpy-如何将向量索引数组转换为掩码?

这里有一个变体:

def create_mask(indices, m):
    mask = np.zeros((len(indices), m), dtype=bool)
    for i, idx in enumerate(indices):
        mask[i, idx] = True
    return mask

用途:

>>> create_mask(indices, 8)
array([[ True, False,  True, False, False, False, False, False],
       [ True, False, False, False, False, False, False, False],
       [False,  True, False, False,  True, False, False,  True]])

这是一条路-

def mask_from_indices(indices, ncols=None):
    # Extract column indices
    col_idx = np.concatenate(indices)

    # If number of cols is not given, infer it based on max column index
    if ncols is None:
        ncols = col_idx.max()+1

    # Length of indices, to be used as no. of rows in o/p
    n = len(indices)

    # Initialize o/p array
    out = np.zeros((n,ncols), dtype=bool)

    # Lengths of each index element that represents each group of col indices
    lens = np.array(list(map(len,indices)))

    # Use np.repeat to generate all row indices
    row_idx = np.repeat(np.arange(len(lens)),lens)

    # Finally use row, col indices to set True values
    out[row_idx,col_idx] = 1
    return out    

样品运行-

In [89]: mask_from_indices(indices)
Out[89]: 
array([[ True, False,  True, False, False, False, False, False],
       [ True, False, False, False, False, False, False, False],
       [False,  True, False, False,  True, False, False,  True]])