numpy：为什么（2,1）数组和垂直矩阵切片的差不是（2,1）数组

Question

Consider the following code:

>>x=np.array([1,3]).reshape(2,1)
array([[1],
   [3]])   
>>M=np.array([[1,2],[3,4]])
array([[1, 2],
   [3, 4]])
>>y=M[:,0]
>>x-y
array([[ 0,  2],
   [-2,  0]])

I would intuitively feel this should give a (2,1) vector of zeros.

I am not saying, however, that this is how it should be done and everything else is stupid. I would simply love if someone could offer some logic that I can remember so things like this don't keep producing bugs in my code.

Note that I am not asking how I can achieve what I want (I could reshape y), but I am hoping to get some deeper understanding of why Python/Numpy works as it does. Maybe I am doing something conceptually wrong?

Answer 1

Look at the shape of y . It is (2,) ; 1d. The source array is (2,2), but you are selecting one column. M[:,0] not only selects the column, but removes that singleton dimension.

So we have for the 2 operations, this change in shape:

M[:,0]: (2,2) => (2,)
x - y: (2,1) (2,) => (2,1), (1,2) => (2,2)

There are various ways of ensuring that y has the shape (2,1). Index with a list/vector, M[:,[0]] ; index with a slice, M[:,:1] . Add a dimension, M[:,0,None] .

Think also what happens when M[0,:] or M[0,0] .

Answer 2

numpy.array indexes such that a single value in any position collapses that dimension, while slicing retains it, even if the slice is only one element wide. This is completely consistent, for any number of dimensions:

>> A = numpy.arange(27).reshape(3, 3, 3)
>> A[0, 0, 0].shape
()

>> A[:, 0, 0].shape
(3,)

>> A[:, :, 0].shape
(3, 3)

>> A[:1, :1, :1].shape
(1, 1, 1)

Notice that every time a single number is used, that dimension is dropped.

You can obtain the semantics you expect by using numpy.matrix , where two single indexes return a order 0 array and all other types of indexing return matrices

>> M = numpy.asmatrix(numpy.arange(9).reshape(3, 3))

>> M[0, 0].shape
()

>> M[:, 0].shape   # This is different from the array
(3, 1)

>> M[:1, :1].shape
(1, 1)

Your example works as you expect when you use numpy.matrix :

>> x = numpy.matrix([[1],[3]])
>> M = numpy.matrix([[1,2],[3,4]])
>> y = M[:, 0]
>> x - y
matrix([[0],
        [0]])