在 TensorFlow 中获取 dynamic_rnn 的最后一个输出

Question

我有一个形状为 [batch, None, dim] 的 3-D 张量，其中第二维（即时间步长）是未知的。我使用 dynamic_rnn 来处理此类输入，如以下代码片段所示：

import numpy as np
import tensorflow as tf

batch = 2
dim = 3
hidden = 4

lengths = tf.placeholder(dtype=tf.int32, shape=[batch])
inputs = tf.placeholder(dtype=tf.float32, shape=[batch, None, dim])
cell = tf.nn.rnn_cell.GRUCell(hidden)
cell_state = cell.zero_state(batch, tf.float32)
output, _ = tf.nn.dynamic_rnn(cell, inputs, lengths, initial_state=cell_state)

实际上，运行剪下了一些实际数字，我得到了一些合理的结果：

inputs_ = np.asarray([[[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3]],
                    [[6, 6, 6], [7, 7, 7], [8, 8, 8], [9, 9, 9]]],
                    dtype=np.int32)
lengths_ = np.asarray([3, 1], dtype=np.int32)

with tf.Session() as sess:
    sess.run(tf.global_variables_initializer())
    output_ = sess.run(output, {inputs: inputs_, lengths: lengths_})
    print(output_)

输出为：

[[[ 0.          0.          0.          0.        ]
  [ 0.02188676 -0.01294564  0.05340237 -0.47148666]
  [ 0.0343586  -0.02243731  0.0870839  -0.89869428]
  [ 0.          0.          0.          0.        ]]

 [[ 0.00284752 -0.00315077  0.00108094 -0.99883419]
  [ 0.          0.          0.          0.        ]
  [ 0.          0.          0.          0.        ]
  [ 0.          0.          0.          0.        ]]]

有没有办法用动态 RNN 的 上次相关输出 获得形状 [batch, 1, hidden] 的 3-D 张量？谢谢！

Answer 1

实际上，解决方案并没有那么难。我实现了以下代码：

slices = []
for index, l in enumerate(tf.unstack(lengths)):
    slice = tf.slice(rnn_out, begin=[index, l - 1, 0], size=[1, 1, 3])
    slices.append(slice)
last = tf.concat(0, slices)

因此，完整的代码段如下：

import numpy as np
import tensorflow as tf

batch = 2
dim = 3
hidden = 4

lengths = tf.placeholder(dtype=tf.int32, shape=[batch])
inputs = tf.placeholder(dtype=tf.float32, shape=[batch, None, dim])
cell = tf.nn.rnn_cell.GRUCell(hidden)
cell_state = cell.zero_state(batch, tf.float32)
output, _ = tf.nn.dynamic_rnn(cell, inputs, lengths, initial_state=cell_state)

inputs_ = np.asarray([[[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3]],
                    [[6, 6, 6], [7, 7, 7], [8, 8, 8], [9, 9, 9]]],
                    dtype=np.int32)
lengths_ = np.asarray([3, 1], dtype=np.int32)

slices = []
for index, l in enumerate(tf.unstack(lengths)):
    slice = tf.slice(output, begin=[index, l - 1, 0], size=[1, 1, 3])
    slices.append(slice)
last = tf.concat(0, slices)

with tf.Session() as sess:
    sess.run(tf.global_variables_initializer())
    outputs = sess.run([output, last], {inputs: inputs_, lengths: lengths_})
    print 'RNN output:'
    print(outputs[0])
    print
    print 'last relevant output:'
    print(outputs[1])

并且输出：

RNN output:
[[[ 0.          0.          0.          0.        ]
 [-0.06667092 -0.09284072  0.01098599 -0.03676109]
 [-0.09101103 -0.19828682  0.03546784 -0.08721405]
 [ 0.          0.          0.          0.        ]]

[[-0.00025157 -0.05704876  0.05527233 -0.03741353]
 [ 0.          0.          0.          0.        ]
 [ 0.          0.          0.          0.        ]
 [ 0.          0.          0.          0.        ]]]

last relevant output:
[[[-0.09101103 -0.19828682  0.03546784]]

 [[-0.00025157 -0.05704876  0.05527233]]]

Answer 2

这就是 gather_nd 的用途！

def extract_axis_1(data, ind):
    """
    Get specified elements along the first axis of tensor.
    :param data: Tensorflow tensor that will be subsetted.
    :param ind: Indices to take (one for each element along axis 0 of data).
    :return: Subsetted tensor.
    """

    batch_range = tf.range(tf.shape(data)[0])
    indices = tf.stack([batch_range, ind], axis=1)
    res = tf.gather_nd(data, indices)

    return res

你的情况：

output = extract_axis_1(output, lengths - 1)

现在output是一个维度为[batch_size, num_cells]的张量。

Answer 3

来自以下两个来源，

http://www.wildml.com/2016/08/rnns-in-tensorflow-a-practical-guide-and-undocumented-features/

outputs, last_states = tf.nn.dynamic_rnn(
cell=cell,
dtype=tf.float64,
sequence_length=X_lengths,
inputs=X)

或https://github.com/ageron/handson-ml/blob/master/14_recurrent_neural_networks.ipynb,

显然可以从 dynamic_rnn 调用的第二个输出中直接提取 last_states。它将为您提供跨 all 层的 last_states（在 LSTM 中，它由 LSTMStateTuple 组成），而输出包含 last[= 中的所有状态23=]层.

