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numpy.dot
numpy.dot(a, b, out=None)-
Dot product of two arrays. Specifically,
- If both
aandbare 1-D arrays, it is inner product of vectors (without complex conjugation). - If both
aandbare 2-D arrays, it is matrix multiplication, but usingmatmulora @ bis preferred. - If either
aorbis 0-D (scalar), it is equivalent tomultiplyand usingnumpy.multiply(a, b)ora * bis preferred. - If
ais an N-D array andbis a 1-D array, it is a sum product over the last axis ofaandb. If
ais an N-D array andbis an M-D array (whereM>=2), it is a sum product over the last axis ofaand the second-to-last axis ofb: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 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: -
output : ndarray -
Returns the dot product of
aandb. Ifaandbare both scalars or both 1-D arrays then a scalar is returned; otherwise an array is returned. Ifoutis given, then it is returned.
Raises: - ValueError
-
If the last dimension of
ais not the same size as the second-to-last dimension ofb.
See also
Examples
>>> np.dot(3, 4) 12Neither 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 - If both
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https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.dot.html