标签：exe wro 需要 alc color start tor 如何 values
你被要求搭建一个Deep Learning model来检测欺诈，每当有人付款，你想知道是否该支付可能是欺诈，例如该用户的账户可能已经被黑客掉。
但是，反向传播实现起来非常有挑战，并且有时有一些bug，因为这是一个mission-critical应用，你公司老板想让十分确定，你实现的反向传播是正确的。你需要用“gradient checking”来证明你的反向传播是正确的。
# Packages
import numpy as np
from testCases import *
from gc_utils import sigmoid, relu, dictionary_to_vector, vector_to_dictionary, gradients_to_vector
Backpropagation 计算梯度(the gradients) \(\frac{\partial J}{\partial \theta}\)， \(\theta\)代表着模型的参数，\(J\) 是使用前向传播和你的loss function来计算的。
前向传播十分容易，因此你使用计算 \(J\) 的代码 来确认计算 \(\frac{\partial J}{\partial \theta}\) 的代码
我们来看一下derivative (or gradient)的定义：
接下来：
我们要使用公式(1) 和一个很小的数 \(\varepsilon\) 来保证你计算 \(\frac{\partial J}{\partial \theta}\) 的代码是正确的。
只考虑一元线性函数 \(J(\theta) = \theta x\). The model contains only a single real-valued parameter \(\theta\), and takes \(x\) as input.
You will implement code to compute \(J(.)\) and its derivative \(\frac{\partial J}{\partial \theta}\). You will then use gradient checking to make sure your derivative computation for \(J\) is correct.
上图展示了关键的计算步骤: 首先开始于 \(x\), 随后评估 \(J(x)\) ("forward propagation"). 然后计算 the derivative \(\frac{\partial J}{\partial \theta}\) ("backward propagation").
Exercise: 实现这个简单函数的 "forward propagation" and "backward propagation" . I.e., 计算 \(J(.)\) ("forward propagation") 和 它关于 \(\theta\) 的导数("backward propagation"), 在两个函数里。
# GRADED FUNCTION: forward_propagation
def forward_propagation(x, theta):
"""
Implement the linear forward propagation (compute J) presented in Figure 1 (J(theta) = theta * x)
Arguments:
x -- a real-valued input
theta -- our parameter, a real number as well
Returns:
J -- the value of function J, computed using the formula J(theta) = theta * x
"""
### START CODE HERE ### (approx. 1 line)
J = theta * x
### END CODE HERE ###
return J
测试：
x, theta = 2, 4
J = forward_propagation(x, theta)
print ("J = " + str(J))
J = 8
Exercise: 现在，实现图1中反向传播（导数计算）步骤：计算 \(J(\theta) = \theta x\) 关于 \(\theta\) 的导数. 用 \(dtheta = \frac { \partial J?}{ \partial \theta} = x\) 来保存你做的计算。
# GRADED FUNCTION: backward_propagation
def backward_propagation(x, theta):
"""
Computes the derivative of J with respect to theta (see Figure 1).
Arguments:
x -- a real-valued input
theta -- our parameter, a real number as well
Returns:
dtheta -- the gradient of the cost with respect to theta
"""
### START CODE HERE ### (approx. 1 line)
dtheta = x
### END CODE HERE ###
return dtheta
测试
x, theta = 2, 4
dtheta = backward_propagation(x, theta)
print ("dtheta = " + str(dtheta))
dtheta = 2
Exercise: 为了显示 backward_propagation() 函数是正确计算 the gradient \(\frac{\partial J}{\partial \theta}\), 让我们实现 gradient checking.
Instructions:
你需要三个步骤来计算这个公式：
# GRADED FUNCTION: gradient_check
def gradient_check(x, theta, epsilon = 1e-7):
"""
Implement the backward propagation presented in Figure 1.
Arguments:
x -- a real-valued input
theta -- our parameter, a real number as well
epsilon -- tiny shift to the input to compute approximated gradient with formula(1)
Returns:
difference -- difference (2) between the approximated gradient and the backward propagation gradient
"""
# Compute gradapprox using left side of formula (1). epsilon is small enough, you don‘t need to worry about the limit.
### START CODE HERE ### (approx. 5 lines)
thetaplus = theta + epsilon
thetaminus = theta - epsilon
J_plus = forward_propagation(x, thetaplus)
J_minus = forward_propagation(x, thetaminus)
gradapprox = (J_plus - J_minus) / (2. * epsilon)
### END CODE HERE ###
# Check if gradapprox is close enough to the output of backward_propagation()
### START CODE HERE ### (approx. 1 line)
grad = backward_propagation(x, theta)
### END CODE HERE ###
### START CODE HERE ### (approx. 1 line)
numerator = np.linalg.norm(grad - gradapprox)
denominator = np.linalg.norm(grad) + np.linalg.norm(gradapprox)
difference = numerator / denominator
### END CODE HERE ###
if difference < 1e-7:
print ("The gradient is correct!")
