在numpy中最简单的方法是索引矩阵的向量,每行不包括一个索引?[副本]

一种方法是使用 masking / boolean-indexing -

mask = np.ones(M.shape,dtype=bool)
mask[np.arange(len(k)),k] = 0
out = M[mask].reshape(len(M),-1)

或者,我们可以用 broadcasting mask -

np.not_equal.outer(k,np.arange(M.shape[1]))
# or k[:,None]!=np.arange(M.shape[1])

因此,给我们一个单行/紧凑型版本-

out = M[k[:,None]!=np.arange(M.shape[1])].reshape(len(M),-1)

要排除每行多个索引,请编辑第一个方法的高级索引部分-

def exclude_multiple(M,*klist):
    k = np.stack(klist).T
    mask = np.ones(M.shape,dtype=bool)
    mask[np.arange(len(k))[:,None],k] = 0
    out = M[mask].reshape(len(M),-1)
    return out

In [185]: M = np.arange(40).reshape(4,10)

In [186]: exclude_multiple(M,[1,3,2,0],[4,5,8,1])
Out[186]: 
array([[ 0,  2,  3,  5,  6,  7,  8,  9],
       [10, 11, 12, 14, 16, 17, 18, 19],
       [20, 21, 23, 24, 25, 26, 27, 29],
       [32, 33, 34, 35, 36, 37, 38, 39]])

改进@Divakar的答案,将其扩展到每行零个或多个排除索引:

def excluding(A, *klist):
    """ 
    excludes column k from each row of A, for each k in klist 
    (make sure the index vectors have no common elements)
    """
    mask = np.ones(A.shape,dtype=bool)
    for k in klist:
        mask[np.arange(len(k)),k] = 0
    return A[mask].reshape(len(A),-1)

测试:

M = np.arange(40).reshape(4,10)
excluding(M,[1,3,2,0],[4,5,8,1])

返回

array([[ 0,  2,  3,  5,  6,  7,  8,  9],
       [10, 11, 12, 14, 16, 17, 18, 19],
       [20, 21, 23, 24, 25, 26, 27, 29],
       [32, 33, 34, 35, 36, 37, 38, 39]])