tensordot的想法非常简单-我们输入数组和相应的轴，沿着这些轴进行求和约简。在输出中移除参与求和缩减的轴，并且来自输入阵列的所有剩余轴在输出中被 * 展开 * 为不同的轴，以保持输入阵列被馈送的顺序。
让我们来看看几个具有一个和两个轴的求和约简的示例案例，并交换输入位置，看看输出中的顺序是如何保持的。

一、一轴求和归约

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)`

情况#2（与情况#1相同，但输入被交换馈送）：

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、两轴求和归约

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，以便将np.dot应用于2个2d数组。然后，它交换并重新成形为目标。实验可能比解释容易。这里没有特殊的Tensor数学，只是将dot扩展到更高的维度。tensor表示数组的维数大于2d。如果您已经熟悉einsum，那么将结果与之进行比较是最简单的。
样本测试，在1对轴上求和

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，在两个上求和。

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，我不认为有另一种组合。

上面的答案很棒，对我理解tensordot有很大帮助。但它们并不显示操作背后的实际数学。这就是为什么我在TF2中为自己做了等效操作，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)\t\t- ((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)\t\t- ((() 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)}\t\t tf.einsum('i,i', a, b)\t\t- ((the last 1 axes of a), (the first 1 axes of b))")
print(f"{tf.tensordot(a, b, ((0,), (0,)))}\t\t tf.einsum('i,i', a, b)\t\t- ((0th axis of a), (0th axis of b))")
print(f"{tf.tensordot(a, b, (0,0))}\t\t tf.einsum('i,i', a, b)\t\t- ((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)\t\t- ((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)\t\t- ((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)\t\t- ((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))

除了上面的答案，如果把np.tensordot分解成嵌套循环，会更容易理解：
假设：

import numpy as np
a = np.arange(24).reshape(2,3,4)
b = np.arange(30).reshape(3,5,2)

那他就是

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]]])

并且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Tensor中具有相同维度的轴（这里是2和3）可以通过它们的和来减少。并且参数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]])