6

我一直在考虑单词序列的 0-padding 以及如何将 0-padding 转换为 Embedding 层。乍一看,人们会认为您也想保持嵌入 = 0.0。但是,Embeddinglayer in 会keras为任何输入标记生成随机值,并且没有办法强制它生成 0.0。注意,mask_zero做一些不同的事情,我已经检查过了。

有人可能会问,为什么要担心这一点,即使嵌入不是 0.0,只要它们相同,代码似乎也能正常工作。所以我想出了一个例子,虽然有点做作,将嵌入设置为 0.0 的填充标记为 0 会有所不同。

我使用了 20 个新闻组数据集from sklearn.datasets import fetch_20newsgroups。我做了一些最小的预处理:删除标点符号、停用词和数字。我from keras.preprocessing.sequence import pad_sequences用于 0 填充。我将大约 18K 的帖子分成训练和验证集,训练/验证的比例 = 4/1。我创建了一个简单的 1 密集隐藏层网络,输入是嵌入的扁平序列:

    EMBEDDING_DIM = 300
    MAX_SEQUENCE_LENGTH = 1100
    layer_size = 25
    dropout = 0.3
    sequence_input = Input(shape=(MAX_SEQUENCE_LENGTH,), dtype='int32', name='dnn_input')
    embedding_layer = Embedding(len(word_index) + 1, EMBEDDING_DIM, input_length=MAX_SEQUENCE_LENGTH, name = 'embedding_dnn')
    embedded_sequences = embedding_layer(sequence_input)
    x = Flatten(name='flatten_dnn')(embedded_sequences)
    x = Dense(layer_size, activation='relu', name ='hidden_dense_dnn')(x)
    x = Dropout(dropout, name='dropout')(x)
    preds = Dense(num_labels, activation='softmax', name = 'output_dnn')(x)

    model = Model(sequence_input, preds)
    model.compile(loss='categorical_crossentropy',optimizer='adam',metrics=['accuracy'])

该模型有大约 14M 的可训练参数(这个例子有点做作,正如我已经提到的)。当我训练它

    earlystop = EarlyStopping(monitor='val_loss', patience=5)
    history = model.fit(x_train, y_train, validation_data=(x_test, y_test), epochs=30, batch_size=BATCH_SIZE, callbacks=[earlystop])

看起来算法在 4 个时期都在努力寻找摆脱“随机性”的方法:

Train on 15048 samples, validate on 3798 samples
Epoch 1/30
15048/15048 [==============================] - 58s 4ms/step - loss: 3.1118 - acc: 0.0519 - val_loss: 2.9894 - val_acc: 0.0534
Epoch 2/30
15048/15048 [==============================] - 56s 4ms/step - loss: 2.9820 - acc: 0.0556 - val_loss: 2.9827 - val_acc: 0.0527
Epoch 3/30
15048/15048 [==============================] - 55s 4ms/step - loss: 2.9712 - acc: 0.0626 - val_loss: 2.9718 - val_acc: 0.0579
Epoch 4/30
15048/15048 [==============================] - 55s 4ms/step - loss: 2.9259 - acc: 0.0756 - val_loss: 2.8363 - val_acc: 0.0874
Epoch 5/30
15048/15048 [==============================] - 56s 4ms/step - loss: 2.7092 - acc: 0.1390 - val_loss: 2.3251 - val_acc: 0.2796
...
Epoch 13/30
15048/15048 [==============================] - 56s 4ms/step - loss: 0.0698 - acc: 0.9807 - val_loss: 0.5010 - val_acc: 0.8736

最终精度约为 0.87

print ('Best validation accuracy is ', max(history.history['val_acc']))
Best validation accuracy is  0.874934175379845

但是,当我将填充的 0 的嵌入显式设置为 0.0

def myMask(x):
    mask= K.greater(x,0) #will return boolean values
    mask= K.cast(mask, dtype=K.floatx()) 
    return mask
layer_size = 25
dropout = 0.3
sequence_input = Input(shape=(MAX_SEQUENCE_LENGTH,), dtype='int32', name='dnn_input')
embedding_layer = Embedding(len(word_index) + 1, EMBEDDING_DIM, input_length=MAX_SEQUENCE_LENGTH, name = 'embedding_dnn')
embedded_sequences = embedding_layer(sequence_input)
y = Lambda(myMask, output_shape=(MAX_SEQUENCE_LENGTH,))(sequence_input)
y = Reshape(target_shape=(MAX_SEQUENCE_LENGTH,1))(y)
merge_layer = Multiply(name = 'masked_embedding_dnn')([embedded_sequences,y])
x = Flatten(name='flatten_dnn')(merge_layer)
x = Dense(layer_size, activation='relu', name ='hidden_dense_dnn')(x)
x = Dropout(dropout, name='dropout')(x)
preds = Dense(num_labels, activation='softmax', name = 'output_dnn')(x)

model = Model(sequence_input, preds)
model.compile(loss='categorical_crossentropy',optimizer='adam',metrics=['accuracy'])

