用有限差分方法检查神经网络梯度不起作用

Question

经过整整一周的打印报表、维度分析、重构和大声讨论代码，我可以说我完全陷入困境了。

我的成本 function 产生的梯度与有限差分产生的梯度相差太远。

我已经确认我的成本 function 为正则化输入产生正确的成本，而不是。 这是function的成本：

def nnCost(nn_params, X, y, lambda_, input_layer_size, hidden_layer_size, num_labels):
  # reshape parameter/weight vectors to suit network size
  Theta1 = np.reshape(nn_params[:hidden_layer_size * (input_layer_size + 1)], (hidden_layer_size, (input_layer_size + 1)))
  Theta2 = np.reshape(nn_params[(hidden_layer_size * (input_layer_size+1)):], (num_labels, (hidden_layer_size + 1)))

  if lambda_ is None:
    lambda_ = 0

  # grab number of observations
  m = X.shape[0]
  
  # init variables we must return
  cost = 0
  Theta1_grad = np.zeros(Theta1.shape)
  Theta2_grad = np.zeros(Theta2.shape)

  # one-hot encode the vector y
  y_mtx = pd.get_dummies(y.ravel()).to_numpy() 

  ones = np.ones((m, 1))
  X = np.hstack((ones, X))
  
  # layer 1
  a1 = X
  z2 = Theta1@a1.T
  # layer 2
  ones_l2 = np.ones((y.shape[0], 1))
  a2 = np.hstack((ones_l2, sigmoid(z2.T)))
  z3 = Theta2@a2.T
  # layer 3
  a3 = sigmoid(z3)

  reg_term = (lambda_/(2*m)) * (np.sum(np.sum(np.multiply(Theta1, Theta1))) + np.sum(np.sum(np.multiply(Theta2,Theta2))) - np.subtract((Theta1[:,0].T@Theta1[:,0]),(Theta2[:,0].T@Theta2[:,0])))
  cost = (1/m) * np.sum((-np.log(a3).T * (y_mtx) - np.log(1-a3).T * (1-y_mtx))) + reg_term
  
  # BACKPROPAGATION
  # δ3 equals the difference between a3 and the y_matrix
  d3 = a3 - y_mtx.T
  # δ2 equals the product of δ3 and Θ2 (ignoring the Θ2 bias units) multiplied element-wise by the g′() of z2 (computed back in Step 2).
  d2 = Theta2[:,1:].T@d3 * sigmoidGradient(z2)
  # Δ1 equals the product of δ2 and a1.
  Delta1 = d2@a1
  Delta1 /= m
  # Δ2 equals the product of δ3 and a2.
  Delta2 = d3@a2
  Delta2 /= m
  
  reg_term1 = (lambda_/m) * np.append(np.zeros((Theta1.shape[0],1)), Theta1[:,1:], axis=1)
  reg_term2 = (lambda_/m) * np.append(np.zeros((Theta2.shape[0],1)), Theta2[:,1:], axis=1)
  
  Theta1_grad = Delta1 + reg_term1
  Theta2_grad = Delta2 + reg_term2
  
  grad = np.append(Theta1_grad.ravel(), Theta2_grad.ravel())
  
  return cost, grad

这是检查渐变的代码。 我已经完成了每一行，在这里我想不出任何可以改变的东西。 它似乎处于正常工作状态。

def checkNNGradients(lambda_):
  """
  Creates a small neural network to check the backpropagation gradients. 
  Credit: Based on the MATLAB code provided by Dr. Andrew Ng, Stanford Univ.
  
  Input: Regularization parameter, lambda, as int or float.
  
  Output: Analytical gradients produced by backprop code and the numerical gradients (computed
  using computeNumericalGradient). These two gradient computations should result in 
  very similar values. 
  """

  input_layer_size = 3
  hidden_layer_size = 5
  num_labels = 3
  m = 5

  # generate 'random' test data
  Theta1 = debugInitializeWeights(hidden_layer_size, input_layer_size)
  Theta2 = debugInitializeWeights(num_labels, hidden_layer_size)

  # reusing debugInitializeWeights to generate X
  X  = debugInitializeWeights(m, input_layer_size - 1)
  y  = np.ones(m) + np.remainder(np.range(m), num_labels)


  # unroll parameters
  nn_params = np.append(Theta1.ravel(), Theta2.ravel())
  costFunc = lambda p: nnCost(p, X, y, lambda_, input_layer_size, hidden_layer_size, num_labels)
    
  cost, grad = costFunc(nn_params)
    
  numgrad = computeNumericalGradient(costFunc, nn_params)

  # examine the two gradient computations; two columns should be very similar. 
  print('The columns below should be very similar.\n')
   
  # Credit: http://stackoverflow.com/a/27663954/583834
  print('{:<25}{}'.format('Numerical Gradient', 'Analytical Gradient'))
  for numerical, analytical in zip(numgrad, grad):
    print('{:<25}{}'.format(numerical, analytical))


  # If you have a correct implementation, and assuming you used EPSILON = 0.0001 
  # in computeNumericalGradient.m, then diff below should be less than 1e-9
  diff = np.linalg.norm(numgrad-grad)/np.linalg.norm(numgrad+grad)
  print(diff)
  print("\n")
  print('If your backpropagation implementation is correct, then \n' \
          'the relative difference will be small (less than 1e-9). \n' \
          '\nRelative Difference: {:.10f}'.format(diff))

The check function generates its own data using a debugInitializeWeights function (so there's the reproducible example; just run that and it will call the other functions), and then calls the function that calculates the gradient using finite differences. 两者都在下面。

def debugInitializeWeights(fan_out, fan_in):
  """
  Initializes the weights of a layer with fan_in
  incoming connections and fan_out outgoing connections using a fixed
  strategy.