Answer 4

好的 - 所以，看起来实际上是一个更简单的解决方案。正如@Shao Tang 和@Rahul 所提到的，执行此操作的首选方法是访问最终单元格状态。原因如下：

如果您查看 GRUCell 源代码（下方），您会发现单元保持的“状态”实际上是隐藏的权重本身。因此，当 tf.nn.dynamic_rnn returns 最终状态时，它实际上返回了您感兴趣的最终隐藏权重。为了证明这一点，我只是调整了您的设置并得到了结果：

GRUCell 调用（rnn_cell_impl.py）：

def call(self, inputs, state):
"""Gated recurrent unit (GRU) with nunits cells."""
if self._gate_linear is None:
      bias_ones = self._bias_initializer
if self._bias_initializer is None:
        bias_ones = init_ops.constant_initializer(1.0, dtype=inputs.dtype)
with vs.variable_scope("gates"):  # Reset gate and update gate.
self._gate_linear = _Linear(
            [inputs, state],
2 * self._num_units,
True,
bias_initializer=bias_ones,
kernel_initializer=self._kernel_initializer)
    value = math_ops.sigmoid(self._gate_linear([inputs, state]))
    r, u = array_ops.split(value=value, num_or_size_splits=2, axis=1)
    r_state = r * state
if self._candidate_linear is None:
with vs.variable_scope("candidate"):
self._candidate_linear = _Linear(
            [inputs, r_state],
self._num_units,
True,
bias_initializer=self._bias_initializer,
kernel_initializer=self._kernel_initializer)
    c = self._activation(self._candidate_linear([inputs, r_state]))
    new_h = u * state + (1 - u) * c
return new_h, new_h

解决方案：

import numpy as np
import tensorflow as tf

batch = 2
dim = 3
hidden = 4

lengths = tf.placeholder(dtype=tf.int32, shape=[batch])
inputs = tf.placeholder(dtype=tf.float32, shape=[batch, None, dim])
cell = tf.nn.rnn_cell.GRUCell(hidden)
cell_state = cell.zero_state(batch, tf.float32)
output, state = tf.nn.dynamic_rnn(cell, inputs, lengths, initial_state=cell_state)

inputs_ = np.asarray([[[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3]],
                    [[6, 6, 6], [7, 7, 7], [8, 8, 8], [9, 9, 9]]],
                    dtype=np.int32)
lengths_ = np.asarray([3, 1], dtype=np.int32)

with tf.Session() as sess:
    sess.run(tf.global_variables_initializer())
    output_, state_ = sess.run([output, state], {inputs: inputs_, lengths: lengths_})
    print (output_)
    print (state_)

输出：

[[[ 0.          0.          0.          0.        ]
  [-0.24305521 -0.15512943  0.06614969  0.16873555]
  [-0.62767833 -0.30741733  0.14819752  0.44313088]
  [ 0.          0.          0.          0.        ]]

 [[-0.99152333 -0.1006391   0.28767768  0.76360202]
  [ 0.          0.          0.          0.        ]
  [ 0.          0.          0.          0.        ]
  [ 0.          0.          0.          0.        ]]]
[[-0.62767833 -0.30741733  0.14819752  0.44313088]
 [-0.99152333 -0.1006391   0.28767768  0.76360202]]

对于使用 LSTMCell（另一个流行选项）的其他读者来说，事情的工作方式略有不同。 LSTMCell 以不同的方式维护状态 - 单元状态是元组或实际单元状态和隐藏状态的串联版本。因此，要访问最终的隐藏权重，您可以在单元初始化期间设置 (is_state_tuple 到 True)，最终状态将是一个元组：（最终单元状态，最终隐藏权重）。所以，在这种情况下，

_, (_, h) = tf.nn.dynamic_rnn(单元格, 输入, 长度, initial_state=cell_state)

会给你最后的权重。

参考资料： https://github.com/tensorflow/tensorflow/blob/438604fc885208ee05f9eef2d0f2c630e1360a83/tensorflow/python/ops/rnn_cell_impl.py#L308 https://github.com/tensorflow/tensorflow/blob/438604fc885208ee05f9eef2d0f2c630e1360a83/tensorflow/python/ops/rnn_cell_impl.py#L415

在 TensorFlow 中获取 dynamic_rnn 的最后一个输出

Get the last output of a dynamic_rnn in TensorFlow

python

neural-network

deep-learning

tensorflow

recurrent-neural-network