【发布时间】:2018-05-14 10:57:24
【问题描述】:
我正在尝试训练一个基本的循环神经网络(多对多)。输入数据有一个特征(一个 sin 函数),标签是二进制的:(1 和 2)
我可以使用 MSE 损失函数对其进行训练,但是当我尝试将其替换为 xentropy 时遇到了一些问题。
这是我目前拥有的适用于非离散标签的代码:
import matplotlib.pyplot as plt
import tensorflow as tf
import numpy as np
n_steps = 20
n_inputs = 1
n_neurons = 100
n_outputs = 1
learning_rate = 0.001
# Create data
fs = 1000 # sample rate
f = 2 # the frequency of the signal
x = np.arange(fs) # the points on the x axis for plotting
# training features
dfX = np.array([ np.sin(2*np.pi*f * (i/fs)) for i in x])
#labels
dfX2 = np.array([ np.sin(2*np.pi*f * (i/fs)) for i in x])
dfX2[dfX2 < 0] = 1
dfX2[dfX2 > 0] = 2
# RNN
X = tf.placeholder(tf.float32, [None, n_steps, n_inputs])
y = tf.placeholder(tf.float32, [None, n_steps, n_outputs])
cell = tf.contrib.rnn.OutputProjectionWrapper( tf.contrib.rnn.BasicRNNCell(num_units=n_neurons, activation=tf.nn.relu),output_size=n_outputs)
outputs, states = tf.nn.dynamic_rnn(cell, X, dtype=tf.float32)
loss = tf.reduce_mean(tf.square(outputs - y))
optimizer = tf.train.AdamOptimizer(learning_rate=learning_rate)
training_op = optimizer.minimize(loss)
init = tf.global_variables_initializer()
n_iterations = 1000
batch_size = 200
n_epouchs=10
with tf.Session() as sess:
init.run()
for epouch in range(n_epouchs):
for iteration in range(n_iterations-200):
X_batch= dfX[iteration:iteration+batch_size]
X_batch= X_batch.reshape(-1,n_steps,n_inputs)
y_batch= dfX2[(iteration+1):(iteration+batch_size+1)]
y_batch= y_batch.reshape(-1,n_steps,n_outputs)
sess.run(training_op, feed_dict={X: X_batch, y: y_batch})
if iteration % 100 == 0:
mse = loss.eval(feed_dict={X: X_batch, y: y_batch})
print(iteration,"--",epouch, "\tMSE:", mse)
X_new1= dfX[37:37+batch_size]
X_new1= X_new1.reshape(-1,n_steps,n_inputs)
y_pred1 = sess.run(outputs, feed_dict={X: X_new1})
我正在尝试将损失函数更改为类似这样(因为我想预测离散标签):
xentropy = tf.nn.sparse_softmax_cross_entropy_with_logits(labels=y, logits=outputs)
loss = tf.reduce_mean(xentropy)
但它不起作用
【问题讨论】:
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如果我的回答解决了您的问题,请告诉我,谢谢。
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我没有抱歉....
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我已经修复了几个问题,也测试了代码。
标签: python tensorflow rnn