【问题标题】:Calculating Multivariate regression using TensorFlow使用 TensorFlow 计算多元回归
【发布时间】:2018-03-17 19:29:38
【问题描述】:

我正在尝试在 tensorflow 中实现多元回归,其中我有 192 个示例,其中包含 6 个特征和一个输出变量。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。

# 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()

【问题讨论】:

    标签: tensorflow regression shape tensor


    【解决方案1】:

    请在您的pred 等式中使用tf.matmul 而不是tf.multiplytf.multiply 进行元素乘法因此,它将生成与train_X 相同维度的矩阵,而tf.matmul 将进行矩阵乘法,并将根据实际矩阵乘法规则生成结果矩阵。

    我不确定您的数据是什么。添加随机数据,然后更改代码以满足所有维度要求。如果您可以帮助我了解您的意图,那将有助于更好地了解问题。

    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()`
    

    【讨论】:

    • 谢谢拉奇特。我使用了它但没有用,我收到这条消息:ValueError:尺寸必须相等,但对于输入形状为 [?,6]、[1、 6].
    • @Hamid 不确定您的输入数据,但我在您的代码中使用了随机数据。
    • 非常感谢您提供的有用的 cmets。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]) ,..." 。现在它可以工作了,但是当我增加 training_epochs 时,我的成本越来越高(培训成本 = 4.81842e+28)。
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