machine-learning - TensorFlow:多次评估测试集但得到不同的精度

Question

我使用CNN训练了MNIST模型，但是当我在训练后用测试数据检查模型的准确性时，我发现我的准确性会提高。这是代码。

BATCH_SIZE = 50
LR = 0.001              # learning rate
mnist = input_data.read_data_sets('./mnist', one_hot=True)  # they has been normalized to range (0,1)
test_x = mnist.test.images[:2000]
test_y = mnist.test.labels[:2000]

def new_cnn(imageinput, inputshape):
    weights = tf.Variable(tf.truncated_normal(inputshape, stddev = 0.1),name = 'weights')
    biases = tf.Variable(tf.constant(0.05, shape = [inputshape[3]]),name = 'biases')
    layer = tf.nn.conv2d(imageinput, weights, strides = [1, 1, 1, 1], padding = 'SAME')
    layer = tf.nn.relu(layer)
    return weights, layer

tf_x = tf.placeholder(tf.float32, [None, 28 * 28])
image = tf.reshape(tf_x, [-1, 28, 28, 1])              # (batch, height, width, channel)
tf_y = tf.placeholder(tf.int32, [None, 10])            # input y

# CNN
weights1, layer1 = new_cnn(image, [5, 5, 1, 32])
pool1 = tf.layers.max_pooling2d(
    layer1,
    pool_size=2,
    strides=2,
)           # -> (14, 14, 32)
weight2, layer2 = new_cnn(pool1, [5, 5, 32, 64])    # -> (14, 14, 64)
pool2 = tf.layers.max_pooling2d(layer2, 2, 2)    # -> (7, 7, 64)
flat = tf.reshape(pool2, [-1, 7 * 7 * 64])          # -> (7*7*64, )
hide = tf.layers.dense(flat, 1024, name = 'hide')              # hidden layer
output = tf.layers.dense(hide, 10, name = 'output')
loss = tf.losses.softmax_cross_entropy(onehot_labels=tf_y, logits=output)           # compute cost
accuracy = tf.metrics.accuracy( labels=tf.argmax(tf_y, axis=1), predictions=tf.argmax(output, axis=1),)[1]
train_op = tf.train.AdamOptimizer(LR).minimize(loss)



sess = tf.Session()
init_op = tf.group(tf.global_variables_initializer(), tf.local_variables_initializer()) # the local var is for accuracy
sess.run(init_op)     # initialize var in graph
saver = tf.train.Saver()
for step in range(101):
    b_x, b_y = mnist.train.next_batch(BATCH_SIZE)
    _, loss_ = sess.run([train_op, loss], {tf_x: b_x, tf_y: b_y})
    if step % 50 == 0:
        print(loss_)
        accuracy_, loss2 = sess.run([accuracy, loss], {tf_x: test_x, tf_y: test_y })
        print('Step:', step, '| test accuracy: %f' % accuracy_)

为了简化问题，我只使用 100 次训练迭代。测试集的最终准确率约为0.655000。

但是当我运行以下代码时:

for i in range(5):
  accuracy2 = sess.run(accuracy, {tf_x: test_x, tf_y: test_y })
  print(sess.run(weight2[1,:,0,0])) # To show that the model parameters won't update 
  print(accuracy2)

输出为

[-0.06928255 -0.13498515  0.01266837  0.05656774  0.09438231]
0.725875
[-0.06928255 -0.13498515  0.01266837  0.05656774  0.09438231]
0.7684
[-0.06928255 -0.13498515  0.01266837  0.05656774  0.09438231]
0.79675
[-0.06928255 -0.13498515  0.01266837  0.05656774  0.09438231]
0.817
[-0.06928255 -0.13498515  0.01266837  0.05656774  0.09438231]
0.832187

这让我很困惑，有人可以告诉我出了什么问题吗？感谢您的耐心等待!

Answer 1

tf.metrics.accuracy并不像你想象的那么微不足道。看一下它的文档:

The accuracy function creates two local variables, total and
count that are used to compute the frequency with which predictions matches labels. This frequency is ultimately returned as accuracy: an idempotent operation that simply divides total by count.

Internally, an is_correct operation computes a Tensor with elements 1.0 where the corresponding elements of predictions and labels match and 0.0 otherwise. Then update_op increments total with the reduced sum of the product of weights and is_correct, and it increments count with the reduced sum of weights.

For estimation of the metric over a stream of data, the function creates an update_op operation that updates these variables and returns the accuracy.

...

Returns:

• accuracy: A Tensor representing the accuracy, the value of total divided by count.
• update_op: An operation that increments the total and count variables appropriately and whose value matches accuracy.

请注意，它返回一个元组，并且您获取第二项，即update_op。连续调用 update_op 被视为数据流，这不是您想要做的(因为训练期间的每个评估都会影响 future 的评估)。事实上，这个运行指标是 pretty counter-intuitive .

您的解决方案是使用简单的精度计算。将此行更改为:

accuracy = tf.reduce_mean(tf.cast(tf.equal(tf.argmax(tf_y, axis=1), tf.argmax(output, axis=1)), tf.float32))

您将获得稳定的准确度计算。