我一直在尝试使用LSTM自动编码器获得矢量序列的矢量表示，以便我可以使用SVM或其他此类监督算法对序列进行分类。大量的数据使我无法使用完全连接的密集层进行分类。
我输入的最短大小为7个时间步长，最长序列为356个时间步长。因此，我用零填充了较短的序列，以获得形状为最终的x_train（1326、356、8），其中1326是训练样本的数量，而8是一个时间步长的维度。我正在尝试使用给定的LSTM自动编码器将这些序列编码为单个向量。
model.add(Masking(mask_value=0.0, input_shape=(max_len, 8)))
model.add(LSTM(100, activation='relu'))
model.add(RepeatVector(max_len))
model.add(LSTM(8, activation='relu', return_sequences=True))
model.compile(optimizer='adam', loss='mse')
model.fit(x_train, x_train, batch_size=32, callbacks=[chk], epochs=1000, validation_split=0.05, shuffle=True)
我试图掩盖零填充的结果，但是RepeatVector（）层可能会阻碍该过程。因此，一段时间后，均方误差损失变为nan。任何人都可以帮助我解决如何仅将相关时间步长包括在内来计算损失函数而忽略其他时间步长吗？
答案 0 :(得分：1)
Keras中的每个层都有一个input_mask和output_mask，在示例中，在第一个LSTM层之后（当return_sequence = False时），该掩码已经丢失。让我在下面的示例中对此进行解释，并展示两种在LSTM自动编码器中实现屏蔽的解决方案。
time_steps = 3
n_features = 2
input_layer = tfkl.Input(shape=(time_steps, n_features))
# I want to mask the timestep where all the feature values are 1 (usually we pad by 0)
x = tfk.layers.Masking(mask_value=1)(input_layer)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = tfkl.LSTM(2, return_sequences=False)(x)
x = tfkl.RepeatVector(time_steps)(x)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = tfk.layers.Dense(n_features)(x)
lstm_ae = tfk.models.Model(inputs=input_layer, outputs=x)
lstm_ae.compile(optimizer='adam', loss='mse')
print(lstm_ae.summary())
Model: "model_2"
_________________________________________________________________
Layer (type) Output Shape Param #
=================================================================
input_3 (InputLayer) [(None, 3, 2)] 0
_________________________________________________________________
masking_2 (Masking) (None, 3, 2) 0
_________________________________________________________________
lstm_8 (LSTM) (None, 3, 2) 40
_________________________________________________________________
lstm_9 (LSTM) (None, 2) 40
_________________________________________________________________
repeat_vector_2 (RepeatVecto (None, 3, 2) 0
_________________________________________________________________
lstm_10 (LSTM) (None, 3, 2) 40
_________________________________________________________________
lstm_11 (LSTM) (None, 3, 2) 40
_________________________________________________________________
dense_2 (Dense) (None, 3, 2) 6
=================================================================
Total params: 166
Trainable params: 166
Non-trainable params: 0
_________________________________________________________________
for i, l in enumerate(lstm_ae.layers):
print(f'layer {i}: {l}')
print(f'has input mask: {l.input_mask}')
print(f'has output mask: {l.output_mask}')
layer 0: <tensorflow.python.keras.engine.input_layer.InputLayer object at 0x645b49cf8>
has input mask: None
has output mask: None
layer 1: <tensorflow.python.keras.layers.core.Masking object at 0x645b49c88>
has input mask: None
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 2: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x645b4d0b8>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 3: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x645b4dba8>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: None
layer 4: <tensorflow.python.keras.layers.core.RepeatVector object at 0x645db0390>
has input mask: None
has output mask: None
layer 5: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x6470b5da0>
has input mask: None
has output mask: None
layer 6: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x6471410f0>
has input mask: None
has output mask: None
layer 7: <tensorflow.python.keras.layers.core.Dense object at 0x647dfdf60>
has input mask: None
has output mask: None
如您在上面看到的，第二个LSTM层（return_sequence = False）返回None，这是有道理的，因为时间步丢失了（形状改变了）并且该层不知道如何通过遮罩，您可以还要检查源代码，如果return_sequence = True，则将看到它返回input_mask，否则返回None。当然，另一个问题是RepeatVector层，该层根本不支持显式遮罩，这又是因为形状已更改。除了这个瓶颈部分（第二个LSTM + RepeatVector）之外，模型的其他部分都可以通过掩码，因此我们只需要处理瓶颈部分。
有2种可能的解决方案，我还将基于计算损失进行验证。
# last timestep should be masked because all feature values are 1
x = np.array([1, 2, 1, 2, 1, 1], dtype='float32').reshape(1, 3, 2)
print(x)
array([[[1., 2.],
[1., 2.],
[1., 1.]]], dtype=float32)
y = lstm_ae.predict(x)
print(y)
array([[[0.00020542, 0.00011909],
[0.0007361 , 0.00047323],
[0.00158514, 0.00107504]]], dtype=float32)
