　　梯度检验可以作为检验神经网络是否有bug的一种方法，现神经网络的反向传播算法含有许多细节，在编程实现中很容易出现一些微妙的bug，但往往这些bug并不会影响你的程序运行，而且你的损失函数看样子也在不断变小。但最终，你的程序得出的结果误差将会比那些无bug的程序高出一个数量级，最终的结果可能并不是最优解。

梯度检验的原理

　　梯度检验法是通过一种简单的方法取得近似的梯度，将这个近似的梯度与真正的梯度对比，如果很接近，则认为梯度正确，否则认为梯度有误。

　　将J(θ)和θ放入直角坐标系，下图所示是θ取定值时J(θ)的导数：

　　ε 是一个很小的值：

　　如上图所示：

　　当ε→0时，这趋近于导数的定义：

　　在实际的应用中，θ往往是一个向量，梯度下降算法要求我们对向量中的每一个分量进行偏导数的计算，对于偏导数，我们同样可以用以下式子进行近似计算：

梯度检验代码检验

　　1、构建一个AI模型来判断是否可靠

　　首先我们需要先导入相关的库

import numpy as np
from testCases import *
from gc_utils import sigmoid, relu, dictionary_to_vector, vector_to_dictionary, gradients_to_vector

　　在这里我们所用到的testCases.py与gc_utils.py代码如下：

import numpy as np

def sigmoid(x):
    """
    Compute the sigmoid of x

    Arguments:
    x -- A scalar or numpy array of any size.

    Return:
    s -- sigmoid(x)
    """
    s = 1/(1+np.exp(-x))
    return s

def relu(x):
    """
    Compute the relu of x

    Arguments:
    x -- A scalar or numpy array of any size.

    Return:
    s -- relu(x)
    """
    s = np.maximum(0,x)
    
    return s

def dictionary_to_vector(parameters):
    """
    Roll all our parameters dictionary into a single vector satisfying our specific required shape.
    """
    keys = []
    count = 0
    for key in ["W1", "b1", "W2", "b2", "W3", "b3"]:
        
        # flatten parameter
        new_vector = np.reshape(parameters[key], (-1,1))
        keys = keys + [key]*new_vector.shape[0]
        
        if count == 0:
            theta = new_vector
        else:
            theta = np.concatenate((theta, new_vector), axis=0)
        count = count + 1

    return theta, keys

def vector_to_dictionary(theta):
    """
    Unroll all our parameters dictionary from a single vector satisfying our specific required shape.
    """
    parameters = {}
    parameters["W1"] = theta[:20].reshape((5,4))
    parameters["b1"] = theta[20:25].reshape((5,1))
    parameters["W2"] = theta[25:40].reshape((3,5))
    parameters["b2"] = theta[40:43].reshape((3,1))
    parameters["W3"] = theta[43:46].reshape((1,3))
    parameters["b3"] = theta[46:47].reshape((1,1))

    return parameters

def gradients_to_vector(gradients):
    """
    Roll all our gradients dictionary into a single vector satisfying our specific required shape.
    """
    
    count = 0
    for key in ["dW1", "db1", "dW2", "db2", "dW3", "db3"]:
        # flatten parameter
        new_vector = np.reshape(gradients[key], (-1,1))
        
        if count == 0:
            theta = new_vector
        else:
            theta = np.concatenate((theta, new_vector), axis=0)
        count = count + 1

    return theta

gc_utils

import numpy as np

def gradient_check_n_test_case(): 
    np.random.seed(1)
    x = np.random.randn(4,3)
    y = np.array([1, 1, 0])
    W1 = np.random.randn(5,4) 
    b1 = np.random.randn(5,1) 
    W2 = np.random.randn(3,5) 
    b2 = np.random.randn(3,1) 
    W3 = np.random.randn(1,3) 
    b3 = np.random.randn(1,1) 
    parameters = {"W1": W1,
                  "b1": b1,
                  "W2": W2,
                  "b2": b2,
                  "W3": W3,
                  "b3": b3}

    
    return x, y, parameters

testCases

　　首先我们进行简单的1维的梯度检验，后面再学N维的，便于理解

　　假设我们有一个简单的1维线性函数 $J (θ) = θ x$

$J (θ) = θ x$

上面的流程图显示出了关键的步骤：输入 $x$

# 前向传播
def forward_propagation(x, theta):
 
    J = np.dot(theta, x)
    
    return J

x, theta = 2, 4
J = forward_propagation(x, theta)
print ("J = " + str(J))

