Tensorflow 自动编码器 - 如何计算重构误差？

Posted 2023-02-16

【中文标题】Tensorflow 自动编码器 - 如何计算重构误差？

【英文标题】：Tensorflow Autoencoder - How To Calculate Reconstruction Error?

【发布时间】：2017-09-09 06:51:02

【问题描述】：

我在 Tensorflow 中实现了以下自动编码器，如下所示。它基本上将 MNIST 数字作为输入，学习数据的结构并在其输出处重现输入。

from __future__ import division, print_function, absolute_import

import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt

# Import MNIST data
from tensorflow.examples.tutorials.mnist import input_data
mnist = input_data.read_data_sets("MNIST_data", one_hot=True)

# Parameters
learning_rate = 0.01
training_epochs = 20
batch_size = 256
display_step = 1
examples_to_show = 10

# Network Parameters
n_hidden_1 = 256 # 1st layer num features
n_hidden_2 = 128 # 2nd layer num features
n_input = 784 # MNIST data input (img shape: 28*28)

# tf Graph input (only pictures)
X = tf.placeholder("float", [None, n_input])

weights = 
    'encoder_h1': tf.Variable(tf.random_normal([n_input, n_hidden_1])),
    'encoder_h2': tf.Variable(tf.random_normal([n_hidden_1, n_hidden_2])),
    'decoder_h1': tf.Variable(tf.random_normal([n_hidden_2, n_hidden_1])),
    'decoder_h2': tf.Variable(tf.random_normal([n_hidden_1, n_input])),

biases = 
    'encoder_b1': tf.Variable(tf.random_normal([n_hidden_1])),
    'encoder_b2': tf.Variable(tf.random_normal([n_hidden_2])),
    'decoder_b1': tf.Variable(tf.random_normal([n_hidden_1])),
    'decoder_b2': tf.Variable(tf.random_normal([n_input])),



# Building the encoder
def encoder(x):
    # Encoder Hidden layer with sigmoid activation #1
    layer_1 = tf.nn.sigmoid(tf.add(tf.matmul(x, weights['encoder_h1']),
                                   biases['encoder_b1']))
    # Decoder Hidden layer with sigmoid activation #2
    layer_2 = tf.nn.sigmoid(tf.add(tf.matmul(layer_1, weights['encoder_h2']),
                                   biases['encoder_b2']))
    return layer_2


# Building the decoder
def decoder(x):
    # Encoder Hidden layer with sigmoid activation #1
    layer_1 = tf.nn.sigmoid(tf.add(tf.matmul(x, weights['decoder_h1']),
                                   biases['decoder_b1']))
    # Decoder Hidden layer with sigmoid activation #2
    layer_2 = tf.nn.sigmoid(tf.add(tf.matmul(layer_1, weights['decoder_h2']),
                                   biases['decoder_b2']))
    return layer_2

# Construct model
encoder_op = encoder(X)
decoder_op = decoder(encoder_op)

# Prediction
y_pred = decoder_op
# Targets (Labels) are the input data.
y_true = X

# Define loss and optimizer, minimize the squared error
cost = tf.reduce_mean(tf.pow(y_true - y_pred, 2))
optimizer = tf.train.RMSPropOptimizer(learning_rate).minimize(cost)

# Initializing the variables
init = tf.global_variables_initializer()

# Launch the graph
with tf.Session() as sess:
    sess.run(init)
    total_batch = int(mnist.train.num_examples/batch_size)
    # Training cycle
    for epoch in range(training_epochs):
        # Loop over all batches
        for i in range(total_batch):
            batch_xs, batch_ys = mnist.train.next_batch(batch_size)
            # Run optimization op (backprop) and cost op (to get loss value)
            _, c = sess.run([optimizer, cost], feed_dict=X: batch_xs)
        # Display logs per epoch step
        if epoch % display_step == 0:
            print("Epoch:", '%04d' % (epoch+1),
                  "cost=", ":.9f".format(c))

    print("Optimization Finished!")

    # Applying encode and decode over test set
    encode_decode = sess.run(
        y_pred, feed_dict=X: mnist.test.images[:examples_to_show])
    # Compare original images with their reconstructions
    f, a = plt.subplots(2, 10, figsize=(10, 2))
    for i in range(examples_to_show):
        a[0][i].imshow(np.reshape(mnist.test.images[i], (28, 28)))
        a[1][i].imshow(np.reshape(encode_decode[i], (28, 28)))
    f.show()
    plt.draw()
    plt.waitforbuttonpress()

当我对测试集进行编码和解码时，如何计算每个样本的重构误差（即均方误差/损失）？

换句话说，我想看看自动编码器能够如何重建其输入，以便我可以将自动编码器用作单类分类器。

非常感谢。

巴里

【参考方案1】：

你可以取解码器的输出，取与真实图像的差，取平均值。

假设 y 是解码器的输出，原始测试图像是 x，那么您可以对每个示例执行类似操作并对其取平均值：

tf.square(y-x)

这将是测试集的重建成本。

【讨论】：

为什么要平方差异？