  Input: fan_out, number of outgoing connections for a layer as int; fan_in, number
  of incoming connections for the same layer as int. 
  
  Output: Weight matrix, W, of size(1 + fan_in, fan_out), as the first row of W handles the "bias" terms
  """
  W = np.zeros((fan_out, 1 + fan_in))
  # Initialize W using "sin", this ensures that the values in W are of similar scale;
  # this will be useful for debugging
  W = np.sin(range(1, np.size(W)+1)) / 10 
  return W.reshape(fan_out, fan_in+1)

def computeNumericalGradient(J, nn_params):
  """
  Computes the gradient using "finite differences"
  and provides a numerical estimate of the gradient (i.e.,
  gradient of the function J around theta).
  Credit: Based on the MATLAB code provided by Dr. Andrew Ng, Stanford Univ. 

  Inputs: Cost, J, as computed by nnCost function; Parameter vector, theta.

  Output: Gradient vector using finite differences. Per Dr. Ng, 
  'Sets numgrad(i) to (a numerical approximation of) the partial derivative of 
  J with respect to the i-th input argument, evaluated at theta. (i.e., numgrad(i) should 
  be the (approximately) the partial derivative of J with respect
  to theta(i).)'          
  """
  numgrad = np.zeros(nn_params.shape)
  perturb = np.zeros(nn_params.shape)
  e = .0001
  for i in range(np.size(nn_params)):
      # Set perturbation (i.e., noise) vector
      perturb[i] = e
      # run cost fxn w/ noise added to and subtracted from parameters theta in nn_params
      cost1, grad1 = J((nn_params - perturb))
      cost2, grad2 = J((nn_params + perturb))
      # record the difference in cost function ouputs; this is the numerical gradient
      numgrad[i] = (cost2 - cost1) / (2*e)
      perturb[i] = 0

  return numgrad

该代码不适用于 class。 那个 MOOC 在 MATLAB 并且结束了。 这是给我的。 web 上存在其他解决方案； 事实证明，看着他们是徒劳的。 每个人都有不同的（不可思议的）方法。 所以，我非常需要帮助或奇迹。

编辑/更新：Fortran 在散布矢量影响结果时排序，但我无法让渐变一起移动以更改该选项。

Answer 1

一个想法：我认为你的扰动有点大，是1e-4 。 对于双精度浮点数，它应该更像1e-8 ，即机器精度的根（或者您是否使用单精度？）。

话虽如此，有限差分对于真正的导数来说可能是非常糟糕的近似。 具体来说，numpy 中的浮点计算不是确定性的，正如您似乎已经发现的那样。 在某些情况下，评估中的噪音会抵消许多有效数字。 你看到了什么价值观，你期待什么？

Answer 2

以下所有内容都可以解决我的问题。 对于那些试图将 MATLAB 代码转换为 Python 的人，无论是否来自 Andrew NG 的 Coursera 机器学习课程，这些都是每个人都应该知道的。

1. MATLAB 按照 FORTRAN 的顺序执行所有操作； Python 按照 C 的顺序执行所有操作。 这会影响向量的填充方式，从而影响您的结果。 如果您希望您的答案与您在 MATLAB 中所做的相匹配，您应该始终按 FORTRAN 顺序排列。 查看 文档

2. 以 FORTRAN 顺序获取向量可以像将order='F'作为参数传递给.reshape() 、 .ravel()或.flatten() 。 但是，如果您使用.ravel()通过转置向量然后应用.ravel() function 来实现相同的效果，就像XTravel()一样。

3. 说到.ravel() ， .ravel()和.flatten()函数做的事情不同，可能有不同的用例。 例如，SciPy 优化方法首选.flatten() 。 因此，如果您的等效fminunc不起作用，可能是因为您忘记了.flatten()您的响应向量y 。 请参阅此 Q&A StackOverflow和.ravel()上链接到.flatten()的文档。 更多文档

4. 如果您要将 MATLAB 实时脚本中的代码转换为 Jupyter 笔记本或 Google COLAB，则必须监管您的名称空间。 有一次，我发现我认为正在传递的变量实际上并不是正在传递的变量。 为什么？ Jupyter 和 Colab 笔记本有很多通常不会写的全局变量。

5. 有一个更好的 function 来评估数值梯度和分析梯度之间的差异：Relative Error Comparison np.abs(numerical-analyitical)/(numerical+analytical) 。 在此处阅读CS231另外，请考虑上面接受的帖子。