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我正在熟悉 Tensorflow-Probability，在这里我遇到了一个问题。在训练期间，模型返回 nan 作为损失（可能意味着导致溢出的巨大损失）。由于合成数据的函数形式并不太复杂，而且数据点与参数的比率乍一看并不可怕，至少我想知道问题是什么以及如何纠正它。

代码如下——附有一些可能有用的图片：

# Create and plot 5000 data points

x_train = np.linspace(-1, 2, 5000)[:, np.newaxis]
y_train = np.power(x_train, 3) + 0.1*(2+x_train)*np.random.randn(5000)[:, np.newaxis]

plt.scatter(x_train, y_train, alpha=0.1)
plt.show()

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# Define the prior weight distribution -- all N(0, 1) -- and not trainable

def prior(kernel_size, bias_size, dtype = None):
    
    n = kernel_size + bias_size
    
    prior_model = Sequential([
        
        tfpl.DistributionLambda(
        
            lambda t: tfd.MultivariateNormalDiag(loc = tf.zeros(n)  ,  scale_diag = tf.ones(n)
                                                
                                                ))
        
    ])
    
    return(prior_model)

# Define variational posterior weight distribution -- multivariate Gaussian

def posterior(kernel_size, bias_size, dtype = None):
    
    n = kernel_size + bias_size
    
    posterior_model = Sequential([
        
        tfpl.VariableLayer(tfpl.MultivariateNormalTriL.params_size(n)  , dtype = dtype),   # The parameters of the model are declared Variables that are trainable
        
        tfpl.MultivariateNormalTriL(n)  # The posterior function will return to the Variational layer that will call it a MultivariateNormalTril object that will have as many dimensions
                                        # as the parameters of the Variational Dense Layer.  That means that each parameter will be generated by a distinct Normal Gaussian shifted and scaled
                                        # by a mu and sigma learned from the data, independently of all the other weights.  The output of this Variablelayer will become the input to the
                                        # MultivariateNormalTriL object.
                                        # The shape of the VariableLayer object will be defined by the number of paramaters needed to create the MultivariateNormalTriL object given
                                        # that it will live in a Space of n dimensions (event_size = n).  This number is returned by the tfpl.MultivariateNormalTriL.params_size(n)
        
        
    ])
    
    return(posterior_model)

x_in = Input(shape = (1,))

x = tfpl.DenseVariational(units= 2**4,
                          make_prior_fn=prior,
                          make_posterior_fn=posterior,
                          kl_weight=1/x_train.shape[0],
                          activation='relu')(x_in)

x = tfpl.DenseVariational(units= 2**4,
                          make_prior_fn=prior,
                          make_posterior_fn=posterior,
                          kl_weight=1/x_train.shape[0],
                          activation='relu')(x)

x =    tfpl.DenseVariational(units=tfpl.IndependentNormal.params_size(1),
                          make_prior_fn=prior,
                          make_posterior_fn=posterior,
                          kl_weight=1/x_train.shape[0])(x)

y_out =  tfpl.IndependentNormal(1)(x)

model = Model(inputs = x_in, outputs = y_out)

def nll(y_true, y_pred):
    return -y_pred.log_prob(y_true)

model.compile(loss=nll, optimizer= 'Adam')
model.summary()

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训练模型

history = model.fit(x_train1, y_train1, epochs=500)

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2 回答 2

1

问题似乎出在损失函数中：没有任何指定位置和尺度的独立正态分布的负对数似然会导致未驯服的方差，从而导致最终损失值爆炸。由于您正在尝试变分层，因此您必须对认知不确定性的估计感兴趣，为此，我建议应用恒定方差。

我尝试在以下几行中对您的代码进行一些细微更改：

  1. 首先，最终的输出y_out直接来自最终的变分层，没有任何 IndpendnetNormal 分布层：

    y_out =    tfpl.DenseVariational(units=1,
                  make_prior_fn=prior,
                  make_posterior_fn=posterior,
                  kl_weight=1/x_train.shape[0])(x)

  2. 其次，损失函数现在包含您需要的正态分布的必要计算，但具有静态方差，以避免训练期间损失的爆炸：

     def nll(y_true, y_pred):
                   dist = tfp.distributions.Normal(loc=y_pred, scale=1.0)
                   return tf.reduce_sum(-dist.log_prob(y_true))

  3. 然后以与之前相同的方式编译和训练模型：

     model.compile(loss=nll, optimizer= 'Adam')
 history = model.fit(x_train, y_train, epochs=3000)

  4. 最后，让我们从训练模型中抽取 100 个不同的预测并绘制这些值以可视化模型的认知不确定性：

     predicted = [model(x_train) for _ in range(100)]
 for i, res in enumerate(predicted):
                   plt.plot(x_train, res , alpha=0.1)
 plt.scatter(x_train, y_train, alpha=0.1)
 plt.show()

在 3000 个 epoch 之后，结果如下所示（将训练点数减少到 3000 个而不是 5000 个以加快训练速度）：

在此处输入图像描述

于 2020-11-25T10:32:10.593 回答
1

该模型有 38,589 个可训练参数，但您只有 5,000 个点作为数据；因此，使用如此多的参数是不可能进行有效训练的。

于 2021-09-17T12:34:23.240 回答