Dot product of two arrays. Specifically,
If both a and b are 1-D arrays, it is inner product of vectors
(without complex conjugation).
If both a and b are 2-D arrays, it is matrix multiplication,
but using matmul or a @ b is preferred.
a @ b
If either a or b is 0-D (scalar), it is equivalent to multiply
and using numpy.multiply(a, b) or a * b is preferred.
a * b
If 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])
Output argument. This must have the exact kind that would be returned
if it was not used. In particular, it must have the right type, must be
C-contiguous, and its dtype must be the dtype that would be returned
for dot(a,b). This is a performance feature. Therefore, if these
conditions are not met, an exception is raised, instead of attempting
to be flexible.
Returns the dot product of a and b. If a and b are both
scalars or both 1-D arrays then a scalar is returned; otherwise
an array is returned.
If out is given, then it is returned.
If the last dimension of a is not the same size as
the second-to-last dimension of b.
Complex-conjugating dot product.
Sum products over arbitrary axes.
Einstein summation convention.
‘@’ operator as method with out parameter.
Chained dot product.
>>> np.dot(3, 4)
Neither argument is complex-conjugated:
>>> np.dot([2j, 3j], [2j, 3j])
For 2-D arrays it is the matrix product:
>>> a = [[1, 0], [0, 1]]
>>> b = [[4, 1], [2, 2]]
>>> np.dot(a, b)
>>> 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]
>>> sum(a[2,3,2,:] * b[1,2,:,2])