As Toan suggests, a simple hack would be to just select the rows first, and then select the columns over that.

>>> a[[0,1,3], :]            # Returns the rows you want
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [12, 13, 14, 15]])
>>> a[[0,1,3], :][:, [0,2]]  # Selects the columns you want as well
array([[ 0,  2],
       [ 4,  6],
       [12, 14]])

[Edit] The built-in method: np.ix_

I recently discovered that numpy gives you an in-built one-liner to doing exactly what @Jaime suggested, but without having to use broadcasting syntax (which suffers from lack of readability). From the docs:

Using ix_ one can quickly construct index arrays that will index the cross product. a[np.ix_([1,3],[2,5])] returns the array [[a[1,2] a[1,5]], [a[3,2] a[3,5]]].

So you use it like this:

>>> a = np.arange(20).reshape((5,4))
>>> a[np.ix_([0,1,3], [0,2])]
array([[ 0,  2],
       [ 4,  6],
       [12, 14]])

And the way it works is that it takes care of aligning arrays the way Jaime suggested, so that broadcasting happens properly:

>>> np.ix_([0,1,3], [0,2])
(array([[0],
        [1],
        [3]]), array([[0, 2]]))

Also, as MikeC says in a comment, np.ix_ has the advantage of returning a view, which my first (pre-edit) answer did not. This means you can now assign to the indexed array:

>>> a[np.ix_([0,1,3], [0,2])] = -1
>>> a    
array([[-1,  1, -1,  3],
       [-1,  5, -1,  7],
       [ 8,  9, 10, 11],
       [-1, 13, -1, 15],
       [16, 17, 18, 19]])

If you've got a boolean array you can do direct selection based on that like so:

>>> a = np.array([True, True, True, False, False])
>>> b = np.array([1,2,3,4,5])
>>> b[a]
array([1, 2, 3])

To go along with your initial example you could do the following:

>>> a = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> b = np.array([[False,True,False],[True,False,False],[False,False,True]])
>>> a[b]
array([2, 4, 9])

You can also add in an arange and do direct selection on that, though depending on how you're generating your boolean array and what your code looks like YMMV.

>>> a = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> a[np.arange(len(a)), [1,0,2]]
array([2, 4, 9])

You simply need to transpose it:

L[:,index[1]].T
#             ^ transpose

By using a transpose, the columns are rows and vice versa. So here (you can transpose before the selection, and then use L.T[index[1],:]) we first make the selection and then turn the columns into rows.

This produces:

>>> L[:,index[1]].T
array([[  1, -15, -10,  -5],
       [ -5,  -1,  10,   1],
       [ -5,  -1,  10,   1],
       [-10,  -5,  -1,  15]])

Note that of course behind the curtains there are still some loops that are done. But these are done outside Python and thus are more efficient.