理解为什么 TensorFlow RNN 不学习玩具数据

kbr*_*ose 0 python tensorflow recurrent-neural-network

我正在尝试使用 Tensorflow(r0.10,python 3.5)在玩具分类问题上训练循环神经网络,但我得到的结果令人困惑。

我想将 0 和 1 的序列输入到 RNN 中,并将序列中给定元素的目标类作为序列的当前值和先前值表示的数字,并将其视为二进制数。例如:

input sequence: [0,     0,     1,     0,     1,     1]
binary digits : [-, [0,0], [0,1], [1,0], [0,1], [1,1]]
target class  : [-,     0,     1,     2,     1,     3]
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看起来这是 RNN 应该能够很容易学习的东西,但我的模型只能区分类 [0,2] 和 [1,3]。换句话说,它能够区分当前数字为 0 的类和当前数字为 1 的类。这让我相信 RNN 模型没有正确学习查看序列的先前值。

有几个教程和示例([ 1 ]、[ 2 ]、[ 3 ])演示了如何在张量流中构建和使用循环神经网络(RNN),但在研究它们之后我仍然没有看到我的问题(它没有有助于所有示例都使用文本作为源数据)。

我将数据作为tf.nn.rnn()length 列表输入T,其元素是[batch_size x input_size]序列。由于我的序列是一维的,input_size等于一,所以本质上我相信我正在输入一个长度序列的列表batch_size(我不清楚文档中哪个维度被视为时间维度)。这种理解正确吗?如果是这样的话,那么我不明白为什么 RNN 模型不能正确学习。

很难获得一小部分可以运行我的完整 RNN 的代码,这是我能做的最好的事情(它主要改编自此处的 PTB 模型此处的 char-rnn 模型):

import tensorflow as tf
import numpy as np

input_size = 1
batch_size = 50
T = 2
lstm_size = 5
lstm_layers = 2
num_classes = 4
learning_rate = 0.1

lstm = tf.nn.rnn_cell.BasicLSTMCell(lstm_size, state_is_tuple=True)
lstm = tf.nn.rnn_cell.MultiRNNCell([lstm] * lstm_layers, state_is_tuple=True)

x = tf.placeholder(tf.float32, [T, batch_size, input_size])
y = tf.placeholder(tf.int32, [T * batch_size * input_size])

init_state = lstm.zero_state(batch_size, tf.float32)

inputs = [tf.squeeze(input_, [0]) for input_ in tf.split(0,T,x)]
outputs, final_state = tf.nn.rnn(lstm, inputs, initial_state=init_state)

w = tf.Variable(tf.truncated_normal([lstm_size, num_classes]), name='softmax_w')
b = tf.Variable(tf.truncated_normal([num_classes]), name='softmax_b')

output = tf.concat(0, outputs)

logits = tf.matmul(output, w) + b

probs = tf.nn.softmax(logits)

cost = tf.reduce_mean(tf.nn.seq2seq.sequence_loss_by_example(
    [logits], [y], [tf.ones_like(y, dtype=tf.float32)]
))

optimizer = tf.train.GradientDescentOptimizer(learning_rate)
tvars = tf.trainable_variables()
grads, _ = tf.clip_by_global_norm(tf.gradients(cost, tvars),
                                  10.0)
train_op = optimizer.apply_gradients(zip(grads, tvars))

init = tf.initialize_all_variables()

with tf.Session() as sess:
    sess.run(init)
    curr_state = sess.run(init_state)
    for i in range(3000):
        # Create toy data where the true class is the value represented
        # by the current and previous value treated as binary, i.e.
        train_x = np.random.randint(0,2,(T * batch_size * input_size))
        train_y = train_x + np.concatenate(([0], (train_x[:-1] * 2)))

        # Reshape into T x batch_size x input_size
        train_x = np.reshape(train_x, (T, batch_size, input_size))

        feed_dict = {
            x: train_x, y: train_y
        }
        for j, (c, h) in enumerate(init_state):
            feed_dict[c] = curr_state[j].c
            feed_dict[h] = curr_state[j].h

        fetch_dict = {
            'cost': cost, 'final_state': final_state, 'train_op': train_op
        }

        # Evaluate the graph
        fetches = sess.run(fetch_dict, feed_dict=feed_dict)

        curr_state = fetches['final_state']

        if i % 300 == 0:
            print('step {}, train cost: {}'.format(i, fetches['cost']))

    # Test
    test_x = np.array([[0],[0],[1],[0],[1],[1]]*(T*batch_size*input_size))
    test_x = test_x[:(T*batch_size*input_size),:]
    probs_out = sess.run(probs, feed_dict={
            x: np.reshape(test_x, [T, batch_size, input_size]),
            init_state: curr_state
        })
    # Get the softmax outputs for the points in the sequence
    # that have [0, 0], [0, 1], [1, 0], [1, 1] as their
    # last two values.
    for i in [1, 2, 3, 5]:
        print('{}: [{:.4f} {:.4f} {:.4f} {:.4f}]'.format(
                [1, 2, 3, 5].index(i), *list(probs_out[i,:]))
             )
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这里的最终输出是

