python - tensorflow 概率中的重新参数化 : tf. GradientTape() 不计算相对于分布均值的梯度

Question

在 tensorflow 版本 2.0.0-beta1 中，我正在尝试实现一个 keras 层，它具有从正态随机分布中采样的权重.我想将分布的均值作为可训练参数。

感谢 tensorflow-probability 中已经实现的“重新参数化技巧”，如果我没记错的话，原则上应该可以计算相对于分布均值的梯度。

但是，当我尝试使用 tf.GradientTape() 计算网络输出相对于平均值变量的梯度时，返回的梯度为 None。

我创建了两个最小示例，一个具有确定性权重的层和一个具有随机权重的层。确定性层梯度的梯度按预期计算，但在随机层的情况下梯度为 None。没有错误消息详细说明为什么梯度为 None，我有点卡住了。

最小示例代码:

A:这是确定性网络的最小示例:

import tensorflow as tf; print(tf.__version__)

from tensorflow.keras import backend as K
from tensorflow.keras.layers import Layer,Input
from tensorflow.keras.models import Model
from tensorflow.keras.initializers import RandomNormal
import tensorflow_probability as tfp

import numpy as np

# example data
x_data = np.random.rand(99,3).astype(np.float32)

# # A: DETERMINISTIC MODEL

# 1 Define Layer

class deterministic_test_layer(Layer):

    def __init__(self, output_dim, **kwargs):
        self.output_dim = output_dim
        super(deterministic_test_layer, self).__init__(**kwargs)

    def build(self, input_shape):
        self.kernel = self.add_weight(name='kernel', 
                                      shape=(input_shape[1], self.output_dim),
                                      initializer='uniform',
                                      trainable=True)
        super(deterministic_test_layer, self).build(input_shape)

    def call(self, x):
        return K.dot(x, self.kernel)

    def compute_output_shape(self, input_shape):
        return (input_shape[0], self.output_dim)

# 2 Create model and calculate gradient

x = Input(shape=(3,))
fx = deterministic_test_layer(1)(x)
deterministic_test_model = Model(name='test_deterministic',inputs=[x], outputs=[fx])

print('\n\n\nCalculating gradients for deterministic model: ')

for x_now in np.split(x_data,3):
#     print(x_now.shape)
    with tf.GradientTape() as tape:
        fx_now = deterministic_test_model(x_now)
        grads = tape.gradient(
            fx_now,
            deterministic_test_model.trainable_variables,
        )
        print('\n',grads,'\n')

print(deterministic_test_model.summary())

B:下面的示例非常相似，但我尝试使用随机采样的权重(在 call() 时间随机采样!)代替确定性权重来测试层:

# # B: RANDOM MODEL

# 1 Define Layer

class random_test_layer(Layer):

    def __init__(self, output_dim, **kwargs):
        self.output_dim = output_dim
        super(random_test_layer, self).__init__(**kwargs)

    def build(self, input_shape):
        self.mean_W = self.add_weight('mean_W',
                                      initializer=RandomNormal(mean=0.5,stddev=0.1),
                                      trainable=True)

        self.kernel_dist = tfp.distributions.MultivariateNormalDiag(loc=self.mean_W,scale_diag=(1.,))
        super(random_test_layer, self).build(input_shape)

    def call(self, x):
        sampled_kernel = self.kernel_dist.sample(sample_shape=x.shape[1])
        return K.dot(x, sampled_kernel)

    def compute_output_shape(self, input_shape):
        return (input_shape[0], self.output_dim)

# 2 Create model and calculate gradient

x = Input(shape=(3,))
fx = random_test_layer(1)(x)
random_test_model = Model(name='test_random',inputs=[x], outputs=[fx])

print('\n\n\nCalculating gradients for random model: ')

for x_now in np.split(x_data,3):
#     print(x_now.shape)
    with tf.GradientTape() as tape:
        fx_now = random_test_model(x_now)
        grads = tape.gradient(
            fx_now,
            random_test_model.trainable_variables,
        )
        print('\n',grads,'\n')

print(random_test_model.summary())

预期/实际输出:

A:确定性网络按预期工作，并且计算了梯度。输出是:

2.0.0-beta1



Calculating gradients for deterministic model: 

 [<tf.Tensor: id=26, shape=(3, 1), dtype=float32, numpy=
array([[17.79845  ],
       [15.764006 ],
       [14.4183035]], dtype=float32)>] 


 [<tf.Tensor: id=34, shape=(3, 1), dtype=float32, numpy=
array([[16.22232 ],
       [17.09122 ],
       [16.195663]], dtype=float32)>] 


 [<tf.Tensor: id=42, shape=(3, 1), dtype=float32, numpy=
array([[16.382954],
       [16.074356],
       [17.718027]], dtype=float32)>] 

Model: "test_deterministic"
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
input_1 (InputLayer)         [(None, 3)]               0         
_________________________________________________________________
deterministic_test_layer (de (None, 1)                 3         
=================================================================
Total params: 3
Trainable params: 3
Non-trainable params: 0
_________________________________________________________________
None

B:然而，在类似随机网络的情况下，梯度没有按预期计算(使用重新参数化技巧)。相反，它们是 None。完整的输出是

Calculating gradients for random model: 

 [None] 


 [None] 


 [None] 

Model: "test_random"
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
input_2 (InputLayer)         [(None, 3)]               0         
_________________________________________________________________
random_test_layer (random_te (None, 1)                 1         
=================================================================
Total params: 1
Trainable params: 1
Non-trainable params: 0
_________________________________________________________________
None

谁能指出我这里的问题？

Answer 1

似乎tfp.distributions.MultivariateNormalDiag就其输入参数而言是不可微分的(例如 loc)。在这种特殊情况下，以下内容是等效的:

class random_test_layer(Layer):
    ...

    def build(self, input_shape):
        ...
        self.kernel_dist = tfp.distributions.MultivariateNormalDiag(loc=0, scale_diag=(1.,))
        super(random_test_layer, self).build(input_shape)

    def call(self, x):
        sampled_kernel = self.kernel_dist.sample(sample_shape=x.shape[1]) + self.mean_W
        return K.dot(x, sampled_kernel)

然而，在这种情况下，损失对于 self.mean_W 是可微的。

注意:尽管这种方法可能对您有用，但请注意，调用密度函数 self.kernel_dist.prob 会产生不同的结果，因为我们采用了 loc 外面。