用于二进制分类的 TensorFlow

TensorFlow for binary classification

我正在尝试使 this MNIST example 适应二进制分类。

但是当我将 NLABELSNLABELS=2 更改为 NLABELS=1 时,损失函数总是 returns 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)

我检查了 batch_ys(输入 y)和 _y 的维度,当 NLABELS=1 时它们都是 1xN 矩阵,所以问题似乎是在此之前。也许与矩阵乘法有关?

我实际上在一个真实的项目中遇到了同样的问题,所以任何帮助将不胜感激...谢谢!

原始 MNIST 示例使用 one-hot encoding to represent the labels in the data: this means that if there are NLABELS = 10 classes (as in MNIST), the target output is [1 0 0 0 0 0 0 0 0 0] for class 0, [0 1 0 0 0 0 0 0 0 0] for class 1, etc. The 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 classes 的概率分布,然后将其与 fed- y_.

的价值

如果 NLABELS = 1,这就好像只有一个 class,并且 tf.nn.softmax() 操作会计算 1.0 的概率 [=47] =],导致 0.0 的交叉熵,因为对于所有示例,tf.log(1.0) 都是 0.0

有(至少)两种方法可以尝试二进制 classification:

  1. 最简单的方法是为两个可能的 class 设置 NLABELS = 2,并将训练数据编码为 [1 0] 标签 0 和 [0 1] 对于标签 1。 有关于如何做到这一点的建议。

  2. 您可以将标签保留为整数 01 并使用 tf.nn.sparse_softmax_cross_entropy_with_logits(), as suggested in .

我一直在寻找如何以类似于在 Keras 中完成的方式在 TensorFlow 中实现二元分类的好例子。我没有找到任何东西,但是在深入研究了代码之后,我想我已经弄明白了。我修改了这里的问题以实现一个使用 sigmoid_cross_entropy_with_logits 的解决方案,就像 Keras 在幕后所做的那样。

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)

培训:

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