0: [0.4899 0.0007 0.5080 0.0014]
1: [0.0003 0.5155 0.0009 0.4833]
2: [0.5078 0.0011 0.4889 0.0021]
3: [0.0003 0.5052 0.0009 0.4936]
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这表明它只是学习区分 [0,2] 和 [1,3]。为什么这个模型不学习使用序列中的前一个值?

kbr*_*ose 5

在这篇博客文章的帮助下弄清楚了(它有精彩的输入张量图表)。事实证明,我没有正确理解输入的形状tf.nn.rnn()

假设您有batch_size多个序列。每个序列都有input_size尺寸和长度(选择这些名称是为了与此处T的文档相匹配)。然后,您需要将输入拆分为一个-length 列表,其中每个元素都有 shape 。这意味着您的连续序列将分布在列表的元素。我认为连续的序列将保留在一起,以便列表中的每个元素都是一个序列的示例。tf.nn.rnn() Tbatch_size x input_sizeinputs

回想起来这是有道理的,因为我们希望并行化序列中的每个步骤,所以我们希望运行每个序列的第一步(列表中的第一个元素),然后运行每个序列的第二步(列表中的第二个元素),等等。

代码的工作版本:

import tensorflow as tf
import numpy as np

sequence_size = 50
batch_size = 7
num_features = 1
lstm_size = 5
lstm_layers = 2
num_classes = 4
learning_rate = 0.1

lstm = tf.nn.rnn_cell.BasicLSTMCell(lstm_size, state_is_tuple=True)
lstm = tf.nn.rnn_cell.MultiRNNCell([lstm] * lstm_layers, state_is_tuple=True)

x = tf.placeholder(tf.float32, [batch_size, sequence_size, num_features])
y = tf.placeholder(tf.int32, [batch_size * sequence_size * num_features])

init_state = lstm.zero_state(batch_size, tf.float32)

inputs = [tf.squeeze(input_, [1]) for input_ in tf.split(1,sequence_size,x)]
outputs, final_state = tf.nn.rnn(lstm, inputs, initial_state=init_state)

w = tf.Variable(tf.truncated_normal([lstm_size, num_classes]), name='softmax_w')
b = tf.Variable(tf.truncated_normal([num_classes]), name='softmax_b')

output = tf.reshape(tf.concat(1, outputs), [-1, lstm_size])

logits = tf.matmul(output, w) + b

probs = tf.nn.softmax(logits)

cost = tf.reduce_mean(tf.nn.seq2seq.sequence_loss_by_example(
    [logits], [y], [tf.ones_like(y, dtype=tf.float32)]
))

# Now optimize on that cost
optimizer = tf.train.GradientDescentOptimizer(learning_rate)
tvars = tf.trainable_variables()
grads, _ = tf.clip_by_global_norm(tf.gradients(cost, tvars),
                                  10.0)
train_op = optimizer.apply_gradients(zip(grads, tvars))

init = tf.initialize_all_variables()

with tf.Session() as sess:
    sess.run(init)
    curr_state = sess.run(init_state)
    for i in range(3000):
        # Create toy data where the true class is the value represented
        # by the current and previous value treated as binary, i.e.
        
        train_x = np.random.randint(0,2,(batch_size * sequence_size * num_features))
        train_y = train_x + np.concatenate(([0], (train_x[:-1] * 2)))
        
        # Reshape into T x batch_size x sequence_size
        train_x = np.reshape(train_x, [batch_size, sequence_size, num_features])
        
        feed_dict = {
            x: train_x, y: train_y
        }
        for j, (c, h) in enumerate(init_state):
            feed_dict[c] = curr_state[j].c
            feed_dict[h] = curr_state[j].h
        
        fetch_dict = {
            'cost': cost, 'final_state': final_state, 'train_op': train_op
        }
        
        # Evaluate the graph
        fetches = sess.run(fetch_dict, feed_dict=feed_dict)
        
        curr_state = fetches['final_state']
        
        if i % 300 == 0:
            print('step {}, train cost: {}'.format(i, fetches['cost']))
    
    # Test
    test_x = np.array([[0],[0],[1],[0],[1],[1]]*(batch_size * sequence_size * num_features))
    test_x = test_x[:(batch_size * sequence_size * num_features),:]
    probs_out = sess.run(probs, feed_dict={
            x: np.reshape(test_x, [batch_size, sequence_size, num_features]),
            init_state: curr_state
        })
    # Get the softmax outputs for the points in the sequence
    # that have [0, 0], [0, 1], [1, 0], [1, 1] as their
    # last two values.
    for i in [1, 2, 3, 5]:
        print('{}: [{:.4f} {:.4f} {:.4f} {:.4f}]'.format(
                [1, 2, 3, 5].index(i), *list(probs_out[i,:]))
             )
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