kron(a, b)
Computes the Kronecker product, a composite array made of blocks of the second array scaled by the first.
The function assumes that the number of dimensions of a and b are the same, if necessary prepending the smallest with ones. If a.shape = (r0,r1,..,rN)
and b.shape = (s0,s1,...,sN)
, the Kronecker product has shape (r0*s0, r1*s1, ..., rN*SN)
. The elements are products of elements from a and b, organized explicitly by:
kron(a,b)[k0,k1,...,kN] = a[i0,i1,...,iN] * b[j0,j1,...,jN]
where:
kt = it * st + jt, t = 0,...,N
In the common 2-D case (N=1), the block structure can be visualized:
[[ a[0,0]*b, a[0,1]*b, ... , a[0,-1]*b ], [ ... ... ], [ a[-1,0]*b, a[-1,1]*b, ... , a[-1,-1]*b ]]
Kronecker product of two arrays.
outer
The outer product
>>> np.kron([1,10,100], [5,6,7]) array([ 5, 6, 7, ..., 500, 600, 700])
>>> np.kron([5,6,7], [1,10,100]) array([ 5, 50, 500, ..., 7, 70, 700])
>>> np.kron(np.eye(2), np.ones((2,2))) array([[1., 1., 0., 0.], [1., 1., 0., 0.], [0., 0., 1., 1.], [0., 0., 1., 1.]])
>>> a = np.arange(100).reshape((2,5,2,5))
... b = np.arange(24).reshape((2,3,4))
... c = np.kron(a,b)
... c.shape (2, 10, 6, 20)
>>> I = (1,3,0,2)See :
... J = (0,2,1)
... J1 = (0,) + J # extend to ndim=4
... S1 = (1,) + b.shape
... K = tuple(np.array(I) * np.array(S1) + np.array(J1))
... c[K] == a[I]*b[J] True
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