你好，我有一个关于 Tensorflow 的问题。我训练了一些 LSTM 模型，我可以访问突触连接的权重和偏差，但是我似乎无法访问 LSTM 单元的输入、新输入、输出和遗忘门权重。我可以得到门张量，但是当我在 session 中尝试 .eval() 它们时，我得到错误。我正在为我的网络使用 tensorflow/python/ops/rnn_cell.py 中的 BasicLSTMCell 类
`
class BasicLSTMCell(RNNCell):
"""Basic LSTM recurrent network cell.
The implementation is based on: http://arxiv.org/abs/1409.2329.
We add forget_bias (default: 1) to the biases of the forget gate in order to
reduce the scale of forgetting in the beginning of the training.
It does not allow cell clipping, a projection layer, and does not
use peep-hole connections: it is the basic baseline.
For advanced models, please use the full LSTMCell that follows.
"""
def __init__(self, num_units, forget_bias=1.0, input_size=None,
state_is_tuple=True, activation=tanh):
"""Initialize the basic LSTM cell.
Args:
num_units: int, The number of units in the LSTM cell.
forget_bias: float, The bias added to forget gates (see above).
input_size: Deprecated and unused.
state_is_tuple: If True, accepted and returned states are 2-tuples of
the `c_state` and `m_state`. If False, they are concatenated
along the column axis. The latter behavior will soon be deprecated.
activation: Activation function of the inner states.
"""
if not state_is_tuple:
logging.warn("%s: Using a concatenated state is slower and will soon be "
"deprecated. Use state_is_tuple=True.", self)
if input_size is not None:
logging.warn("%s: The input_size parameter is deprecated.", self)
self._num_units = num_units
self._forget_bias = forget_bias
self._state_is_tuple = state_is_tuple
self._activation = activation
@property
def state_size(self):
return (LSTMStateTuple(self._num_units, self._num_units)
if self._state_is_tuple else 2 * self._num_units)
@property
def output_size(self):
return self._num_units
def __call__(self, inputs, state, scope=None):
"""Long short-term memory cell (LSTM)."""
with vs.variable_scope(scope or type(self).__name__): # "BasicLSTMCell"
# Parameters of gates are concatenated into one multiply for efficiency.
if self._state_is_tuple:
c, h = state
else:
c, h = array_ops.split(1, 2, state)
concat = _linear([inputs, h], 4 * self._num_units, True)
# i = input_gate, j = new_input, f = forget_gate, o = output_gate
i, j, f, o = array_ops.split(1, 4, concat)
new_c = (c * sigmoid(f + self._forget_bias) + sigmoid(i) *
self._activation(j))
new_h = self._activation(new_c) * sigmoid(o)
if self._state_is_tuple:
new_state = LSTMStateTuple(new_c, new_h)
else:
new_state = array_ops.concat(1, [new_c, new_h])
return new_h, new_state
def _get_concat_variable(name, shape, dtype, num_shards):
"""Get a sharded variable concatenated into one tensor."""
sharded_variable = _get_sharded_variable(name, shape, dtype, num_shards)
if len(sharded_variable) == 1:
return sharded_variable[0]
concat_name = name + "/concat"
concat_full_name = vs.get_variable_scope().name + "/" + concat_name + ":0"
for value in ops.get_collection(ops.GraphKeys.CONCATENATED_VARIABLES):
if value.name == concat_full_name:
return value
concat_variable = array_ops.concat(0, sharded_variable, name=concat_name)
ops.add_to_collection(ops.GraphKeys.CONCATENATED_VARIABLES,
concat_variable)
return concat_variable
def _get_sharded_variable(name, shape, dtype, num_shards):
"""Get a list of sharded variables with the given dtype."""
