等式是
$C'_{ijkl} = Q_{im} Q_{jn} C_{mnop} (Q^{-1})_{ok} (Q^{-1})_{pl}$
我能够使用
np.einsum('im,jn,mnop,ok,pl', Q, Q, C, Q_inv, Q_inv)
做这项工作，也期待
np.tensordot(np.tensordot(np.tensordot(Q, np.tensordot(Q, C, axes=[1,1]), axes=[1,0]), Q_inv, axes=[2,0]), Q_inv, axes=[3,0])
工作，但它没有。
具体细节：
C是一个四阶弹性张量：
array([[[[ 552.62389047, -0.28689554, -0.32194701],
[ -0.28689554, 118.89168597, -0.65559912],
[ -0.32194701, -0.65559912, 130.21758722]],
[[ -0.28689554, 166.02923119, -0.00000123],
[ 166.02923119, 0.49494431, -0.00000127],
[ -0.00000123, -0.00000127, -0.57156702]],
[[ -0.32194701, -0.00000123, 165.99413061],
[ -0.00000123, -0.64666809, -0.0000013 ],
[ 165.99413061, -0.0000013 , 0.42997465]]],
[[[ -0.28689554, 166.02923119, -0.00000123],
[ 166.02923119, 0.49494431, -0.00000127],
[ -0.00000123, -0.00000127, -0.57156702]],
[[ 118.89168597, 0.49494431, -0.64666809],
[ 0.49494431, 516.15898907, -0.33132485],
[ -0.64666809, -0.33132485, 140.09010389]],
[[ -0.65559912, -0.00000127, -0.0000013 ],
[ -0.00000127, -0.33132485, 165.98553869],
[ -0.0000013 , 165.98553869, 0.41913346]]],
[[[ -0.32194701, -0.00000123, 165.99413061],
[ -0.00000123, -0.64666809, -0.0000013 ],
[ 165.99413061, -0.0000013 , 0.42997465]],
[[ -0.65559912, -0.00000127, -0.0000013 ],
[ -0.00000127, -0.33132485, 165.98553869],
[ -0.0000013 , 165.98553869, 0.41913346]],
[[ 130.21758722, -0.57156702, 0.42997465],
[ -0.57156702, 140.09010389, 0.41913346],
[ 0.42997465, 0.41913346, 486.62412063]]]])
Q是一个改变x和y坐标的旋转矩阵。
array([[ 0, 1, 0],
[-1, 0, 0],
[ 0, 0, 1]])
Q_inv是
array([[-0., -1., -0.],
[ 1., 0., 0.],
[ 0., 0., 1.]])
np.einsum导致
array([[[[ 516.15898907, -0.49494431, -0.33132485],
[ -0.49494431, 118.89168597, 0.64666809],
[ -0.33132485, 0.64666809, 140.09010389]],
[[ -0.49494431, 166.02923119, 0.00000127],
[ 166.02923119, 0.28689554, -0.00000123],
[ 0.00000127, -0.00000123, 0.57156702]],
[[ -0.33132485, 0.00000127, 165.98553869],
[ 0.00000127, -0.65559912, 0.0000013 ],
[ 165.98553869, 0.0000013 , 0.41913346]]],
[[[ -0.49494431, 166.02923119, 0.00000127],
[ 166.02923119, 0.28689554, -0.00000123],
[ 0.00000127, -0.00000123, 0.57156702]],
[[ 118.89168597, 0.28689554, -0.65559912],
[ 0.28689554, 552.62389047, 0.32194701],
[ -0.65559912, 0.32194701, 130.21758722]],
[[ 0.64666809, -0.00000123, 0.0000013 ],
[ -0.00000123, 0.32194701, 165.99413061],
[ 0.0000013 , 165.99413061, -0.42997465]]],
[[[ -0.33132485, 0.00000127, 165.98553869],
[ 0.00000127, -0.65559912, 0.0000013 ],
[ 165.98553869, 0.0000013 , 0.41913346]],
[[ 0.64666809, -0.00000123, 0.0000013 ],
[ -0.00000123, 0.32194701, 165.99413061],
[ 0.0000013 , 165.99413061, -0.42997465]],
[[ 140.09010389, 0.57156702, 0.41913346],
[ 0.57156702, 130.21758722, -0.42997465],
[ 0.41913346, -0.42997465, 486.62412063]]]])
我认为是正确的，而四个np.tensordot导致
array([[[[ 552.62389047, -0.28689554, 0.32194701],
[ -0.28689554, 118.89168597, 0.65559912],
[ -0.32194701, -0.65559912, -130.21758722]],
[[ -0.28689554, 166.02923119, 0.00000123],
[ 166.02923119, 0.49494431, 0.00000127],
[ -0.00000123, -0.00000127, 0.57156702]],
