python - 在 TensorFlow2 中结合来自不同“网络”的梯度

Question

我正在尝试将一些“网络”组合成一个最终的损失函数。我想知道我所做的是否“合法”，截至目前我似乎无法完成这项工作。我正在使用张量流概率：

主要问题在这里：

# Get gradients of the loss wrt the weights.
gradients = tape.gradient(loss, [m_phis.trainable_weights, m_mus.trainable_weights, m_sigmas.trainable_weights])

# Update the weights of our linear layer.
optimizer.apply_gradients(zip(gradients, [m_phis.trainable_weights, m_mus.trainable_weights, m_sigmas.trainable_weights])

这给了我无渐变并抛出应用渐变：

AttributeError：“列表”对象没有属性“设备”

完整代码：

univariate_gmm = tfp.distributions.MixtureSameFamily(
        mixture_distribution=tfp.distributions.Categorical(probs=phis_true),
        components_distribution=tfp.distributions.Normal(loc=mus_true,scale=sigmas_true)
    )
x = univariate_gmm.sample(n_samples, seed=random_seed).numpy()
dataset = tf.data.Dataset.from_tensor_slices(x) 
dataset = dataset.shuffle(buffer_size=1024).batch(64)  

m_phis = keras.layers.Dense(2, activation=tf.nn.softmax)
m_mus = keras.layers.Dense(2)
m_sigmas = keras.layers.Dense(2, activation=tf.nn.softplus)

def neg_log_likelihood(y, phis, mus, sigmas):
    a = tfp.distributions.Normal(loc=mus[0],scale=sigmas[0]).prob(y)
    b = tfp.distributions.Normal(loc=mus[1],scale=sigmas[1]).prob(y)
    c = np.log(phis[0]*a + phis[1]*b)
    return tf.reduce_sum(-c, axis=-1)

# Instantiate a logistic loss function that expects integer targets.
loss_fn = neg_log_likelihood

# Instantiate an optimizer.
optimizer = tf.keras.optimizers.SGD(learning_rate=1e-3)

# Iterate over the batches of the dataset.
for step, y in enumerate(dataset):
    
    yy = np.expand_dims(y, axis=1)

    # Open a GradientTape.
    with tf.GradientTape() as tape:
        
        # Forward pass.
        phis = m_phis(yy)
        mus = m_mus(yy)
        sigmas = m_sigmas(yy)

        # Loss value for this batch.
        loss = loss_fn(yy, phis, mus, sigmas)

    # Get gradients of the loss wrt the weights.
    gradients = tape.gradient(loss, [m_phis.trainable_weights, m_mus.trainable_weights, m_sigmas.trainable_weights])

    # Update the weights of our linear layer.
    optimizer.apply_gradients(zip(gradients, [m_phis.trainable_weights, m_mus.trainable_weights, m_sigmas.trainable_weights]))

    # Logging.
    if step % 100 == 0:
        print("Step:", step, "Loss:", float(loss))

标签： pythonkerastensorflow2.0

Answer 1

有两个不同的问题需要考虑。

1.梯度是None：

通常，如果在GradientTape. 具体来说，这涉及到函数np.log中的计算。neg_log_likelihood如果你用 替换np.log，tf.math.log梯度应该计算。尽量不要在“内部”张量流组件中使用 numpy 可能是一个好习惯，因为这样可以避免这样的错误。对于大多数 numpy 操作，有一个很好的 tensorflow 替代品。

2.apply_gradients对于多个可训练对象：

这主要与apply_gradients期望的输入有关。你有两个选择：

第一种选择：调用apply_gradients3 次，每次使用不同的可训练对象

optimizer.apply_gradients(zip(m_phis_gradients, m_phis.trainable_weights))
optimizer.apply_gradients(zip(m_mus_gradients, m_mus.trainable_weights))
optimizer.apply_gradients(zip(m_sigmas_gradients, m_sigmas.trainable_weights))

另一种方法是创建一个元组列表，如tensorflow 文档中所示（引用：“grads_and_vars：（梯度，变量）对列表。”）。这意味着调用类似的东西

optimizer.apply_gradients(
   [
      zip(m_phis_gradients, m_phis.trainable_weights),
      zip(m_mus_gradients, m_mus.trainable_weights),
      zip(m_sigmas_gradients, m_sigmas.trainable_weights),
   ]
)

这两个选项都要求您拆分渐变。您可以通过计算梯度并分别索引它们来做到这一点（gradients[0],...），或者您可以简单地单独计算梯度。请注意，这可能需要persistent=True在您的GradientTape.

    # [...]
    # Open a GradientTape.
    with tf.GradientTape(persistent=True) as tape:
        # Forward pass.
        phis = m_phis(yy)
        mus = m_mus(yy)
        sigmas = m_sigmas(yy)

        # Loss value for this batch.
        loss = loss_fn(yy, phis, mus, sigmas)

    # Get gradients of the loss wrt the weights.
    m_phis_gradients = tape.gradient(loss, m_phis.trainable_weights)
    m_mus_gradients = tape.gradient(loss, m_mus.trainable_weights)
    m_sigmas_gradients = tape.gradient(loss, m_sigmas .trainable_weights)

    # Update the weights of our linear layer.
    optimizer.apply_gradients(
        [
            zip(m_phis_gradients, m_phis.trainable_weights),
            zip(m_mus_gradients, m_mus.trainable_weights),
            zip(m_sigmas_gradients, m_sigmas.trainable_weights),
       ]
   )
   # [...]