Question

我试图基于此work using the STS dataset实现句子相似性架构。标签是从0到1的归一化相似度分数，因此它被假定为回归模型。

我的问题是从第一个纪元开始直接损失到NaN。我做错了什么？

我已经尝试更新到最新的keras和theano版本。

我的模型的代码是：

def create_lstm_nn(input_dim):
    seq = Sequential()`
    # embedd using pretrained 300d embedding
    seq.add(Embedding(vocab_size, emb_dim, mask_zero=True, weights=[embedding_weights]))
    # encode via LSTM
    seq.add(LSTM(128))
    seq.add(Dropout(0.3))
    return seq

lstm_nn = create_lstm_nn(input_dim)

input_a = Input(shape=(input_dim,))
input_b = Input(shape=(input_dim,))

processed_a = lstm_nn(input_a)
processed_b = lstm_nn(input_b)

cos_distance = merge([processed_a, processed_b], mode='cos', dot_axes=1)
cos_distance = Reshape((1,))(cos_distance)
distance = Lambda(lambda x: 1-x)(cos_distance)

model = Model(input=[input_a, input_b], output=distance)

# train
rms = RMSprop()
model.compile(loss='mse', optimizer=rms)
model.fit([X1, X2], y, validation_split=0.3, batch_size=128, nb_epoch=20)

我也尝试使用简单的Lambda代替Merge图层，但结果相同。

def cosine_distance(vests):
    x, y = vests
    x = K.l2_normalize(x, axis=-1)
    y = K.l2_normalize(y, axis=-1)
    return -K.mean(x * y, axis=-1, keepdims=True)

def cos_dist_output_shape(shapes):
    shape1, shape2 = shapes
    return (shape1[0],1)

distance = Lambda(cosine_distance, output_shape=cos_dist_output_shape)([processed_a, processed_b])

Answer 1

我没有遇到nan问题，但我的损失不会改变。我发现了这个信息 check this out

def cosine_distance(shapes):
    y_true, y_pred = shapes
    def l2_normalize(x, axis):
        norm = K.sqrt(K.sum(K.square(x), axis=axis, keepdims=True))
        return K.sign(x) * K.maximum(K.abs(x), K.epsilon()) /     K.maximum(norm, K.epsilon())
    y_true = l2_normalize(y_true, axis=-1)
    y_pred = l2_normalize(y_pred, axis=-1)
    return K.mean(1 - K.sum((y_true * y_pred), axis=-1))

使用keras的句子相似度

1 个答案: