为什么Tensorflow的自动微分在loss函数中使用.numpy（）时失败？

Question

我注意到，当损失函数将输入转换为numpy数组以计算输出值时，Tensorflow的自动微分不会给出与有限差分相同的值。下面是该问题的一个最起码的工作示例：

import tensorflow as tf
import numpy as np

def lossFn(inputTensor):
    # Input is a rank-2 square tensor
    return tf.linalg.trace(inputTensor @ inputTensor)

def lossFnWithNumpy(inputTensor):
    # Same function, but converts input to a numpy array before performing the norm
    inputArray = inputTensor.numpy()

    return tf.linalg.trace(inputArray @ inputArray)

N = 2
tf.random.set_seed(0)
randomTensor = tf.random.uniform([N, N])

# Prove that the two functions give the same output; evaluates to exactly zero
print(lossFn(randomTensor) - lossFnWithNumpy(randomTensor)) 

theoretical, numerical = tf.test.compute_gradient(lossFn, [randomTensor])
# These two values match
print(theoretical[0])
print(numerical[0])

theoretical, numerical = tf.test.compute_gradient(lossFnWithNumpy, [randomTensor])
# The theoretical value is [0 0 0 0]
print(theoretical[0])
print(numerical[0])

函数tf.test.compute_gradients使用自动微分计算“理论”梯度，使用有限差分计算数值梯度。如代码所示，如果在损失函数中使用.numpy()，则自动微分不会计算梯度

有人能解释一下原因吗

Answer 1

来自指南：Introduction to Gradients and Automatic Differentiation

The tape can't record the gradient path if the calculation exits TensorFlow. For example:

x = tf.Variable([[1.0, 2.0],
                 [3.0, 4.0]], dtype=tf.float32)

with tf.GradientTape() as tape:   
  x2 = x**2
  # This step is calculated with NumPy   
  y = np.mean(x2, axis=0)
  # Like most ops, reduce_mean will cast the NumPy array to a constant tensor 
  # using `tf.convert_to_tensor`. 
  y = tf.reduce_mean(y,axis=0)

print(tape.gradient(y, x)) 

outputs None

numpy值将在对tf.linalg.trace的调用中被转换回常量张量，而Tensorflow无法在其上计算梯度