- numpy.dot(a, b, out=None)#
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
a @ bis preferred.
If either a or b is 0-D (scalar), it is equivalent to
a * bis preferred.
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])
It uses an optimized BLAS library when possible (see
- outndarray, 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 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.
>>> 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