numpy.dot¶

numpy.
dot
(a, b, out=None)¶ Dot product of two arrays. Specifically,
If both a and b are 1D arrays, it is inner product of vectors (without complex conjugation).
If both a and b are 2D arrays, it is matrix multiplication, but using
matmul
ora @ b
is preferred.If either a or b is 0D (scalar), it is equivalent to
multiply
and usingnumpy.multiply(a, b)
ora * b
is preferred.If a is an ND array and b is a 1D array, it is a sum product over the last axis of a and b.
If a is an ND array and b is an MD array (where
M>=2
), it is a sum product over the last axis of a and the secondtolast axis of b:dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])
Parameters: a : array_like
First argument.
b : array_like
Second argument.
out : ndarray, optional
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 Ccontiguous, 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: output : ndarray
Returns the dot product of a and b. If a and b are both scalars or both 1D arrays then a scalar is returned; otherwise an array is returned. If out is given, then it is returned.
Raises: ValueError
If the last dimension of a is not the same size as the secondtolast dimension of b.
See also
Examples
>>> np.dot(3, 4) 12
Neither argument is complexconjugated:
>>> np.dot([2j, 3j], [2j, 3j]) (13+0j)
For 2D 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