用于二元分类的张量流

Question

我正在尝试将这个MNIST示例改编为二元分类。

但是当我NLABELS从 NLABELS=2 更改为 NLABELS=1 时，损失函数总是返回 0(精度为 1(。

from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
from tensorflow.examples.tutorials.mnist import input_data
import tensorflow as tf
# Import data
mnist = input_data.read_data_sets('data', one_hot=True)
NLABELS = 2
sess = tf.InteractiveSession()
# Create the model
x = tf.placeholder(tf.float32, [None, 784], name='x-input')
W = tf.Variable(tf.zeros([784, NLABELS]), name='weights')
b = tf.Variable(tf.zeros([NLABELS], name='bias'))
y = tf.nn.softmax(tf.matmul(x, W) + b)
# Add summary ops to collect data
_ = tf.histogram_summary('weights', W)
_ = tf.histogram_summary('biases', b)
_ = tf.histogram_summary('y', y)
# Define loss and optimizer
y_ = tf.placeholder(tf.float32, [None, NLABELS], name='y-input')
# More name scopes will clean up the graph representation
with tf.name_scope('cross_entropy'):
    cross_entropy = -tf.reduce_mean(y_ * tf.log(y))
    _ = tf.scalar_summary('cross entropy', cross_entropy)
with tf.name_scope('train'):
    train_step = tf.train.GradientDescentOptimizer(10.).minimize(cross_entropy)
with tf.name_scope('test'):
    correct_prediction = tf.equal(tf.argmax(y, 1), tf.argmax(y_, 1))
    accuracy = tf.reduce_mean(tf.cast(correct_prediction, tf.float32))
    _ = tf.scalar_summary('accuracy', accuracy)
# Merge all the summaries and write them out to /tmp/mnist_logs
merged = tf.merge_all_summaries()
writer = tf.train.SummaryWriter('logs', sess.graph_def)
tf.initialize_all_variables().run()
# Train the model, and feed in test data and record summaries every 10 steps
for i in range(1000):
    if i % 10 == 0:  # Record summary data and the accuracy
        labels = mnist.test.labels[:, 0:NLABELS]
        feed = {x: mnist.test.images, y_: labels}
        result = sess.run([merged, accuracy, cross_entropy], feed_dict=feed)
        summary_str = result[0]
        acc = result[1]
        loss = result[2]
        writer.add_summary(summary_str, i)
        print('Accuracy at step %s: %s - loss: %f' % (i, acc, loss)) 
   else:
        batch_xs, batch_ys = mnist.train.next_batch(100)
        batch_ys = batch_ys[:, 0:NLABELS]
        feed = {x: batch_xs, y_: batch_ys}
    sess.run(train_step, feed_dict=feed)

我已经检查了batch_ys(输入y(和_y的维度，它们都是 1xN NLABELS=1矩阵，所以问题似乎在此之前。也许与矩阵乘法有关？

我实际上在实际项目中遇到了同样的问题，因此任何帮助将不胜感激......谢谢！

Answer 1

原始的 MNIST 示例使用独热编码来表示数据中的标签：这意味着如果有NLABELS = 10类(如 MNIST(，则类 0 的目标输出[1 0 0 0 0 0 0 0 0 0]，类 1 的目标输出[0 1 0 0 0 0 0 0 0 0]，依此类推。tf.nn.softmax()运算符将tf.matmul(x, W) + b计算的对数转换为跨不同输出类的概率分布，然后将其与y_的输入值进行比较。

如果NLABELS = 1，这就像只有一个类，并且tf.nn.softmax()运算将计算该类的1.0概率，导致0.0的交叉熵，因为所有示例都0.0 tf.log(1.0)。

对于二元分类，您可以尝试(至少(两种方法：

1. 最简单的方法是为两个可能的类设置 NLABELS = 2，并将训练数据编码为标签 0 的[1 0]和标签 1 的[0 1]。这个答案对如何做到这一点有一个建议。

