2017-10-06 92 views
1

我想在张量流中实现多元回归,其中我有192个具有6个特征和一个输出变量的示例。从我的模型中,我得到一个矩阵(192,6),而它应该是(192,1)。有人知道我的代码有什么问题吗?我在下面提供了我的代码。使用TensorFlow计算多元回归

# Parameters 
learning_rate = 0.0001 
training_epochs = 50 
display_step = 5 

train_X = Data_ABX3[0:192, 0:6] 
train_Y = Data_ABX3[0:192, [24]] 


# placeholders for a tensor that will be always fed. 
X = tf.placeholder('float', shape = [None, 6]) 
Y = tf.placeholder('float', shape = [None, 1]) 


# Training Data 

n_samples = train_Y.shape[0] 


# Set model weights 
W = tf.cast(tf.Variable(rng.randn(1, 6), name="weight"), tf.float32) 
b = tf.Variable(rng.randn(), name="bias") 

# Construct a linear model 
pred = tf.add(tf.multiply(X, W), b) 

# Mean squared error 
cost = tf.reduce_sum(tf.pow(pred-Y, 2))/(2*n_samples) 
# Gradient descent 
# Note, minimize() knows to modify W and b because Variable objects are  trainable=True by default 
optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost) 

# Accuracy 
# #accuracy = tf.contrib.metrics.streaming_accuracy(Y, pred) 

# Initialize the variables (i.e. assign their default value) 
init = tf.global_variables_initializer() 

# Start training 
with tf.Session() as sess: 

    # Run the initializer 
    sess.run(init) 

    # Fit all training data 
    for epoch in range(training_epochs): 
     #for (x, y) in zip(train_X, train_Y): 
     sess.run(optimizer, feed_dict={X: train_X, Y: train_Y}) 

     # Display logs per epoch step 
     if (epoch+1) % display_step == 0: 
      c = sess.run(cost, feed_dict={X: train_X, Y:train_Y}) 
      print("Epoch:", '%04d' % (epoch+1), "cost=", "{:.9f}".format(c), \ 
       "W=", sess.run(W), "b=", sess.run(b)) 

    print("Optimization Finished!") 
    #training_cost = 0 
    #for (x, y) in zip(train_X, train_Y): 
    #  tr_cost = sess.run(cost, feed_dict={X: x, Y: y}) 
    #  training_cost += tr_cost 
    training_cost = sess.run(cost, feed_dict={X: train_X, Y: train_Y}) 
    print("Training cost=", training_cost, "W=", sess.run(W), "b=", sess.run(b), '\n') 

    # Graphic display 
    plt.plot(train_Y, train_X * sess.run(W) + sess.run(b), label='Fitted line') 
    plt.legend() 
    plt.show() 

回答

1

请您pred方程使用tf.matmul代替tf.multiplytf.multiply做元素明智乘法因此,它将生成一个与train_X相同维度的矩阵,而tf.matmul将执行矩阵乘法,并且将基于实际矩阵乘法规则生成结果矩阵。

我不确定你的数据是什么。添加随机数据,然后更改代码以满足所有维度要求。如果你能帮助我的意图,这将有助于更好地看到问题。

编辑

import numpy as np 
import tensorflow as tf 
import matplotlib.pyplot as plt 
# Parameters 
learning_rate = 0.0001 
training_epochs = 50 
display_step = 5 

Data_ABX3 = np.random.random((193, 8)).astype('f') 

train_X = Data_ABX3[0:192, 0:6] 
train_Y = Data_ABX3[0:192, [7]] 


# placeholders for a tensor that will be always fed. 
X = tf.placeholder('float32', shape = [None, 6]) 
Y = tf.placeholder('float32', shape = [None, 1]) 

# Training Data 
n_samples = train_Y.shape[0] 

# Set model weights 
W = tf.cast(tf.Variable(np.random.randn(6, 1), name="weight"), tf.float32) 
b = tf.Variable(np.random.randn(), name="bias") 

mult_node = tf.matmul(X, W) 
print(mult_node.shape) 
# Construct a linear model 
pred = tf.add(tf.matmul(X, W), b) 

# Mean squared error 
cost = tf.reduce_sum(tf.pow(pred-Y, 2))/(2*n_samples) 
# Gradient descent 
# Note, minimize() knows to modify W and b because Variable objects are    trainable=True by default 
optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost) 

# Accuracy 
# #accuracy = tf.contrib.metrics.streaming_accuracy(Y, pred) 

# Initialize the variables (i.e. assign their default value) 
init = tf.global_variables_initializer() 

# Start training 
with tf.Session() as sess: 

# Run the initializer 
sess.run(init) 

# Fit all training data 
for epoch in range(training_epochs): 
    #for (x, y) in zip(train_X, train_Y): 
    sess.run(optimizer, feed_dict={X: train_X, Y: train_Y}) 

    # Display logs per epoch step 
    if (epoch+1) % display_step == 0: 
     c = sess.run(cost, feed_dict={X: train_X, Y:train_Y}) 
     print("Epoch:", '%04d' % (epoch+1), "cost=", "{:.9f}".format(c), \ 
      "W=", sess.run(W), "b=", sess.run(b)) 

print("Optimization Finished!") 
#training_cost = 0 
#for (x, y) in zip(train_X, train_Y): 
#  tr_cost = sess.run(cost, feed_dict={X: x, Y: y}) 
#  training_cost += tr_cost 
training_cost = sess.run(cost, feed_dict={X: train_X, Y: train_Y}) 
print("Training cost=", training_cost, "W=", sess.run(W), "b=", sess.run(b), '\n') 

line = sess.run(tf.add(tf.matmul(train_X, W), b)) 
# Graphic display 
plt.plot(train_Y, line, label='Fitted line') 
plt.legend() 
plt.show()` 
+0

谢谢Rachit。我使用它并且没有工作,我得到这个消息:ValueError:尺寸必须相等,但对于'MatMul'(op:'MatMul'),其输入形状为[?,6],[1, 6]。 – Hamid

+0

@Hamid不知道你的输入数据,但我用你的代码随机数据。 –

+0

非常感谢您的有用评论。我更改了代码的这些部分:“Data_ABX3 = numpy.loadtxt(file,dtype ='float32',...”以及“W = tf.cast(tf.Variable(tf.zeros([6,1])) ,...“。现在它正在工作,但我得到越来越高的成本(培训成本= 4.81842e + 28),同时我增加了training_epochs。 – Hamid