J = 8

# 反向传播
def backward_propagation(x, theta):
    
    # 这个函数的导数就是x，这是由微积分公式得来的，如果你没有学过微积分，没有关系，不用弄明白为什么。重点不在于此。
    dtheta = x 
    
    return dtheta

x, theta = 2, 4
dtheta = backward_propagation(x, theta)
print ("dtheta = " + str(dtheta))

dtheta = 2

　　下面我们将用梯度检验来确认上面反向传播计算出来的梯度dtheta是正确的。主要步骤如下:

def gradient_check(x, theta, epsilon=1e-7):

    # 利用前向传播计算出一个梯度
    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)         
    
    # 利用反向传播也计算出一个梯度
    grad = backward_propagation(x, theta)

    # 对比两个梯度相差多远
    numerator = np.linalg.norm(grad - gradapprox)                    
    denominator = np.linalg.norm(grad) + np.linalg.norm(gradapprox)   
    difference = numerator / denominator                              
    
    if difference < 1e-7:
        print("反向传播是正确的!")
    else:
        print("反向传播有问题！")
    
    return difference

x, theta = 2, 4
difference = gradient_check(x, theta)
print("difference = " + str(difference))

反向传播是正确的!
difference = 2.919335883291695e-10
但是通常情况下，神经网络的成本函数不仅仅只有一个1维的参数。在神经网络模型中， $θ$

上图展示了你的支付可靠度预测模型的前向传播和反向传播流程，下面为前向传播和反向传播的代码实现：

def forward_propagation_n(X, Y, 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)

    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

def backward_propagation_n(X, Y, cache):    
    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  # ~~
    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) # ~~
    
    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

　　下面进行多维度梯度检验：

　　多维检验中的 $θ$

　转化代码如下：

# 友情赠送向量转换为字典
def vector_to_dictionary(theta):

    parameters = {}
    parameters["W1"] = theta[:20].reshape((5,4))
    parameters["b1"] = theta[20:25].reshape((5,1))
    parameters["W2"] = theta[25:40].reshape((3,5))
    parameters["b2"] = theta[40:43].reshape((3,1))
    parameters["W3"] = theta[43:46].reshape((1,3))
    parameters["b3"] = theta[46:47].reshape((1,1))

    return parameters


def gradients_to_vector(gradients):
  
    count = 0
    for key in ["dW1", "db1", "dW2", "db2", "dW3", "db3"]:
        # flatten parameter
        new_vector = np.reshape(gradients[key], (-1,1))
        
        if count == 0:
            theta = new_vector
        else:
            theta = np.concatenate((theta, new_vector), axis=0)
        count = count + 1

    return theta

向量转换为字典

　代码如下：

def gradient_check_n(parameters, gradients, X, Y, epsilon=1e-7):
     
    parameters_values, _ = dictionary_to_vector(parameters)
    grad = gradients_to_vector(gradients)
    num_parameters = parameters_values.shape[0]
    J_plus = np.zeros((num_parameters, 1))
    J_minus = np.zeros((num_parameters, 1))
    gradapprox = np.zeros((num_parameters, 1))

    # 计算gradapprox
    for i in range(num_parameters):
        thetaplus =  np.copy(parameters_values)                                      
        thetaplus[i][0] = thetaplus[i][0] + epsilon                                 
        J_plus[i], _ =  forward_propagation_n(X, Y, vector_to_dictionary(thetaplus))  
        
        thetaminus = np.copy(parameters_values)                                     
        thetaminus[i][0] = thetaminus[i][0] - epsilon                                      
        J_minus[i], _ = forward_propagation_n(X, Y, vector_to_dictionary(thetaminus)) 
        
        gradapprox[i] = (J_plus[i] - J_minus[i]) / (2 * epsilon)
    
    numerator = np.linalg.norm(grad - gradapprox)                                
    denominator = np.linalg.norm(grad) + np.linalg.norm(gradapprox)               
    difference = numerator / denominator                                         

    if difference > 2e-7:
        print("\033[93m" + "反向传播有问题! difference = " + str(difference) + "\033[0m")
    else:
        print("\033[92m" + "反向传播很完美! 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)

[93m反向传播有问题! difference = 0.2850931566540251[0m

注意：

梯度检验是很缓慢的。通过 $\frac{\partial J}{\partial θ} \approx \frac{J (θ + ε) - J (θ - ε)}{2 ε}$
梯度检验是无法与dropout共存的。所以在执行梯度检验时，要把dropout关掉，检验完毕后再开启。

**本次实战编程需要记住的几点**:

梯度检验通过用前向传播的方式求出一个梯度，然后将其与反向传播求出的梯度进行对比来判断梯度是否正确
梯度检验很浪费计算力。所以只在需要验证代码是否正确时才开启。确认代码没有问题后，就关闭掉梯度检验。

参考：https://gitee.com/bijingrui1997/deep_learning_notes/blob/master/

posted on 2021-08-24 13:38 Hzzhbest 阅读(343) 评论(0) 编辑收藏举报

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