特征收缩与Numpy Dot

时间:2018-02-07 21:45:07

标签: python c++ numpy eigen3 tensor

大家好我在Numpy有以下张量点产品:

import numpy as np

tensorA = np.array([[[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]]])

tensorB = np.array([[1,2],
                       [1,2],
                       [1,2]])

print tensorA.dot(tensorB)

它给出了以下答案:

[[[  6  12]
  [ 15  30]
  [ 24  48]]

 [[ 33  66]
  [ 42  84]
  [ 51 102]]

 [[ 60 120]
  [ 69 138]
  [ 78 156]]]

然而,当我在C ++ Eigen中做同样的事情时:

Eigen::Tensor<float, 3> tensorA(3,3,3);
tensorA.setValues({{{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}}});

Eigen::Tensor<float, 2> tensorB(3,2);
tensorB.setValues({{1,2},
                   {1,2},
                   {1,2}});

// Compute the traditional matrix product
Eigen::array<Eigen::IndexPair<float>, 1> product_dims = { Eigen::IndexPair<float>(0, 1) };
Eigen::Tensor<float, 3> AB = tensorA.contract(tensorB, product_dims);

我明白了:

D: 3 R: 3 C: 2
[[12.000     24.000     ]
[15.000     30.000     ]
[18.000     36.000     ]
]
R: 3 C: 2
[[39.000     78.000     ]
[42.000     84.000     ]
[45.000     90.000     ]
]
R: 3 C: 2
[[66.000     132.000     ]
[69.000     138.000     ]
[72.000     144.000     ]
]

为什么会这样。我想要一个张量点产品,相当于numpy给我的东西。它与c ++中的product_dims参数有关吗?或者是否涉及其他一些错误?基本上,它需要将深度分量乘以3次。

1 个答案:

答案 0 :(得分:1)

我无法帮助您使用c ++代码,但我可以用numpy术语确定发生了什么:

In [1]: A=np.arange(1,28).reshape(3,3,3)
In [3]: A
Out[3]: 
array([[[ 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]]])
In [5]: B=np.repeat([[1,2]],3, axis=0)
In [6]: B
Out[6]: 
array([[1, 2],
       [1, 2],
       [1, 2]])

你的点 - 记住A的最后一个,在第二个到最后一个B:

In [7]: A.dot(B)
Out[7]: 
array([[[  6,  12],
        [ 15,  30],
        [ 24,  48]],

       [[ 33,  66],
        [ 42,  84],
        [ 51, 102]],

       [[ 60, 120],
        [ 69, 138],
        [ 78, 156]]])

使用einsum索引这很清楚(至少对我而言):

In [8]: np.einsum('ijk,kl',A,B)    # notice the k pair
Out[8]: 
array([[[  6,  12],
        [ 15,  30],
        [ 24,  48]],

       [[ 33,  66],
        [ 42,  84],
        [ 51, 102]],

       [[ 60, 120],
        [ 69, 138],
        [ 78, 156]]])

但如果我将einsum更改为使用第二个到最后两个,我会得到你的c ++结果(我认为):

In [9]: np.einsum('ijk,jl',A,B)    # notice the j pair
Out[9]: 
array([[[ 12,  24],
        [ 15,  30],
        [ 18,  36]],

       [[ 39,  78],
        [ 42,  84],
        [ 45,  90]],

       [[ 66, 132],
        [ 69, 138],
        [ 72, 144]]])