如何在 Tensorflow 2.0 中使用 gradient_override_map? [英] How to use gradient_override_map in Tensorflow 2.0?

问题描述

我正在尝试将 gradient_override_map 与 Tensorflow 2.0 一起使用.文档中有一个 示例，我也将在此处用作示例.

I'm trying to use gradient_override_map with Tensorflow 2.0. There is an example in the documentation, which I will use as the example here as well.

在 2.0 中，GradientTape 可用于计算梯度如下:

In 2.0, GradientTape can be used to compute gradients as follows:

import tensorflow as tf
print(tf.version.VERSION)  # 2.0.0-alpha0

x = tf.Variable(5.0)
with tf.GradientTape() as tape:
    s_1 = tf.square(x)
print(tape.gradient(s_1, x))

还有 tf.custom_gradient 装饰器，可用于为 new 函数定义渐变(同样，使用 文档中的示例):

There is also the tf.custom_gradient decorator, which can be used to define the gradient for a new function (again, using the example from the docs):

import tensorflow as tf
print(tf.version.VERSION)  # 2.0.0-alpha

@tf.custom_gradient
def log1pexp(x):
    e = tf.exp(x)

    def grad(dy):
        return dy * (1 - 1 / (1 + e))

    return tf.math.log(1 + e), grad

x = tf.Variable(100.)

with tf.GradientTape() as tape:
    y = log1pexp(x)

print(tape.gradient(y, x))

但是，我想替换tf.square 等标准函数的渐变.我尝试使用以下代码:

However, I would like to replace the gradient for standard functions such as tf.square. I tried to use the following code:

@tf.RegisterGradient("CustomSquare")
def _custom_square_grad(op, grad):
  return tf.constant(0)

with tf.Graph().as_default() as g:
    x = tf.Variable(5.0)
    with g.gradient_override_map({"Square": "CustomSquare"}):
        with tf.GradientTape() as tape:
            s_2 = tf.square(x, name="Square")

    with tf.compat.v1.Session() as sess:
        sess.run(tf.compat.v1.global_variables_initializer())            
        print(sess.run(tape.gradient(s_2, x)))

但是，有两个问题:梯度替换似乎不起作用(它被评估为 10.0 而不是 0.0)，我需要求助于 session.run() 来执行图表.有没有办法在原生"TensorFlow 2.0 中实现这一点?

However, there are two issues: The gradient replacement does not seem to work (it is evaluated to 10.0 instead of 0.0) and I need to resort to session.run() to execute the graph. Is there a way to achieve this in "native" TensorFlow 2.0?

在 TensorFlow 1.12.0 中，以下生成所需的输出:

In TensorFlow 1.12.0, the following produces the desired output:

import tensorflow as tf
print(tf.__version__)  # 1.12.0

@tf.RegisterGradient("CustomSquare")
def _custom_square_grad(op, grad):
  return tf.constant(0)

x = tf.Variable(5.0)

g = tf.get_default_graph()
with g.gradient_override_map({"Square": "CustomSquare"}):
    s_2 = tf.square(x, name="Square")
grad = tf.gradients(s_2, x)

with tf.Session() as sess:
  sess.run(tf.global_variables_initializer())
  print(sess.run(grad))

推荐答案

TensorFlow 2.0 中没有内置机制来覆盖作用域内内置运算符的所有梯度.但是，如果您能够修改对内置运算符的每次调用的调用站点，则可以使用 tf.custom_gradient 装饰器，如下所示:

There is no built-in mechanism in TensorFlow 2.0 to override all gradients for a built-in operator within a scope. However, if you are able to modify the call-site for each call to the built-in operator, you can use the tf.custom_gradient decorator as follows:

@tf.custom_gradient
def custom_square(x):
  def grad(dy):
    return tf.constant(0.0)
  return tf.square(x), grad

with tf.Graph().as_default() as g:
  x = tf.Variable(5.0)
  with tf.GradientTape() as tape:
    s_2 = custom_square(x)

  with tf.compat.v1.Session() as sess:
    sess.run(tf.compat.v1.global_variables_initializer())            
    print(sess.run(tape.gradient(s_2, x)))

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