两个3dim矩阵的numpy枚举的Theano版本 [英] Theano version of a numpy einsum for two 3dim matrices
本文介绍了两个3dim矩阵的numpy枚举的Theano版本的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有两个3dim numpy矩阵，我想根据一个轴做一个点积，而不在theano中使用循环.一个带有示例数据的numpy解决方案将是:
I have two 3dim numpy matrices and I want to do a dot product according to one axis without using a loop in theano. a numpy solution with sample data would be like:
a=[ [[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]],
[[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]],
[ [ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]],
[ [ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]],
[[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]],
[[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0.]],
[[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
[ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
[ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]]]
b=[[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]],
[[ 0, 0, 1, 0, 0.],
[ 1, 0, 0, 0, 0.],
[ 0, 0, 0, 0, 0.],
[ 0, 1, 0, 0, 0.]]]
dt = np.dtype(np.float32)
a=np.asarray(a,dtype=dt)
b=np.asarray(b,dtype=dt)
print(a.shape)
print(b.shape)
其中，"a"的形状为(7，4，15)，"b"的形状为(7，4，5). "c"定义为"a"和"b"的点积:
where "a", has the shape of (7, 4, 15) and "b", has the shape of (7, 4, 5). "c", is defined as dot product of "a" and "b":
c = np.einsum('ijk,ijl->ilk',a,b)
我正在寻找此示例的theano实现来计算"c".
I am looking for a theano implementation of this example to calculate "c".
有什么想法吗?
推荐答案
要结束此问题:
import theano as th
import then.Tensor as T
ta = T.tensor3('a')
tb = T.tensor3('b')
tc = T.batched_tensordot(ta, tb, axes=[[1],[1]])
......
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