【问题标题】:How to extract the cell state and hidden state from an RNN model in tensorflow?如何从张量流中的RNN模型中提取细胞状态和隐藏状态?
【发布时间】:2016-10-26 21:41:55
【问题描述】:

我是 TensorFlow 新手,难以理解 RNN 模块。我正在尝试从 LSTM 中提取隐藏/单元状态。 对于我的代码,我使用来自https://github.com/aymericdamien/TensorFlow-Examples 的实现。

# tf Graph input
x = tf.placeholder("float", [None, n_steps, n_input])
y = tf.placeholder("float", [None, n_classes])

# Define weights
weights = {'out': tf.Variable(tf.random_normal([n_hidden, n_classes]))}
biases = {'out': tf.Variable(tf.random_normal([n_classes]))}

def RNN(x, weights, biases):
    # Prepare data shape to match `rnn` function requirements
    # Current data input shape: (batch_size, n_steps, n_input)
    # Required shape: 'n_steps' tensors list of shape (batch_size, n_input)

    # Permuting batch_size and n_steps
    x = tf.transpose(x, [1, 0, 2])
    # Reshaping to (n_steps*batch_size, n_input)
    x = tf.reshape(x, [-1, n_input])
    # Split to get a list of 'n_steps' tensors of shape (batch_size, n_input)
    x = tf.split(0, n_steps, x)

    # Define a lstm cell with tensorflow
    #with tf.variable_scope('RNN'):
    lstm_cell = rnn_cell.BasicLSTMCell(n_hidden, forget_bias=1.0, state_is_tuple=True)

    # Get lstm cell output
        outputs, states = rnn.rnn(lstm_cell, x, dtype=tf.float32)

    # Linear activation, using rnn inner loop last output
    return tf.matmul(outputs[-1], weights['out']) + biases['out'], states

pred, states = RNN(x, weights, biases)

# Define loss and optimizer
cost = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(pred, y))
optimizer = tf.train.AdamOptimizer(learning_rate=learning_rate).minimize(cost)

# Evaluate model
correct_pred = tf.equal(tf.argmax(pred,1), tf.argmax(y,1))
accuracy = tf.reduce_mean(tf.cast(correct_pred, tf.float32))
# Initializing the variables
init = tf.initialize_all_variables()

现在我想为预测中的每个时间步提取单元格/隐藏状态。状态存储在 (c,h) 形式的 LSTMStateTuple 中,我可以通过评估 print states 找到它。但是,尝试调用 print states.c.eval()(根据文档应该给我张量 states.c 中的值)会产生一个错误,指出我的变量没有初始化,即使我在预测某些事情后立即调用它。代码在这里:

# Launch the graph
with tf.Session() as sess:
    sess.run(init)
    step = 1
    # Keep training until reach max iterations
    for v in tf.get_collection(tf.GraphKeys.VARIABLES, scope='RNN'):
        print v.name
    while step * batch_size < training_iters:
        batch_x, batch_y = mnist.train.next_batch(batch_size)
        # Reshape data to get 28 seq of 28 elements
        batch_x = batch_x.reshape((batch_size, n_steps, n_input))
        # Run optimization op (backprop)
        sess.run(optimizer, feed_dict={x: batch_x, y: batch_y})

        print states.c.eval()
        # Calculate batch accuracy
        acc = sess.run(accuracy, feed_dict={x: batch_x, y: batch_y})

        step += 1
    print "Optimization Finished!"

错误信息是

InvalidArgumentError: You must feed a value for placeholder tensor 'Placeholder' with dtype float
     [[Node: Placeholder = Placeholder[dtype=DT_FLOAT, shape=[], _device="/job:localhost/replica:0/task:0/cpu:0"]()]]

tf.all_variables() 中也看不到这些状态,只有经过训练的矩阵/偏置张量(如此处所述:Tensorflow: show or save forget gate values in LSTM)。我不想从头开始构建整个 LSTM,因为我在 states 变量中有状态,我只需要调用它。

【问题讨论】:

    标签: python tensorflow lstm


    【解决方案1】:

    您可以像收集准确性一样简单地收集states 的值。

    我猜,pred, states, acc = sess.run(pred, states, accuracy, feed_dict={x: batch_x, y: batch_y}) 应该可以正常工作。

    【讨论】:

    • 非常感谢!我不知道它是这样工作的。对于将来偶然发现这个问题的人来说,有两个小调整:1)参数应该在一个数组中 2)我分配的变量的名称必须与以前分配的变量不同。对我有用的语法是preds, stat, acc = sess.run([pred, states, accuracy], feed_dict={x: batch_x, y: batch_y})
    【解决方案2】:

    我不得不不同意 user3480922 的回答。代码:

    outputs, states = rnn.rnn(lstm_cell, x, dtype=tf.float32)
    

    为了能够在预测中提取每个 time_step 的隐藏状态,您必须使用输出。因为输出具有每个 time_step 的隐藏状态值。但是,我不确定我们是否也可以存储每个 time_step 的单元状态值。因为 states 元组提供了单元状态值,但仅用于最后一个 time_step。

    例如,在以下具有 5 个 time_steps 的示例中,输出[4,:,:], time_step = 0,...,4 具有 time_step=4 的隐藏状态值,而状态元组 h 仅具有time_step=4 的隐藏状态值。状态元组 c 虽然在 time_step=4 处具有单元格值。

      outputs = [[[ 0.0589103 -0.06925126 -0.01531546 0.06108122]
      [ 0.00861215 0.06067181 0.03790079 -0.04296958]
      [ 0.00597713 0.03916606 0.02355802 -0.0277683 ]]
    
      [[ 0.06252582 -0.07336216 -0.01607122 0.05024602]
      [ 0.05464711 0.03219429 0.06635305 0.00753127]
      [ 0.05385715 0.01259535 0.0524035 0.01696803]]
    
      [[ 0.0853352 -0.06414541 0.02524283 0.05798233]
      [ 0.10790729 -0.05008117 0.03003334 0.07391824]
      [ 0.10205664 -0.04479517 0.03844892 0.0693808 ]]
    
      [[ 0.10556188 0.0516542 0.09162509 -0.02726674]
      [ 0.11425048 -0.00211394 0.06025286 0.03575509]
      [ 0.11338984 0.02839304 0.08105748 0.01564003]]
    
      **[[ 0.10072514 0.14767936 0.12387902 -0.07391471]
      [ 0.10510238 0.06321315 0.08100517 -0.00940042]
      [ 0.10553667 0.0984127 0.10094948 -0.02546882]]**]
      states = LSTMStateTuple(c=array([[ 0.23870754, 0.24315512, 0.20842518, -0.12798975],
      [ 0.23749796, 0.10797793, 0.14181322, -0.01695861],
      [ 0.2413336 , 0.16692916, 0.17559692, -0.0453596 ]], dtype=float32), h=array(**[[ 0.10072514, 0.14767936, 0.12387902, -0.07391471],
      [ 0.10510238, 0.06321315, 0.08100517, -0.00940042],
      [ 0.10553667, 0.0984127 , 0.10094948, -0.02546882]]**, dtype=float32))
    

    【讨论】:

      【解决方案3】:

      关于您的假设的一条评论:“状态”确实只有上一个时间步的“隐藏状态”和“记忆单元”的值。

      “输出”包含您想要的每个时间步的“隐藏状态”(输出的大小是 [batch_size, seq_len, hidden_​​size]。所以我假设您想要“输出”变量,而不是“状态”。见documentation

      【讨论】:

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