Numpy 向量和矩阵加法

Question

What is the best way to perform normal vector addition, where one of the operands is an nx 1 matrix?

Why do I care? Sometimes, a function that should return a vector returns an nx 1 matrix (because the function would equivalently work element-wise on a matrix). When I want to further work with the returned "vector", I always have to reshape - there must be a better way.

For example:

v = np.zeros(shape=(2,1))
w = np.array([1,1])
print('{}, {}'.format(v.shape,w.shape))

Prints: (2, 1), (2,)

print(v+w)

[[1. 1.] [1. 1.]]

print(v+w.reshape((2,1)))

[[1.] [1.]] (the desired output!)

Answer 1

If w has the desired shape of the result (eg (2,) ), and v has the same size (eg (2,1) , or (2,) ), this is safe and easy:

w + v.reshape(w.shape)

Less generally, if all you want is to get rid of the last dimension, knowing it is of length 1, you can do:

w + v[..., 0]

Answer 2

Sounds a bit like you are coming from MATLAB where everything is 2d (scalars size is (1,1)) and the trailing dimension is outermost. Or a linear algebra that treats 'vectors' as single column matrices.

In numpy , 0 and 1d arrays are just a normal as 2d. A shape like (n,) is common. By the rules of broadcasting adding a leading dimension is automatic (1,n), but adding a trailing dimension requires user action. That a[:,None] is most idiomatic, though not the only option.

The v+w broadcasting logic is

(2,1) + (2,) => (2,1) + (1,2) => (2,2)

The auto-leading logic avoids ambiguity (what should happen if you try to add a (2,) to a (3,)?). And since leading dimensions are 'outer-most' it makes most sense to expand in that direction. MATLAB on the other hand 'naturally' expends and contracts the trailing dimensions.

So to some degree or other a (n,1) shape is more awkward in numpy , though it is still relatively easy to create.

Another example of an auto leading dimension:

In [129]: np.atleast_2d(np.arange(3)).shape
Out[129]: (1, 3)

On the other hand expand_dims lets us add dimensions all over the place

In [132]: np.expand_dims(np.arange(3),(0,2,3)).shape
Out[132]: (1, 3, 1, 1)