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..。对这个问题的回答有资格获得+300的声誉奖励。
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关于这个问题:
如果需要更多细节,COLAB笔记本可以分享,尽管我相信一切都在这里。
在一个
整整一周
打印语句、维度分析、重构和大声讨论代码,我可以说
我完全卡住了
..。
我的代价函数产生的梯度与有限差产生的梯度太远了。
我已经确认我的成本函数为正规化的投入产生了正确的成本,而不是。下面是成本函数:
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。我已经检查了每一行代码,在这里我想不到任何需要修改的地方。它似乎处于工作状态。
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))check函数使用
函数(这是一个可重现的例子;只要运行它,它就会调用其他函数),然后调用使用有限差分计算梯度的函数。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该代码不适用于类。那个MOOC是在MATLAB上的,现在已经结束了。这是给我的。其他的解决方案也存在于网络上;查看这些解决方案被证明是徒劳无功的。每个人都有不同的(难以理解的)方法。所以,我真的需要帮助或者奇迹。
编辑/更新:当分解向量影响结果时,Fortran排序,但我无法让梯度一起移动更改该选项。
发布于 2021-02-28 22:19:51
一种想法是:我认为你的干扰有点大,
..。对于双精度浮点数,它应该更像
,即机器精度的根(或者您使用的是单精度?!)。
也就是说,有限差分可以是
非常
对真实导数的不好的近似。具体地说,numpy中的浮点计算并不是确定性的,您似乎已经发现了这一点。在某些情况下,评估中的噪声可能会抵消许多有效数字。你看到了什么价值,你期待什么?
https://stackoverflow.com/questions/66320255
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