else:
print ("The gradient is wrong!")
return difference
x, theta = 2, 4
difference = gradient_check(x, theta)
print("difference = " + str(difference))
The gradient is correct!
difference = 2.919335883291695e-10
上述计算检验正确。即，可以正确的计算反向传播。
现在，你的 cost function \(J\) has more than a single 1D input。当你训练一个神经网络，\(\theta\) 事实上由multiple matrices \(W^{[l]}\) and biases \(b^{[l]}\)组成，知道如何 梯度检验 高维度输入 非常重要。
下图描述了你的欺诈检测的前向和反向传播的模型：
下面实现forward propagation and backward propagation.
def forward_propagation_n(X, Y, parameters):
"""
Implements the forward propagation (and computes the cost) presented in Figure 3.
Arguments:
X -- training set for m examples
Y -- labels for m examples
parameters -- python dictionary containing your parameters "W1", "b1", "W2", "b2", "W3", "b3":
W1 -- weight matrix of shape (5, 4)
b1 -- bias vector of shape (5, 1)
W2 -- weight matrix of shape (3, 5)
b2 -- bias vector of shape (3, 1)
W3 -- weight matrix of shape (1, 3)
b3 -- bias vector of shape (1, 1)
Returns:
cost -- the cost function (logistic cost for one example)
"""
# retrieve parameters
m = X.shape[1]
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
W3 = parameters["W3"]
b3 = parameters["b3"]
# LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID
Z1 = np.dot(W1, X) + b1
A1 = relu(Z1)
Z2 = np.dot(W2, A1) + b2
A2 = relu(Z2)
Z3 = np.dot(W3, A2) + b3
A3 = sigmoid(Z3)
# Cost
logprobs = np.multiply(-np.log(A3),Y) + np.multiply(-np.log(1 - A3), 1 - Y)
cost = 1./m * np.sum(logprobs)
cache = (Z1, A1, W1, b1, Z2, A2, W2, b2, Z3, A3, W3, b3)
return cost, cache
Now, run backward propagation.
def backward_propagation_n(X, Y, cache):
"""
Implement the backward propagation presented in figure 2.
Arguments:
X -- input datapoint, of shape (input size, 1)
Y -- true "label"
cache -- cache output from forward_propagation_n()
Returns:
gradients -- A dictionary with the gradients of the cost with respect to each parameter, activation and pre-activation variables.
"""
m = X.shape[1]
(Z1, A1, W1, b1, Z2, A2, W2, b2, Z3, A3, W3, b3) = cache
dZ3 = A3 - Y
dW3 = 1./m * np.dot(dZ3, A2.T)
db3 = 1./m * np.sum(dZ3, axis=1, keepdims=True)
dA2 = np.dot(W3.T, dZ3)
dZ2 = np.multiply(dA2, np.int64(A2 > 0))
dW2 = 1./m * np.dot(dZ2, A1.T) * 2 # 这里故意写错
db2 = 1./m * np.sum(dZ2, axis=1, keepdims = True)
dA1 = np.dot(W2.T, dZ2)
dZ1 = np.multiply(dA1, np.int64(A1 > 0))
dW1 = 1./m * np.dot(dZ1, X.T)
db1 = 4./m * np.sum(dZ1, axis=1, keepdims = True) # 这里故意写错
gradients = {"dZ3": dZ3, "dW3": dW3, "db3": db3,
"dA2": dA2, "dZ2": dZ2, "dW2": dW2, "db2": db2,
"dA1": dA1, "dZ1": dZ1, "dW1": dW1, "db1": db1}
return gradients
下面进行梯度检验来确保你的梯度是正确的.
gradient checking 如何工作？.
As in 1) and 2), you want to compare "gradapprox" to the gradient computed by backpropagation. The formula is still:
但是, \(\theta\) 不再是标量. 而是一个叫 "parameters"的字典. 下面实现一个 "dictionary_to_vector()"(字典转向量). 它将"parameters" dictionary 转换成名为"values"的vector, 通过 reshaping all parameters (W1, b1, W2, b2, W3, b3) into vectors and concatenating(连接) them 获得.
The inverse function is "vector_to_dictionary"(向量转字典) which outputs back the "parameters" dictionary.
We have also converted the "gradients" dictionary into a vector "grad" using gradients_to_vector(). You don‘t need to worry about that.
Exercise: Implement gradient_check_n().
Instructions: 这里的伪代码(pseudo-code)将帮助你实现梯度检测(the gradient check).
