在我学会了如何使用einsum之后，我现在试图了解np.tensordot是如何工作的。
但是，我有点迷茫，尤其是关于参数axes的各种可能性。
为了理解它，因为我从未练习过张量微积分，我使用以下示例：
A = np.random.randint(2, size=(2, 3, 5))
B = np.random.randint(2, size=(3, 2, 4))
在这种情况下，有哪些不同的可能np.tensordot以及如何手动计算它？
tensordot的想法非常简单 - 我们输入数组和相应的轴，这些轴将沿着这些轴进行求和减少。参与求和约简的轴在输出中被删除，输入数组中的所有剩余轴作为输出中的不同轴展开，保持输入数组的进给顺序。
让我们看几个具有一个和两个求和减少轴的示例案例，并交换输入位置，看看顺序是如何在输出中保持的。
我。求和减少的一个轴
输入：
In [7]: A = np.random.randint(2, size=(2, 6, 5))
...: B = np.random.randint(2, size=(3, 2, 4))
...:
案例#1：
In [9]: np.tensordot(A, B, axes=((0),(1))).shape
Out[9]: (6, 5, 3, 4)
A : (2, 6, 5) -> reduction of axis=0
B : (3, 2, 4) -> reduction of axis=1
Output : `(2, 6, 5)`, `(3, 2, 4)` ===(2 gone)==> `(6,5)` + `(3,4)` => `(6,5,3,4)`
2（与情况 #1 相同，但输入是交换的）：
In [8]: np.tensordot(B, A, axes=((1),(0))).shape
Out[8]: (3, 4, 6, 5)
B : (3, 2, 4) -> reduction of axis=1
A : (2, 6, 5) -> reduction of axis=0
Output : `(3, 2, 4)`, `(2, 6, 5)` ===(2 gone)==> `(3,4)` + `(6,5)` => `(3,4,6,5)`.
二、总和约简的两个轴心
输入：
In [11]: A = np.random.randint(2, size=(2, 3, 5))
...: B = np.random.randint(2, size=(3, 2, 4))
...:
案例#1：
In [12]: np.tensordot(A, B, axes=((0,1),(1,0))).shape
Out[12]: (5, 4)
A : (2, 3, 5) -> reduction of axis=(0,1)
B : (3, 2, 4) -> reduction of axis=(1,0)
Output : `(2, 3, 5)`, `(3, 2, 4)` ===(2,3 gone)==> `(5)` + `(4)` => `(5,4)`
案例#2：
In [14]: np.tensordot(B, A, axes=((1,0),(0,1))).shape
Out[14]: (4, 5)
B : (3, 2, 4) -> reduction of axis=(1,0)
A : (2, 3, 5) -> reduction of axis=(0,1)
Output : `(3, 2, 4)`, `(2, 3, 5)` ===(2,3 gone)==> `(4)` + `(5)` => `(4,5)`
我们可以将其扩展到尽可能多的轴。
tensordot交换轴并重塑输入，以便它可以np.dot应用于 2 个 2D 数组。然后，它会交换并重新调整回目标。实验可能比解释更容易。没有特殊的张量数学，只是扩展dot以在更高维度上工作。 tensor 仅表示超过 2d 的数组。 如果您已经对einsum感到满意，那么将结果与此进行比较将是最简单的。
在 1 对轴上求和的示例测试
In [823]: np.tensordot(A,B,[0,1]).shape
Out[823]: (3, 5, 3, 4)
In [824]: np.einsum('ijk,lim',A,B).shape
Out[824]: (3, 5, 3, 4)
In [825]: np.allclose(np.einsum('ijk,lim',A,B),np.tensordot(A,B,[0,1]))
Out[825]: True
另一个，召唤两个。
In [826]: np.tensordot(A,B,[(0,1),(1,0)]).shape
Out[826]: (5, 4)
In [827]: np.einsum('ijk,jim',A,B).shape
Out[827]: (5, 4)
In [828]: np.allclose(np.einsum('ijk,jim',A,B),np.tensordot(A,B,[(0,1),(1,0)]))
Out[828]: True
我们可以对(1,0)对做同样的事情。 考虑到维度的混合，我认为没有另一种组合。
上面的
答案很棒，对我理解tensordot有很大帮助。但它们并没有显示操作背后的实际数学。这就是为什么我在 TF 2 中为自己做了等效的操作，并决定在这里分享它们：
a = tf.constant([1,2.])
b = tf.constant([2,3.])
print(f"{tf.tensordot(a, b, 0)}t tf.einsum('i,j', a, b)tt- ((the last 0 axes of a), (the first 0 axes of b))")
print(f"{tf.tensordot(a, b, ((),()))}t tf.einsum('i,j', a, b)tt- ((() axis of a), (() axis of b))")
print(f"{tf.tensordot(b, a, 0)}t tf.einsum('i,j->ji', a, b)t- ((the last 0 axes of b), (the first 0 axes of a))")
print(f"{tf.tensordot(a, b, 1)}tt tf.einsum('i,i', a, b)tt- ((the last 1 axes of a), (the first 1 axes of b))")
print(f"{tf.tensordot(a, b, ((0,), (0,)))}tt tf.einsum('i,i', a, b)tt- ((0th axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (0,0))}tt tf.einsum('i,i', a, b)tt- ((0th axis of a), (0th axis of b))")
[[2. 3.]