具有相同数量参数的模型会立即摆脱“随机性”:

Train on 15048 samples, validate on 3798 samples
Epoch 1/30
15048/15048 [==============================] - 64s 4ms/step - loss: 2.4356 - acc: 0.3060 - val_loss: 1.2424 - val_acc: 0.7754
Epoch 2/30
15048/15048 [==============================] - 61s 4ms/step - loss: 0.6973 - acc: 0.8267 - val_loss: 0.5240 - val_acc: 0.8797
...
Epoch 10/30
15048/15048 [==============================] - 61s 4ms/step - loss: 0.0496 - acc: 0.9881 - val_loss: 0.4176 - val_acc: 0.8944

最终获得更好的精度,约为 0.9。

同样,这是一个有些人为的例子,但它仍然表明将那些“填充”嵌入保持在 0.0 可能是有益的。

我在这里错过了什么吗?如果我没有遗漏任何东西,那么 Keras 不提供开箱即用功能的原因是什么?

更新

@DanielMöller 我试过你的建议:

layer_size = 25
dropout = 0.3
init = RandomUniform(minval=0.0001, maxval=0.05, seed=None)
constr = NonNeg()



sequence_input = Input(shape=(MAX_SEQUENCE_LENGTH,), dtype='int32', name='dnn_input')
embedding_layer = Embedding(len(word_index) + 1, 
                            EMBEDDING_DIM, 
                            input_length=MAX_SEQUENCE_LENGTH, 
                            name = 'embedding_dnn', 
                            embeddings_initializer=init,
                            embeddings_constraint=constr)

embedded_sequences = embedding_layer(sequence_input)
y = Lambda(myMask, output_shape=(MAX_SEQUENCE_LENGTH,))(sequence_input)
y = Reshape(target_shape=(MAX_SEQUENCE_LENGTH,1))(y)
merge_layer = Multiply(name = 'masked_embedding_dnn')([embedded_sequences,y])
x = Flatten(name='flatten_dnn')(merge_layer)
x = Dense(layer_size, activation='relu', name ='hidden_dense_dnn')(x)
x = Dropout(dropout, name='dropout')(x)
preds = Dense(num_labels, activation='softmax', name = 'output_dnn')(x)

model = Model(sequence_input, preds)
model.compile(loss='categorical_crossentropy',optimizer='adam',metrics=['accuracy'])

不幸的是,网络陷入了“随机性”:

Train on 15197 samples, validate on 3649 samples
Epoch 1/30
15197/15197 [==============================] - 60s 4ms/step - loss: 3.1354 - acc: 0.0505 - val_loss: 2.9943 - val_acc: 0.0496
....
Epoch 24/30
15197/15197 [==============================] - 60s 4ms/step - loss: 2.9905 - acc: 0.0538 - val_loss: 2.9907 - val_acc: 0.0496

我也尝试过没有NonNeg()约束,结果相同。

4

1 回答 1

4

好吧,您正在消除与填充步骤相关的权重梯度的计算。

如果你有太多的填充步骤,那么关于填充值的嵌入权重将参与大量计算,并将与其他权重显着竞争。但是训练这些权重是浪费计算,换句话说肯定会干扰。

还要考虑一下,例如,一些填充权重的值可能介于有意义单词的值之间。因此,增加权重可能会使它与另一个词相似,而实际上并非如此。而且也在减少……

这些额外的计算、对损失和梯度计算的额外贡献等将产生更多的计算需求和更多的障碍。这就像在数据中间有很多垃圾。

另请注意,这些零点直接进入密集层,这也将消除许多密集权重的梯度。这可能会过度拟合较长的序列,尽管与较短的序列相比它们很少。


出于好奇,如果你这样做会发生什么?

from keras.initializers import RandomUniform
from keras.constraints import NonNeg

init = RandomUniform(minval=0.0001, maxval=0.05, seed=None)
constr = NonNeg()


......
embedding_layer = Embedding(len(word_index) + 1, 
                            EMBEDDING_DIM, 
                            input_length=MAX_SEQUENCE_LENGTH, 
                            name = 'embedding_dnn', 
                            embeddings_initializer=init,
                            embeddings_constraint=constr)
..........
于 2019-06-27T21:26:45.653 回答