[[ -0.32194701, -0.00000123, -165.99413061],
[ -0.00000123, -0.64666809, 0.0000013 ],
[ 165.99413061, -0.0000013 , -0.42997465]]],
[[[ -0.28689554, 166.02923119, 0.00000123],
[ 166.02923119, 0.49494431, 0.00000127],
[ -0.00000123, -0.00000127, 0.57156702]],
[[ 118.89168597, 0.49494431, 0.64666809],
[ 0.49494431, 516.15898907, 0.33132485],
[ -0.64666809, -0.33132485, -140.09010389]],
[[ -0.65559912, -0.00000127, 0.0000013 ],
[ -0.00000127, -0.33132485, -165.98553869],
[ -0.0000013 , 165.98553869, -0.41913346]]],
[[[ 0.32194701, 0.00000123, 165.99413061],
[ 0.00000123, 0.64666809, -0.0000013 ],
[-165.99413061, 0.0000013 , 0.42997465]],
[[ 0.65559912, 0.00000127, -0.0000013 ],
[ 0.00000127, 0.33132485, 165.98553869],
[ 0.0000013 , -165.98553869, 0.41913346]],
[[-130.21758722, 0.57156702, 0.42997465],
[ 0.57156702, -140.09010389, 0.41913346],
[ -0.42997465, -0.41913346, 486.62412063]]]])
注意负数大数字。
答案 0 :(得分：3)
方法＃1
一种方法是使用np.tensordot获得与np.einsum相同的结果，但不是在一步中并且在可信赖的 {{3}的帮助下} -
# Get broadcasted elementwise multiplication between two versions of Q.
# This corresponds to "np.einsum('im,jn,..', Q, Q)" producing "'ijmn""
# broadcasted version of elementwise multiplications between Q's.
Q_ext = Q[:,None,:,None]*Q[:,None,:]
# Similarly for Q_inv : For "np.einsum('..ok,pl', Q_inv, Q_inv)" get "'opkl'"
# broadcasted version of elementwise multiplications between Q_inv's.
Q_inv_ext = Q_inv[:,None,:,None]*Q_inv[:,None,:]
# Perform "np.einsum('im,jn,mnop,ok,pl', Q, Q, C)" with "np.tensordot".
# Notice that we are using the last two axes from 'Q_ext', so "axes=[2,3]"
# and first two from 'C', so "axes=[0,1]" for it.
# These axes would be reduced by the dot-product, leaving us with 'ijop'.
parte1 = np.tensordot(Q_ext,C,axes=([2,3],[0,1]))
# Do it one more time to perform "np.einsum('ijop,ok,pl', parte1,Q_inv,Q_inv)"
# to reduce dimensions represented by 'o,p', leaving us with 'ijkl'.
# To confirm, compare the following against original einsum approach :
# "np.einsum('im,jn,mnop,ok,pl->ijkl', Q, Q, C, Q_inv, Q_inv)"
out = np.tensordot(parte1,Q_inv_ext,axes=([2,3],[0,1]))
方法＃2
如果您希望避免broadcasting支持再使用两个np.tensordot个实例，则可以这样做 -
# Perform "np.einsum('jn,mnop', Q, C). Notice how, Q is represented by 'jn'
# and C by 'mnop'. We need to reduce the 'm' dimension, i.e. reduce 'axes=1'
# from Q and `axes=1` from C corresponding to `n' in each of the inputs.
# Thus, 'jn' + 'mnop' => 'jmop' after 'n' is reduced and order is maintained.
Q_C1 = np.tensordot(Q,C,axes=([1],[1]))
# Perform "np.einsum('im,jn,mnop', Q, Q, C). We need to use Q and Q_C1.
# Q is 'im' and Q_C1 is 'jmop'. Thus, again we need to reduce 'axes=1'
# from Q and `axes=1` from Q_C1 corresponding to `m' in each of the inputs.
# Thus, 'im' + 'jmop' => 'ijop' after 'm' is reduced and order is maintained.
parte1 = np.tensordot(Q,Q_C1,axes=([1],[1]))
# Use the same philosophy to get the rest of the einsum equivalent,
# but use parte1 and go right and use Q_inv
out = np.tensordot(np.tensordot(parte1,Q_inv,axes=([2],[0])),Q_inv,axes=([2],[0]))
np.tensordot的诀窍是跟踪axes参数减少的维度以及折叠维度与其余输入的对齐方式。尺寸。