2. 您可以将标签保留为整数0和1并使用tf.nn.sparse_softmax_cross_entropy_with_logits()，如本答案所示。

Answer 2

我一直在寻找如何在TensorFlow中以类似于在Keras中实现二元分类的方式实现的好例子。 我没有找到任何代码，但是在挖掘了一下代码之后，我想我已经弄清楚了。 我在这里修改了问题，以实现一个解决方案，该解决方案使用sigmoid_cross_entropy_with_logits Keras 在后台的方式。

from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
from tensorflow.examples.tutorials.mnist import input_data
import tensorflow as tf
# Import data
mnist = input_data.read_data_sets('data', one_hot=True)
NLABELS = 1
sess = tf.InteractiveSession()
# Create the model
x = tf.placeholder(tf.float32, [None, 784], name='x-input')
W = tf.get_variable('weights', [784, NLABELS],
                    initializer=tf.truncated_normal_initializer()) * 0.1
b = tf.Variable(tf.zeros([NLABELS], name='bias'))
logits = tf.matmul(x, W) + b
# Define loss and optimizer
y_ = tf.placeholder(tf.float32, [None, NLABELS], name='y-input')
# More name scopes will clean up the graph representation
with tf.name_scope('cross_entropy'):
    #manual calculation : under the hood math, don't use this it will have gradient problems
    # entropy = tf.multiply(tf.log(tf.sigmoid(logits)), y_) + tf.multiply((1 - y_), tf.log(1 - tf.sigmoid(logits)))
    # loss = -tf.reduce_mean(entropy, name='loss')
    entropy = tf.nn.sigmoid_cross_entropy_with_logits(labels=y_, logits=logits)
    loss = tf.reduce_mean(entropy, name='loss')
with tf.name_scope('train'):
    # Using Adam instead
    # train_step = tf.train.GradientDescentOptimizer(learning_rate=0.001).minimize(loss)
    train_step = tf.train.AdamOptimizer(learning_rate=0.002).minimize(loss)
with tf.name_scope('test'):
    preds = tf.cast((logits > 0.5), tf.float32)
    correct_prediction = tf.equal(preds, y_)
    accuracy = tf.reduce_mean(tf.cast(correct_prediction, tf.float32))
tf.initialize_all_variables().run()
# Train the model, and feed in test data and record summaries every 10 steps
for i in range(2000):
    if i % 100 == 0:  # Record summary data and the accuracy
        labels = mnist.test.labels[:, 0:NLABELS]
        feed = {x: mnist.test.images, y_: labels}
        result = sess.run([loss, accuracy], feed_dict=feed)
        print('Accuracy at step %s: %s - loss: %f' % (i, result[1], result[0]))
    else:
        batch_xs, batch_ys = mnist.train.next_batch(100)
        batch_ys = batch_ys[:, 0:NLABELS]
        feed = {x: batch_xs, y_: batch_ys}
    sess.run(train_step, feed_dict=feed)

训练：

Accuracy at step 0: 0.7373 - loss: 0.758670
Accuracy at step 100: 0.9017 - loss: 0.423321
Accuracy at step 200: 0.9031 - loss: 0.322541
Accuracy at step 300: 0.9085 - loss: 0.255705
Accuracy at step 400: 0.9188 - loss: 0.209892
Accuracy at step 500: 0.9308 - loss: 0.178372
Accuracy at step 600: 0.9453 - loss: 0.155927
Accuracy at step 700: 0.9507 - loss: 0.139031
Accuracy at step 800: 0.9556 - loss: 0.125855
Accuracy at step 900: 0.9607 - loss: 0.115340
Accuracy at step 1000: 0.9633 - loss: 0.106709
Accuracy at step 1100: 0.9667 - loss: 0.099286
Accuracy at step 1200: 0.971 - loss: 0.093048
Accuracy at step 1300: 0.9714 - loss: 0.087915
Accuracy at step 1400: 0.9745 - loss: 0.083300
Accuracy at step 1500: 0.9745 - loss: 0.079019
Accuracy at step 1600: 0.9761 - loss: 0.075164
Accuracy at step 1700: 0.9768 - loss: 0.071803
Accuracy at step 1800: 0.9777 - loss: 0.068825
Accuracy at step 1900: 0.9788 - loss: 0.066270