Question

我有2个3D形状张量A（batch_size，dim1，dim2）和B大小（batch_size，dim2，dim3）。我想执行以下计算：

C = np.zeros((batch_size, dim1, dim3))
for i in range(batch_size):
    c[i] = A[i].dot(B[i])

我尝试了C = tf.matmul(A,B)，但收到了错误：

tensorflow.python.framework.errors_impl.InvalidArgumentError：Shape 必须是等级2，但对于'MatMul'（op：'MatMul'）的输入是3级形状：[10,100,100]，[10,100]

在TensorFlow中有没有有效的方法来实现它？谢谢！

Answer 1

让我们先用numpy：

来做

import numpy as np
batch_size = 5
dim1 = 7
dim2 = 2
dim3 = 3

A = np.random.rand(batch_size, dim1, dim2)
B = np.random.rand(batch_size, dim2, dim3)  
C = np.zeros((batch_size, dim1, dim3)) 

for i in range(batch_size):                         
    c[i] = A[i].dot(B[i])
print(np.einsum('ikl,ilm->ikm',A,B)) - C # prints a zero array

现在让我们使用tensorflow：

import tensorflow as tf
a = tf.constant(A) # reusing numpy arrays
b = tf.constant(B)
op = tf.einsum('ikl,ilm->ikm',a,b)
with tf.Session() as sess:       
    print(sess.run(op) - C) # prints a zero array

现在根据这个answer，看起来性能应该与天真的实现非常相似。

在Tensorflow中将3D张量与矩阵相乘的有效方法？

1 个答案: