Question

我有一个损失值/函数，我想计算所有关于张量 f （大小为n）的二阶导数。我设法使用了两次tf.gradients，但是当第二次应用它时，它会对第一个输入中的导数求和（请参阅我的代码中的 second_derivatives ）。

我还设法检索了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，f [0,0]）[0] 可以检索例如关于f_0的二阶导数，但似乎张量流没有' t允许从张量的一个片段中导出。

Answer 1

tf.gradients([f1,f2,f3],...)计算f=f1+f2+f3的渐变此外，区分x[0]是有问题的，因为x[0]指的是一个新的Slice节点，它不是您丢失的祖先，因此对它的衍生将是{{1} }。您可以使用None将pack粘合到x[0], x[1], ...中来解决问题，并使您的损失取决于xx而不是xx。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看起来就像这样。

你会看到

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()

Answer 2

现在考虑tf.hessians，

tf.hessians(loss, f)

https://www.tensorflow.org/api_docs/python/tf/hessians

Answer 3

以下函数在Tensorflow 2.0中计算二阶导数（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

这是一个使用函数f(x) = ln(r)的示例用法，其中x是3D坐标，而r是半径是球坐标：

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)

它将看起来像这样：

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算子（即通过对黑森矩阵的对角线求和）并与我们选择的函数2 / r^2的理论答案进行比较来检查实现的正确性：

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)

他们同意四舍五入以内。

Answer 4

如果没有必要使用Tensorflow来计算二阶导数，则可以执行以下操作：

import sympy as sp
x =sp.symbols('x', real=True)
loss = x**2 - 3*x + 1
f = loss
a = sp.diff(f,x)
b = sp.diff(a)   
print ("First derivative of: loss = f ** 2 - 3 * f + 1 respect to f is: ", a,"and the 
second derivative is: ",b)

结果为：

First derivative of: loss = f ** 2 - 3 * f + 1 respect to f is:  2*x - 3 and the second derivative is:  2

如果要计算粗麻布，则可以遵循以下示例：

import sympy as sp
x,y,z = sp.symbols('x y z', real=True)
f = (x**2)*y*z

Jacobian = np.array([sp.diff(f,x),sp.diff(f,y),sp.diff(f,z)])
Hessian = 

np.matrix([[sp.diff(sp.diff(f,x),x),sp.diff(sp.diff(f,y),x),sp.diff(sp.diff(f,z),x)],

[sp.diff(sp.diff(f,x),y),sp.diff(sp.diff(f,y),y),sp.diff(sp.diff(f,z),y)],

[sp.diff(sp.diff(f,x),z),sp.diff(sp.diff(f,z),y),sp.diff(sp.diff(f,z),z)]]) 

print ("La Jacobian es:", Jacobian)
print ("La Hessian es:\n",Hessian)

结果为：

La Jacobian es: [[2*x*y*z x**2*z x**2*y]] 
La Hessian es:
[[2*y*z 2*x*z 2*x*y]
 [2*x*z 0 x**2]
 [2*x*y x**2 0]]

如何在Tensorflow中计算所有二阶导数（只有Hessian矩阵的对角线）？

4 个答案: