numpy.dot(a, b, out=None)
Dot product of two arrays. Specifically,
a
and b
are 1-D arrays, it is inner product of vectors (without complex conjugation). a
and b
are 2-D arrays, it is matrix multiplication, but using matmul
or a @ b
is preferred. a
or b
is 0-D (scalar), it is equivalent to multiply
and using numpy.multiply(a, b)
or a * b
is preferred. a
is an N-D array and b
is a 1-D array, it is a sum product over the last axis of a
and b
. If a
is an N-D array and b
is an M-D array (where M>=2
), it is a sum product over the last axis of a
and the second-to-last axis of b
:
dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])
Parameters: |
|
---|---|
Returns: |
|
Raises: |
|
See also
>>> np.dot(3, 4) 12
Neither argument is complex-conjugated:
>>> np.dot([2j, 3j], [2j, 3j]) (-13+0j)
For 2-D arrays it is the matrix product:
>>> a = [[1, 0], [0, 1]] >>> b = [[4, 1], [2, 2]] >>> np.dot(a, b) array([[4, 1], [2, 2]])
>>> a = np.arange(3*4*5*6).reshape((3,4,5,6)) >>> b = np.arange(3*4*5*6)[::-1].reshape((5,4,6,3)) >>> np.dot(a, b)[2,3,2,1,2,2] 499128 >>> sum(a[2,3,2,:] * b[1,2,:,2]) 499128
© 2005–2019 NumPy Developers
Licensed under the 3-clause BSD License.
https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.dot.html