我正在尝试训练神经网络来预测两个数字的总和。但我不明白我的模型有什么问题。模型由2个输入,2个隐藏层和1个输出层组成。每1000次迭代我打印测试执行,但结果越来越小。
import numpy as np
import tensorflow as tf
input_size = 2
hidden_size = 3
out_size = 1
def generate_test_data():
inp = 0.5*np.random.rand(10, 2)
oup = np.zeros((10, 1))
for idx, val in enumerate(inp):
oup[idx] = np.array([val[0] + val[1]])
return inp, oup
def create_network():
x = tf.placeholder(tf.float32, [None, input_size])
w01 = tf.Variable(tf.truncated_normal([input_size, hidden_size], stddev=0.1))
y1 = tf.sigmoid(tf.matmul(tf.sigmoid(x), w01))
w12 = tf.Variable(tf.truncated_normal([hidden_size, out_size], stddev=0.1))
y2 = tf.sigmoid(tf.matmul(y1, w12))
y_ = tf.placeholder(tf.float32, [None, out_size])
return x, y_, y2
def train(x, y_, y2):
cross_entropy = tf.reduce_mean(
tf.nn.softmax_cross_entropy_with_logits(labels=y_, logits=y2)
)
train_step = tf.train.GradientDescentOptimizer(0.5).minimize(cross_entropy)
sess = tf.InteractiveSession()
tf.global_variables_initializer().run()
# Train
for i in range(100000):
batch_xs, batch_ys = generate_test_data()
sess.run(train_step, feed_dict={x: batch_xs, y_: batch_ys})
# Test
if i % 1000 == 0:
out_batch = sess.run(y2, {x: batch_xs})
inx = 0
print(batch_xs[inx][0], " + ", batch_xs[inx][1], " = ", out_batch[inx][0])
(x, y_, y2) = create_network()
train(x, y_, y2)
每1000次迭代输出:
0.37301352864927173 + 0.28949461772342683 = 0.49111518
0.050899466843458474 + 0.006174158992116541 = 0.0025260744
0.3974852369427063 + 0.22402098418952499 = 0.00090828544
0.15735921047969498 + 0.39645077887600294 = 0.0005903727
0.23560825884336228 + 0.29010766384718145 = 0.0004317883
0.4250063393420791 + 0.24181166029062096 = 0.00031525563
= smaller and smaller
答案 0 :(得分:3)
交叉熵丢失用于分类问题,而您的任务显然是回归。计算出的cross_entropy
值没有意义,因此就是结果。
将您的损失更改为:
cross_entropy = tf.reduce_mean(
tf.nn.l2_loss(y_ - y2)
)
......你会看到更明智的结果。
答案 1 :(得分:0)
Maxim,非常感谢。现在它的工作。
import numpy as np
import tensorflow as tf
input_size = 2
hidden_size = 3
out_size = 1
def generate_test_data():
inp = 0.5*np.random.rand(10, 2)
oup = np.zeros((10, 1))
for idx, val in enumerate(inp):
oup[idx] = np.array([val[0] + val[1]])
return inp, oup
def create_network():
x = tf.placeholder(tf.float32, [None, input_size])
w01 = tf.Variable(tf.truncated_normal([input_size, hidden_size], stddev=0.1))
y1 = tf.matmul(x, w01)
w12 = tf.Variable(tf.truncated_normal([hidden_size, out_size], stddev=0.1))
y2 = tf.matmul(y1, w12)
y_ = tf.placeholder(tf.float32, [None, out_size])
return x, y_, y2
def train(x, y_, y2):
cross_entropy = tf.reduce_mean(
tf.nn.l2_loss(y_ - y2)
)
train_step = tf.train.GradientDescentOptimizer(0.5).minimize(cross_entropy)
sess = tf.InteractiveSession()
tf.global_variables_initializer().run()
# Train
for i in range(100000):
batch_xs, batch_ys = generate_test_data()
sess.run(train_step, feed_dict={x: batch_xs, y_: batch_ys})
# Test
if i % 2000 == 0:
out_batch = sess.run(y2, {x: batch_xs})
inx = 0
print(batch_xs[inx][0], " + ", batch_xs[inx][1], " = ", out_batch[inx][0], "|", batch_xs[inx][0] + batch_xs[inx][1])
(x, y_, y2) = create_network()
train(x, y_, y2)
答案 2 :(得分:0)
如果您考虑将每个数字预测为您在“0123456789”中预测值的分类问题,则可以使用交叉熵作为损失。有关参考,请参阅Keras - Addition RNN Example。
但正如马克西姆所说,它不应该用于回归问题。