我正在尝试在Keras中训练自动编码器以进行信号处理，但是我却以某种方式失败了。
我的输入是128帧长度的段，用于6个小节（acceleration_x / y / z，gyro_x / y / z），因此数据集的总体形状为(22836, 128, 6)，其中22836是样本大小。
这是我用于自动编码器的示例代码：
X_train, X_test, Y_train, Y_test = load_dataset()
# reshape the input, whose size is (22836, 128, 6)
X_train = X_train.reshape(X_train.shape[0], np.prod(X_train.shape[1:]))
X_test = X_test.reshape(X_test.shape[0], np.prod(X_test.shape[1:]))
# now the shape will be (22836, 768)
### MODEL ###
input_shape = [X_train.shape[1]]
X_input = Input(input_shape)
x = Dense(1000, activation='sigmoid', name='enc0')(X_input)
encoded = Dense(350, activation='sigmoid', name='enc1')(x)
x = Dense(1000, activation='sigmoid', name='dec0')(encoded)
decoded = Dense(input_shape[0], activation='sigmoid', name='dec1')(x)
model = Model(inputs=X_input, outputs=decoded, name='autoencoder')
model.compile(optimizer='rmsprop', loss='mean_squared_error')
print(model.summary())
model.summary()的输出是Model summary
_________________________________________________________________
Layer (type) Output Shape Param #
=================================================================
input_55 (InputLayer) (None, 768) 0
_________________________________________________________________
enc0 (Dense) (None, 1000) 769000
_________________________________________________________________
enc1 (Dense) (None, 350) 350350
_________________________________________________________________
dec1 (Dense) (None, 1000) 351000
_________________________________________________________________
dec0 (Dense) (None, 768) 768768
=================================================================
Total params: 2,239,118
Trainable params: 2,239,118
Non-trainable params: 0
培训通过
# train the model
history = model.fit(x = X_train, y = X_train,
epochs=5,
batch_size=32,
validation_data=(X_test, X_test))
在这里，我只是想学习生成的身份函数：
Train on 22836 samples, validate on 5709 samples
Epoch 1/5
22836/22836 [==============================] - 27s 1ms/step - loss: 0.9481 - val_loss: 0.8862
Epoch 2/5
22836/22836 [==============================] - 24s 1ms/step - loss: 0.8669 - val_loss: 0.8358
Epoch 3/5
22836/22836 [==============================] - 25s 1ms/step - loss: 0.8337 - val_loss: 0.8146
Epoch 4/5
22836/22836 [==============================] - 25s 1ms/step - loss: 0.8164 - val_loss: 0.7960
Epoch 5/5
22836/22836 [==============================] - 25s 1ms/step - loss: 0.8004 - val_loss: 0.7819
在这一点上，为了理解其性能，我检查了一些真实输入与预测输入的关系图：
prediction = model.predict(X_test)
for i in np.random.randint(0, 100, 7):
pred = prediction[i, :].reshape(128,6)
# getting only values for acceleration_x
pred = pred[:, 0]
true = X_test[i, :].reshape(128,6)
# getting only values for acceleration_x
true = true[:, 0]
# plot original and reconstructed
fig, (ax1, ax2) = plt.subplots(2, 1, figsize=(20, 6))
ax1.plot(true, color='green')
ax2.plot(pred, color='red')
这些是一些似乎完全错误的图：
除了少数时期（实际上似乎没有什么区别）之外，您对什么地方有什么建议吗？
最佳答案
您的数据不在[0,1]范围内，为什么在上一层使用sigmoid作为激活函数？从最后一层中删除激活功能（最好在先前的层中使用relu）。
还规范化训练数据。您可以使用基于功能的规范化：
X_mean = X_train.mean(axis=0)
X_train -= X_mean
X_std = X_train.std(axis=0)
X_train /= X_std + 1e-8
并且不要忘记在推理时间（即测试）中使用计算出的统计信息（
X_mean和X_std）来规范化测试数据。关于python - 无法在Keras中训练简单的自动编码器，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/51840826/