![](/img/trans.png)
[英]Tensorflow save final state of LSTM in dynamic_rnn for prediction
[英]Hidden states vs. final state returned by Tensorflow's dynamic_rnn
tensorflow.nn.dynamic_rnn
创建一个递归神经网络给定的cell
,它是RNNCell
一个实例,并返回一对包括:
outputs
:RNN输出Tensor state
:最终状态 这是一个玩具递归神经网络及其输出[*]:
import numpy as np
import tensorflow as tf
dim = 3
hidden = 4
lengths = tf.placeholder(dtype=tf.int32, shape=[None])
inputs = tf.placeholder(dtype=tf.float32, shape=[None, None, dim])
cell = tf.nn.rnn_cell.LSTMCell(hidden, state_is_tuple=True)
output, final_state = tf.nn.dynamic_rnn(
cell, inputs, lengths, dtype=tf.float32)
inputs_ = np.asarray([[[0, 0, 0], [1, 1, 1], [2, 2, 2]],
[[6, 6, 6], [7, 7, 7], [8, 8, 8]],
[[9,9,9], [10,10,10], [11,11,11]]],
dtype=np.int32)
lengths_ = np.asarray([3, 1, 2], dtype=np.int32)
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
output_, final_state_ = sess.run(
[output, final_state],
{inputs: inputs_, lengths: lengths_})
print('hidden states:')
print(output_)
print('final state :')
print(final_state_)
输出:
hidden states:
[[[ 0.0000000e+00 0.0000000e+00 0.0000000e+00 0.0000000e+00]
[-3.0096283e-02 1.6747195e-01 2.3113856e-02 -4.5677904e-02]
[-6.0795926e-02 3.5036778e-01 6.0140129e-02 -1.6039203e-01]]
[[-2.1957003e-03 8.1749000e-02 1.2620161e-02 -2.8342882e-01]
[ 0.0000000e+00 0.0000000e+00 0.0000000e+00 0.0000000e+00]
[ 0.0000000e+00 0.0000000e+00 0.0000000e+00 0.0000000e+00]]
[[-1.7376180e-04 2.7789388e-02 3.1011081e-03 -3.5858861e-01]
[-2.5059914e-04 4.5771234e-02 4.5708413e-03 -6.5035087e-01]
[ 0.0000000e+00 0.0000000e+00 0.0000000e+00 0.0000000e+00]]]
final state :
LSTMStateTuple(
c=array([[-1.0705842e-01, 5.2945197e-01, 1.5602852e-01, -2.5641304e-01],
[-3.3140955e-03, 8.6112522e-02, 7.2794281e-02, -3.6088336e-01],
[-3.4701003e-04, 4.6147645e-02, 6.7321308e-02, -8.6465287e-01]],
dtype=float32),
h=array([[-6.0795926e-02, 3.5036778e-01, 6.0140129e-02, -1.6039203e-01],
[-2.1957003e-03, 8.1749000e-02, 1.2620161e-02, -2.8342882e-01],
[-2.5059914e-04, 4.5771234e-02, 4.5708413e-03, -6.5035087e-01]],
dtype=float32))
我的理解如下:
c
组分)的最后一个细胞状态以及每个序列的最后隐藏状态( h
组分); 因此,我不应该在结局状态的h
分量和每个序列的最后隐藏状态中获得相同的值吗?
[*] Code很大程度上受到了这篇文章的启发
最终状态组件的h
包含LSTM的最后一个输出 ,而在dynamic_rnn的情况下,它考虑了您作为参数给出的长度 ( lengths
)
正如您在示例中看到的那样, final_state.h[0]
等于output[0][2]
,因为第一个示例的长度是3, final_state.h[1]
等于output[1][0]
因为你的第二个例子的长度是一个等等。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.