[4. 6.]] tf.einsum('i,j', a, b) - ((the last 0 axes of a), (the first 0 axes of b))
[[2. 3.]
[4. 6.]] tf.einsum('i,j', a, b) - ((() axis of a), (() axis of b))
[[2. 4.]
[3. 6.]] tf.einsum('i,j->ji', a, b) - ((the last 0 axes of b), (the first 0 axes of a))
8.0 tf.einsum('i,i', a, b) - ((the last 1 axes of a), (the first 1 axes of b))
8.0 tf.einsum('i,i', a, b) - ((0th axis of a), (0th axis of b))
8.0 tf.einsum('i,i', a, b) - ((0th axis of a), (0th axis of b))
对于(2,2)形状：
a = tf.constant([[1,2],
[-2,3.]])
b = tf.constant([[-2,3],
[0,4.]])
print(f"{tf.tensordot(a, b, 0)}t tf.einsum('ij,kl', a, b)t- ((the last 0 axes of a), (the first 0 axes of b))")
print(f"{tf.tensordot(a, b, (0,0))}t tf.einsum('ij,ik', a, b)t- ((0th axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (0,1))}t tf.einsum('ij,ki', a, b)t- ((0th axis of a), (1st axis of b))")
print(f"{tf.tensordot(a, b, 1)}t tf.matmul(a, b)tt- ((the last 1 axes of a), (the first 1 axes of b))")
print(f"{tf.tensordot(a, b, ((1,), (0,)))}t tf.einsum('ij,jk', a, b)t- ((1st axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (1, 0))}t tf.matmul(a, b)tt- ((1st axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, 2)}t tf.reduce_sum(tf.multiply(a, b))t- ((the last 2 axes of a), (the first 2 axes of b))")
print(f"{tf.tensordot(a, b, ((0,1), (0,1)))}t tf.einsum('ij,ij->', a, b)tt- ((0th axis of a, 1st axis of a), (0th axis of b, 1st axis of b))")
[[[[-2. 3.]
[ 0. 4.]]
[[-4. 6.]
[ 0. 8.]]]
[[[ 4. -6.]
[-0. -8.]]
[[-6. 9.]
[ 0. 12.]]]] tf.einsum('ij,kl', a, b) - ((the last 0 axes of a), (the first 0 axes of b))
[[-2. -5.]
[-4. 18.]] tf.einsum('ij,ik', a, b) - ((0th axis of a), (0th axis of b))
[[-8. -8.]
[ 5. 12.]] tf.einsum('ij,ki', a, b) - ((0th axis of a), (1st axis of b))
[[-2. 11.]
[ 4. 6.]] tf.matmul(a, b) - ((the last 1 axes of a), (the first 1 axes of b))
[[-2. 11.]
[ 4. 6.]] tf.einsum('ij,jk', a, b) - ((1st axis of a), (0th axis of b))
[[-2. 11.]
[ 4. 6.]] tf.matmul(a, b) - ((1st axis of a), (0th axis of b))
16.0 tf.reduce_sum(tf.multiply(a, b)) - ((the last 2 axes of a), (the first 2 axes of b))
16.0 tf.einsum('ij,ij->', a, b) - ((0th axis of a, 1st axis of a), (0th axis of b, 1st axis of b))
除了上述答案之外，如果将np.tensordot分解为嵌套循环，则更容易理解：
假设：
import numpy as np
a = np.arange(24).reshape(2,3,4)
b = np.arange(30).reshape(3,5,2)
那么 a 是
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]]])
而 b 是
array([[[ 0, 1],
[ 2, 3],
[ 4, 5],
[ 6, 7],
[ 8, 9]],
[[10, 11],
[12, 13],
[14, 15],
[16, 17],
[18, 19]],
[[20, 21],
[22, 23],
[24, 25],
[26, 27],
[28, 29]]])
让
c = np.tensordot(a, b, axes=([0,1],[2,0]))
相当于
c = np.zeros((4,5))
for i in range(4):
for j in range(5):
for p in range(2):
for q in range(3):
c[i,j] += a[p,q,i] * b[q,j,p]
两个张量中具有相同维度的轴（这里是 2 和 3）可以减少它们的总和。参数 axes=（[0,1]，[2,0]） 与 axes=（[1,0]，[0,2]） 相同。
最后的 c 是
array([[ 808, 928, 1048, 1168, 1288],
[ 871, 1003, 1135, 1267, 1399],
[ 934, 1078, 1222, 1366, 1510],
[ 997, 1153, 1309, 1465, 1621]])