For each i in num_parameters:
J_plus[i]:
np.copy(parameters_values) (深拷贝)forward_propagation_n(x, y, vector_to_dictionary(\(\theta^{+}\) )).J_minus[i]: do the same thing with \(\theta^{-}\)Thus, you get a vector gradapprox, where gradapprox[i] is an approximation of the gradient with respect to parameter_values[i]. You can now compare this gradapprox vector to the gradients vector from backpropagation. Just like for the 1D case (Steps 1‘, 2‘, 3‘), compute:
# GRADED FUNCTION: gradient_check_n
def gradient_check_n(parameters, gradients, X, Y, epsilon = 1e-7):
"""
Checks if backward_propagation_n computes correctly the gradient of the cost output by forward_propagation_n
Arguments:
parameters -- python dictionary containing your parameters "W1", "b1", "W2", "b2", "W3", "b3":
grad -- output of backward_propagation_n, contains gradients of the cost with respect to the parameters.
x -- input datapoint, of shape (input size, 1)
y -- true "label"
epsilon -- tiny shift to the input to compute approximated gradient with formula(1)
Returns:
difference -- difference (2) between the approximated gradient and the backward propagation gradient
"""
# Set-up variables
# print(parameters)
parameters_values, _ = dictionary_to_vector(parameters) # 将字典转换成向量
# print(parameters_values, i) # (W1, b1, W2, b2, .....) (注：此处W1,b1都转换成了向量)
grad = gradients_to_vector(gradients) # 梯度转换成向量
num_parameters = parameters_values.shape[0] # 所有参数个数
J_plus = np.zeros((num_parameters, 1)) # 初始化为 (num, 1)的向量
J_minus = np.zeros((num_parameters, 1))
gradapprox = np.zeros((num_parameters, 1))
# Compute gradapprox
for i in range(num_parameters): # 遍历所有参数，每个参数都求一遍 gradapprox，很费时间
# Compute J_plus[i]. Inputs: "parameters_values, epsilon". Output = "J_plus[i]".
# "_" is used because the function you have to outputs two parameters but we only care about the first one
### START CODE HERE ### (approx. 3 lines)
thetaplus = np.copy(parameters_values)
thetaplus[i, 0] += epsilon
# Step 2
J_plus[i], _ = forward_propagation_n(X, Y, vector_to_dictionary(thetaplus)) # Step 3
### END CODE HERE ###
# Compute J_minus[i]. Inputs: "parameters_values, epsilon". Output = "J_minus[i]".
### START CODE HERE ### (approx. 3 lines)
thetaminus = np.copy(parameters_values) # Step 1
thetaminus[i, 0] -= epsilon # Step 2
J_minus[i], _ = forward_propagation_n(X, Y, vector_to_dictionary(thetaminus)) # Step 3
### END CODE HERE ###
# Compute gradapprox[i]
### START CODE HERE ### (approx. 1 line)
gradapprox[i] = (J_plus[i] - J_minus[i]) / (2. * epsilon)
### END CODE HERE ###
# Compare gradapprox to backward propagation gradients by computing difference.
### START CODE HERE ### (approx. 1 line)
numerator = np.linalg.norm(grad - gradapprox) # Step 1‘
denominator = np.linalg.norm(grad) + np.linalg.norm(gradapprox) # Step 2‘
difference = numerator / denominator
if difference > 1.2e-7:
print ("\033[93m" + "There is a mistake in the backward propagation! difference = " + str(difference) + "\033[0m")
else:
print ("\033[92m" + "Your backward propagation works perfectly fine! difference = " + str(difference) + "\033[0m")
return difference
X, Y, parameters = gradient_check_n_test_case()
cost, cache = forward_propagation_n(X, Y, parameters)
gradients = backward_propagation_n(X, Y, cache)
difference = gradient_check_n(parameters, gradients, X, Y)
可以看出，在 backward_propagation_n代码中有一些错误。
现在，我们修复这个错误，再来运行一下上面代码：
dA2 = np.dot(W3.T, dZ3)
dZ2 = np.multiply(dA2, np.int64(A2 > 0))
# dW2 = 1./m * np.dot(dZ2, A1.T) * 2 # 这里故意写错
dW2 = 1./m * np.dot(dZ2, A1.T) # 修复
db2 = 1./m * np.sum(dZ2, axis=1, keepdims = True)
dA1 = np.dot(W2.T, dZ2)
dZ1 = np.multiply(dA1, np.int64(A1 > 0))
dW1 = 1./m * np.dot(dZ1, X.T)
# db1 = 4./m * np.sum(dZ1, axis=1, keepdims = True) # 这里故意写错
db1 = 1./m * np.sum(dZ1, axis=1, keepdims = True) # 修复
重新运行 backward_propagation_n()
输出：
Your backward propagation works perfectly fine! difference = 1.1885552035482147e-07
Note
Improving DNNs Hyperparameter tuning-Regularization and Optimization（week1）GradientChecking
标签：exe wro 需要 alc color start tor 如何 values
原文地址：https://www.cnblogs.com/douzujun/p/13084271.html