if num_shards > shape[0]:
raise ValueError("Too many shards: shape=%s, num_shards=%d" %
(shape, num_shards))
unit_shard_size = int(math.floor(shape[0] / num_shards))
remaining_rows = shape[0] - unit_shard_size * num_shards
shards = []
for i in range(num_shards):
current_size = unit_shard_size
if i < remaining_rows:
current_size += 1
shards.append(vs.get_variable(name + "_%d" % i, [current_size] + shape[1:],
dtype=dtype))
return shards
`
我可以看到在 def 调用 中使用了 i, j, f, o 门，但是当我 tf.print 它们时，我得到张量，当我尝试在 session 中 .eval() 它们时，我得到错误。我也试过 tf.getVariable 但无法提取权重矩阵。我的问题:有没有办法评估 i、j、f 和 o 门权重/矩阵？
最佳答案
首先，澄清一些混淆:i、j、f 和 o 张量不是权重矩阵；它们是依赖于特定 LSTM 单元输入的中间计算步骤。 LSTM 单元的所有权重都存储在变量 self._kernel 和 self._bias 中，以及常量 self._forget_bias 中。
因此，为了回答您的问题的两种可能解释，我将展示如何在每一步打印 self._kernel 和 self._bias 的值，以及 i、j、f 和 o 张量的值。
假设我们有以下图表:
import numpy as np
import tensorflow as tf
timesteps = 7
num_input = 4
num_units = 3
x_val = np.random.normal(size=(1, timesteps, num_input))
lstm = tf.nn.rnn_cell.BasicLSTMCell(num_units = num_units)
X = tf.placeholder("float", [1, timesteps, num_input])
inputs = tf.unstack(X, timesteps, 1)
outputs, state = tf.contrib.rnn.static_rnn(lstm, inputs, dtype=tf.float32)
如果我们知道它的名字，我们就可以找到任何张量的值。查找张量名称的一种方法是查看 TensorBoard。
init = tf.global_variables_initializer()
graph = tf.get_default_graph()
with tf.Session(graph=graph) as sess:
train_writer = tf.summary.FileWriter('./graph', sess.graph)
sess.run(init)
现在我们可以通过终端命令启动 TensorBoard
tensorboard --logdir=graph --host=localhost
并发现产生 i, j, f, o 张量的操作名为 'rnn/basic_lstm_cell/split'，而 kernel 和bias 称为 'rnn/basic_lstm_cell/kernel' 和 'rnn/basic_lstm_cell/bias':
tf.contrib.rnn.static_rnn 函数调用我们的基本 lstm 单元 7 次，每个时间步调用一次。当 Tensorflow 被要求以相同的名称创建多个操作时，它会为它们添加后缀，如下所示:
rnn/basic_lstm_cell/split，
rnn/basic_lstm_cell/split_1，
...,
rnn/basic_lstm_cell/split_6。
这些是我们的业务名称。
tensorflow 中张量的名称由产生张量的操作的名称组成，后跟一个冒号，后跟产生该张量的操作输出的索引。内核和偏置操作有一个输出，所以张量名称将是
kernel = graph.get_tensor_by_name("rnn/basic_lstm_cell/kernel:0")
bias = graph.get_tensor_by_name("rnn/basic_lstm_cell/bias:0")
拆分操作产生四个输出:i、j、f 和 o，因此这些张量的名称将是:
i_list = []
j_list = []
f_list = []
o_list = []
for suffix in ["", "_1", "_2", "_3", "_4", "_5", "_6"]:
i_list.append(graph.get_tensor_by_name(
"rnn/basic_lstm_cell/split{}:0".format(suffix)
))
j_list.append(graph.get_tensor_by_name(
"rnn/basic_lstm_cell/split{}:1".format(suffix)
))
f_list.append(graph.get_tensor_by_name(
"rnn/basic_lstm_cell/split{}:2".format(suffix)
))
o_list.append(graph.get_tensor_by_name(
"rnn/basic_lstm_cell/split{}:3".format(suffix)
))
现在我们可以找到所有张量的值:
with tf.Session(graph=graph) as sess:
train_writer = tf.summary.FileWriter('./graph', sess.graph)
sess.run(init)
weights = sess.run([kernel, bias])
print("Weights:\n", weights)
i_values, j_values, f_values, o_values = sess.run([i_list, j_list, f_list, o_list],
feed_dict = {X:x_val})
print("i values:\n", i_values)
print("j values:\n", j_values)
print("f_values:\n", f_values)
print("o_values:\n", o_values)
或者，我们可以通过查看图中所有张量的列表来找到张量名称，可以通过以下方式生成:
tensors_per_node = [node.values() for node in graph.get_operations()]
tensor_names = [tensor.name for tensors in tensors_per_node for tensor in tensors]
print(tensor_names)
或者，对于所有操作的较短列表:
print([node.name for node in graph.get_operations()])
第三种方法是读取 source code 并找到哪些名称分配给了哪些张量。
关于python - Tensorflow LSTM 门权重，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42681757/