# the expected loss should be the sum of square error between the first 2 timesteps
# (2 features each timestep) divided by 6. you might expect that this should be
# divided by 4, but in the source code this is actually divided by 6, which doesn't
# matter a lot because only the gradient of loss matter, but not the loss itself.
expected_loss = np.square(x[:, :2, :] - y[:, :2, :]).sum()/6
print(expected_loss)
1.665958086649577
actual_loss_with_masking = lstm_ae.evaluate(x=x, y=x)
print(actual_loss_with_masking)
1.9984053373336792
# the actual loss still includes the last timestep, which means the masking is not # effectively passed to the output layer for calculating the loss
print(np.square(x-y).sum()/6)
1.9984052975972493
# if we provide the sample_weight 0 for each timestep that we want to mask, the
# loss will be ignored correctly
lstm_ae.compile(optimizer='adam', loss='mse', sample_weight_mode='temporal')
sample_weight_array = np.array([1, 1, 0]).reshape(1, 3) # it means to ignore the last timestep
actual_loss_with_sample_weight = lstm_ae.evaluate(x=x, y=x, sample_weight=sample_weight_array)
# the actual loss now is correct
print(actual_loss_with_sample_weight)
1.665958046913147
class lstm_bottleneck(tf.keras.layers.Layer):
def __init__(self, lstm_units, time_steps, **kwargs):
self.lstm_units = lstm_units
self.time_steps = time_steps
self.lstm_layer = tfkl.LSTM(lstm_units, return_sequences=False)
self.repeat_layer = tfkl.RepeatVector(time_steps)
super(lstm_bottleneck, self).__init__(**kwargs)
def call(self, inputs):
# just call the two initialized layers
return self.repeat_layer(self.lstm_layer(inputs))
def compute_mask(self, inputs, mask=None):
# return the input_mask directly
return mask
time_steps = 3
n_features = 2
input_layer = tfkl.Input(shape=(time_steps, n_features))
# I want to mask the timestep where all the feature values are 1 (usually we pad by 0)
x = tfk.layers.Masking(mask_value=1)(input_layer)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = lstm_bottleneck(lstm_units=2, time_steps=3)(x)
# x = tfkl.LSTM(2, return_sequences=False)(x)
# x = tfkl.RepeatVector(time_steps)(x)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = tfkl.LSTM(2, return_sequences=True)(x)
x = tfk.layers.Dense(n_features)(x)
lstm_ae = tfk.models.Model(inputs=input_layer, outputs=x)
lstm_ae.compile(optimizer='adam', loss='mse')
print(lstm_ae.summary())
Model: "model_2"
_________________________________________________________________
Layer (type) Output Shape Param #
=================================================================
input_3 (InputLayer) [(None, 3, 2)] 0
_________________________________________________________________
masking_2 (Masking) (None, 3, 2) 0
_________________________________________________________________
lstm_10 (LSTM) (None, 3, 2) 40
_________________________________________________________________
lstm_bottleneck_3 (lstm_bott (None, 3, 2) 40
_________________________________________________________________
lstm_12 (LSTM) (None, 3, 2) 40
_________________________________________________________________
lstm_13 (LSTM) (None, 3, 2) 40
_________________________________________________________________
dense_2 (Dense) (None, 3, 2) 6
=================================================================
Total params: 166
Trainable params: 166
Non-trainable params: 0
_________________________________________________________________
for i, l in enumerate(lstm_ae.layers):
print(f'layer {i}: {l}')
print(f'has input mask: {l.input_mask}')
print(f'has output mask: {l.output_mask}')
layer 0: <tensorflow.python.keras.engine.input_layer.InputLayer object at 0x64dbf98d0>
has input mask: None
has output mask: None
layer 1: <tensorflow.python.keras.layers.core.Masking object at 0x64dbf9f60>
has input mask: None
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 2: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x64dbf9550>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 3: <__main__.lstm_bottleneck object at 0x64dbf91d0>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 4: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x64e04ca20>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 5: <tensorflow.python.keras.layers.recurrent_v2.LSTM object at 0x64eeb8b00>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
layer 6: <tensorflow.python.keras.layers.core.Dense object at 0x64ef43208>
has input mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
has output mask: Tensor("masking_2/Identity_1:0", shape=(None, 3), dtype=bool)
我们已经看到，掩码现在已成功传递到输出层。我们还将验证损失是否不包括掩盖的时间步长。
# last timestep should be masked because all feature values are 1
x = np.array([1, 2, 1, 2, 1, 1], dtype='float32').reshape(1, 3, 2)
print(x)
array([[[1., 2.],
[1., 2.],
[1., 1.]]], dtype=float32)
y = lstm_ae.predict(x)
print(y)
array([[[ 0.00065455, -0.00294413],
[ 0.00166675, -0.00742249],
[ 0.00166675, -0.00742249]]], dtype=float32)
# the expected loss should be the square error between the first 2 timesteps divided by 6
expected_loss = np.square(x[:, :2, :] - y[:, :2, :]).sum()/6
print(expected_loss)
1.672815163930257
# now the loss is correct with a custom layer
actual_loss_with_masking = lstm_ae.evaluate(x=x, y=x)
print(actual_loss_with_masking)
1.672815203666687