python - tensorflow 神经网络多层感知器回归示例

我正在尝试为多元 REGRESSION (请不要使用MNIST)和 TensorFlow (我刚刚开始学习，对代码很抱歉!)写一个 MLP 。这是我的MWE，在这里我选择使用sklearn的linnerud数据集。 (实际上，我使用的是更大的数据集，在这里我也只使用了一层，因为我想使MWE较小，但如有必要，可以添加)。顺便说一下，我在shuffle = False中使用train_test_split只是因为实际上我正在使用时间序列数据集。
MWE

######################### import stuff ##########################
import numpy as np
import pandas as pd
import tensorflow as tf
from sklearn.datasets import load_linnerud
from sklearn.model_selection import train_test_split


######################## prepare the data ########################
X, y = load_linnerud(return_X_y = True)
X_train, X_test, y_train, y_test = train_test_split(X, y, shuffle = False, test_size = 0.33)


######################## set learning variables ##################
learning_rate = 0.0001
epochs = 100
batch_size = 3


######################## set some variables #######################
x = tf.placeholder(tf.float32, [None, 3], name = 'x')   # 3 features
y = tf.placeholder(tf.float32, [None, 3], name = 'y')   # 3 outputs

# input-to-hidden layer1
W1 = tf.Variable(tf.truncated_normal([3,300], stddev = 0.03), name = 'W1')
b1 = tf.Variable(tf.truncated_normal([300]), name = 'b1')

# hidden layer1-to-output
W2 = tf.Variable(tf.truncated_normal([300,3], stddev = 0.03), name=  'W2')
b2 = tf.Variable(tf.truncated_normal([3]), name = 'b2')


######################## Activations, outputs ######################
# output hidden layer 1
hidden_out = tf.nn.relu(tf.add(tf.matmul(x, W1), b1))

# total output
y_ = tf.nn.relu(tf.add(tf.matmul(hidden_out, W2), b2))


####################### Loss Function  #########################
mse = tf.losses.mean_squared_error(y, y_)


####################### Optimizer      #########################
optimizer = tf.train.GradientDescentOptimizer(learning_rate = learning_rate).minimize(mse)


###################### Initialize, Accuracy and Run #################
# initialize variables
init_op = tf.global_variables_initializer()

# accuracy for the test set
accuracy = tf.reduce_mean(tf.square(tf.subtract(y, y_))) # or could use tf.losses.mean_squared_error

#run
with tf.Session() as sess:
     sess.run(init_op)
     total_batch = int(len(y_train) / batch_size)
     for epoch in range(epochs):
         avg_cost = 0
         for i in range(total_batch):
              batch_x, batch_y =  X_train[i*batch_size:min(i*batch_size + batch_size, len(X_train)), :], y_train[i*batch_size:min(i*batch_size + batch_size, len(y_train)), :]
              _, c = sess.run([optimizer, mse], feed_dict = {x: batch_x, y: batch_y})
              avg_cost += c / total_batch
         print('Epoch:', (epoch+1), 'cost =', '{:.3f}'.format(avg_cost))
     print(sess.run(mse, feed_dict = {x: X_test, y:y_test}))

这会打印出这样的内容

...
Epoch: 98 cost = 10992.617
Epoch: 99 cost = 10992.592
Epoch: 100 cost = 10992.566
11815.1

所以很明显这里出了问题。我怀疑问题可能出在成本函数/准确性或我使用批处理的方式上，但我无法完全弄清楚。

最佳答案

据我所知，该模型是学习中的。我尝试调整一些超参数(最重要的是-学习率和隐藏层大小)，并获得了更好的结果。这是完整的代码:

######################### import stuff ##########################
import numpy as np
import pandas as pd
import tensorflow as tf
from sklearn.datasets import load_linnerud
from sklearn.model_selection import train_test_split

######################## prepare the data ########################
X, y = load_linnerud(return_X_y=True)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.33, shuffle=False)

######################## set learning variables ##################
learning_rate = 0.0005
epochs = 2000
batch_size = 3

######################## set some variables #######################
x = tf.placeholder(tf.float32, [None, 3], name='x')  # 3 features
y = tf.placeholder(tf.float32, [None, 3], name='y')  # 3 outputs

# hidden layer 1
W1 = tf.Variable(tf.truncated_normal([3, 10], stddev=0.03), name='W1')
b1 = tf.Variable(tf.truncated_normal([10]), name='b1')

# hidden layer 2
W2 = tf.Variable(tf.truncated_normal([10, 3], stddev=0.03), name='W2')
b2 = tf.Variable(tf.truncated_normal([3]), name='b2')

######################## Activations, outputs ######################
# output hidden layer 1
hidden_out = tf.nn.relu(tf.add(tf.matmul(x, W1), b1))

# total output
y_ = tf.nn.relu(tf.add(tf.matmul(hidden_out, W2), b2))

####################### Loss Function  #########################
mse = tf.losses.mean_squared_error(y, y_)

####################### Optimizer      #########################
optimizer = tf.train.GradientDescentOptimizer(learning_rate=learning_rate).minimize(mse)

###################### Initialize, Accuracy and Run #################
# initialize variables
init_op = tf.global_variables_initializer()

# accuracy for the test set
accuracy = tf.reduce_mean(tf.square(tf.subtract(y, y_)))  # or could use tf.losses.mean_squared_error

# run
with tf.Session() as sess:
  sess.run(init_op)
  total_batch = int(len(y_train) / batch_size)
  for epoch in range(epochs):
    avg_cost = 0
    for i in range(total_batch):
      batch_x, batch_y = X_train[i * batch_size:min(i * batch_size + batch_size, len(X_train)), :], \
                         y_train[i * batch_size:min(i * batch_size + batch_size, len(y_train)), :]
      _, c = sess.run([optimizer, mse], feed_dict={x: batch_x, y: batch_y})
      avg_cost += c / total_batch
    if epoch % 10 == 0:
      print 'Epoch:', (epoch + 1), 'cost =', '{:.3f}'.format(avg_cost)
  print sess.run(mse, feed_dict={x: X_test, y: y_test})

输出:

Epoch: 1901 cost = 173.914
Epoch: 1911 cost = 171.928
Epoch: 1921 cost = 169.993
Epoch: 1931 cost = 168.110
Epoch: 1941 cost = 166.277
Epoch: 1951 cost = 164.492
Epoch: 1961 cost = 162.753
Epoch: 1971 cost = 161.061
Epoch: 1981 cost = 159.413
Epoch: 1991 cost = 157.808
482.433

我认为您可以进一步调整它，但是由于数据太小而没有意义。
虽然我没有尝试过正则化，但是我确定您需要使用L2 reg或dropout来避免过度拟合。