tensorflow 计算两个序列的co-attention矩阵

序列1的每个元素和序列2的每个元素​​feature dim计算​​最终排成一个矩阵∈Rn1×n2

import tensorflow as tf
import numpy as np

n1 = tf.ones([10,3],dtype=tf.int32)*2 # 序列1,长度10, feature dim 3
n2 = tf.ones([2,3],dtype=tf.int32)*3 # 序列2,长度2, feature dim 3

ta1 = tf.TensorArray(tf.int32, tf.constant(2))

n1_tile = tf.tile(tf.expand_dims(n1,0),[2,1,1]) # 2个n1
n2_tile = tf.tile(tf.expand_dims(n2,0),[10,1,1]) # 10个n2

final = []

for i in range(2):
    final.append([])
    for j in range(10):
        element = n1_tile[i][j]*n2_tile[j][i]
        final[i].append(element)

final = tf.stack(final)

sess=tf.Session()
# feature dim 按位相乘，feature dim求和后才是co-attention matrix
print(sess.run(final))
print(np.shape(sess.run(final)))

# feature dim 按位相乘，feature dim求和后才是co-attention matrix
final_compare = n1_tile*tf.transpose(n2_tile,[1,0,2])
print(sess.run(final_compare))
print(np.shape(sess.run(final_compare)))

#直接求co-attention matrix，feature dim 按位相乘再求和，∈Rn1×n2
final_compare2 = tf.matmul(n1,n2,adjoint_b=True)
print(sess.run(final_compare2))
print(np.shape(sess.run(final_compare2)))

PyTorch实现的更多参考​​https://github.com/allenai/allennlp/blob/master/allennlp/modules/matrix_attention.py​​