如何计算Tensorflow中的所有二阶导数(只有Hessian矩阵的对角线)？

Question

我有一个损失值/函数，我想计算关于张量f ( n)的所有二阶导数。我成功地使用了tf.gradients两次，但是当第二次应用它时，它会在第一个输入中对派生项进行求和(请参阅我的代码中的second_derivatives )。

此外，我设法检索Hessian矩阵，但我只想计算它的对角线，以避免额外的计算。

import tensorflow as tf
import numpy as np

f = tf.Variable(np.array([[1., 2., 0]]).T)
loss = tf.reduce_prod(f ** 2 - 3 * f + 1)

first_derivatives = tf.gradients(loss, f)[0]

second_derivatives = tf.gradients(first_derivatives, f)[0]

hessian = [tf.gradients(first_derivatives[i,0], f)[0][:,0] for i in range(3)]

model = tf.initialize_all_variables()
with tf.Session() as sess:
    sess.run(model)
    print "\nloss\n", sess.run(loss)
    print "\nloss'\n", sess.run(first_derivatives)
    print "\nloss''\n", sess.run(second_derivatives)
    hessian_value = np.array(map(list, sess.run(hessian)))
    print "\nHessian\n", hessian_value

我的想法是，tf.gradients(first_derivatives，f0，0)会努力检索关于f_0的二阶导数，但似乎tensorflow不允许从张量的一个片段中导出。

Answer 1

现在想想tf.hessians，

tf.hessians(loss, f)

docs/python/tf/hessians

Answer 2

tf.gradients([f1,f2,f3],...)也计算f=f1+f2+f3的梯度，对于x[0]的区分也是有问题的，因为x[0]引用的是一个新的Slice节点，它不是您丢失的祖先，因此与它相关的导数将是None。您可以使用pack将x[0], x[1], ...粘合到xx中，从而使您的损失依赖于xx而不是x。另一种方法是为单个组件使用单独的变量，在这种情况下，计算Hessian会类似于这样。

def replace_none_with_zero(l):
  return [0 if i==None else i for i in l] 

tf.reset_default_graph()

x = tf.Variable(1.)
y = tf.Variable(1.)
loss = tf.square(x) + tf.square(y)
grads = tf.gradients([loss], [x, y])
hess0 = replace_none_with_zero(tf.gradients([grads[0]], [x, y]))
hess1 = replace_none_with_zero(tf.gradients([grads[1]], [x, y]))
hessian = tf.pack([tf.pack(hess0), tf.pack(hess1)])
sess = tf.InteractiveSession()
sess.run(tf.initialize_all_variables())
print hessian.eval()

你终究会明白的

[[ 2.  0.]
 [ 0.  2.]]

Answer 3

以下函数在Tensorflow 2.0中计算第二导数( Hessian矩阵的对角线)：

%tensorflow_version 2.x  # Tells Colab to load TF 2.x
import tensorflow as tf

def calc_hessian_diag(f, x):
    """
    Calculates the diagonal entries of the Hessian of the function f
    (which maps rank-1 tensors to scalars) at coordinates x (rank-1
    tensors).
    
    Let k be the number of points in x, and n be the dimensionality of
    each point. For each point k, the function returns

      (d^2f/dx_1^2, d^2f/dx_2^2, ..., d^2f/dx_n^2) .

    Inputs:
      f (function): Takes a shape-(k,n) tensor and outputs a
          shape-(k,) tensor.
      x (tf.Tensor): The points at which to evaluate the Laplacian
          of f. Shape = (k,n).
    
    Outputs:
      A tensor containing the diagonal entries of the Hessian of f at
      points x. Shape = (k,n).
    """
    # Use the unstacking and re-stacking trick, which comes
    # from https://github.com/xuzhiqin1990/laplacian/
    with tf.GradientTape(persistent=True) as g1:
        # Turn x into a list of n tensors of shape (k,)
        x_unstacked = tf.unstack(x, axis=1)
        g1.watch(x_unstacked)

        with tf.GradientTape() as g2:
            # Re-stack x before passing it into f
            x_stacked = tf.stack(x_unstacked, axis=1) # shape = (k,n)
            g2.watch(x_stacked)
            f_x = f(x_stacked) # shape = (k,)
        
        # Calculate gradient of f with respect to x
        df_dx = g2.gradient(f_x, x_stacked) # shape = (k,n)
        # Turn df/dx into a list of n tensors of shape (k,)
        df_dx_unstacked = tf.unstack(df_dx, axis=1)

    # Calculate 2nd derivatives
    d2f_dx2 = []
    for df_dxi,xi in zip(df_dx_unstacked, x_unstacked):
        # Take 2nd derivative of each dimension separately:
        #   d/dx_i (df/dx_i)
        d2f_dx2.append(g1.gradient(df_dxi, xi))
    
    # Stack 2nd derivates
    d2f_dx2_stacked = tf.stack(d2f_dx2, axis=1) # shape = (k,n)
    
    return d2f_dx2_stacked

下面是一个使用f(x) = ln(r)函数的例子，其中x是3D坐标，r是半径是球形坐标：

f = lambda q : tf.math.log(tf.math.reduce_sum(q**2, axis=1))
x = tf.random.uniform((5,3))

d2f_dx2 = calc_hessian_diag(f, x)
print(d2f_dx2)

遗嘱看起来是这样的：

tf.Tensor(
[[ 1.415968    1.0215727  -0.25363517]
 [-0.67299247  2.4847088   0.70901346]
 [ 1.9416015  -1.1799507   1.3937857 ]
 [ 1.4748447   0.59702784 -0.52290654]
 [ 1.1786096   0.07442689  0.2396735 ]], shape=(5, 3), dtype=float32)

我们可以通过计算拉普拉斯(即通过计算赫森矩阵的对角线)来检验实现的正确性，并与我们选择的函数2 / r^2的理论答案进行比较。

print(tf.reduce_sum(d2f_dx2, axis=1)) # Laplacian from summing above results
print(2./tf.math.reduce_sum(x**2, axis=1)) # Analytic expression for Lapalcian

我得到以下信息：

tf.Tensor([2.1839054 2.5207298 2.1554365 1.5489659 1.49271  ], shape=(5,), dtype=float32)
tf.Tensor([2.1839058 2.5207298 2.1554365 1.5489662 1.4927098], shape=(5,), dtype=float32)

他们同意在四舍五入误差范围内。