吴恩达深度学习课程deeplearning.ai课程作业：Class 2 Week 1 3.Gradient Checking
吳恩達deeplearning.ai課程作業(yè)，自己寫的答案。
補充說明：
1. 評論中總有人問為什么直接復制這些notebook運行不了？請不要直接復制粘貼，不可能運行通過的，這個只是notebook中我們要自己寫的那部分，要正確運行還需要其他py文件，請自己到GitHub上下載完整的。這里的部分僅僅是參考用的，建議還是自己按照提示一點一點寫，如果實在卡住了再看答案。個人覺得這樣才是正確的學習方法，況且作業(yè)也不算難。
2. 關于評論中有人說我是抄襲，注釋還沒別人詳細，復制下來還運行不過。答復是：做伸手黨之前，請先搞清這個作業(yè)是干什么的。大家都是從GitHub上下載原始的作業(yè)，然后根據(jù)代碼前面的提示（通常會指定函數(shù)和公式）來編寫代碼，而且后面還有expected output供你比對，如果程序正確，結果一般來說是一樣的。請不要無腦噴，說什么跟別人的答案一樣的。說到底，我們要做的就是，看他的文字部分，根據(jù)提示在代碼中加入部分自己的代碼。我們自己要寫的部分只有那么一小部分代碼。
3. 由于實在很反感無腦噴子，故禁止了下面的評論功能，請見諒。如果有問題，請私信我，在力所能及的范圍內(nèi)會盡量幫忙。
Gradient Checking
Welcome to the final assignment for this week! In this assignment you will learn to implement and use gradient checking.
You are part of a team working to make mobile payments available globally, and are asked to build a deep learning model to detect fraud–whenever someone makes a payment, you want to see if the payment might be fraudulent, such as if the user’s account has been taken over by a hacker.
But backpropagation is quite challenging to implement, and sometimes has bugs. Because this is a mission-critical application, your company’s CEO wants to be really certain that your implementation of backpropagation is correct. Your CEO says, “Give me a proof that your backpropagation is actually working!” To give this reassurance, you are going to use “gradient checking”.
Let’s do it!
# Packages import numpy as np from testCases import * from gc_utils import sigmoid, relu, dictionary_to_vector, vector_to_dictionary, gradients_to_vector1) How does gradient checking work?
Backpropagation computes the gradients ?J?θ?J?θ, where θθ denotes the parameters of the model. JJ is computed using forward propagation and your loss function.
Because forward propagation is relatively easy to implement, you’re confident you got that right, and so you’re almost 100% sure that you’re computing the cost JJ correctly. Thus, you can use your code for computing JJ to verify the code for computing ?J?θ?J?θ.
Let’s look back at the definition of a derivative (or gradient):
If you’re not familiar with the “limε→0limε→0” notation, it’s just a way of saying “when εε is really really small.”
We know the following:
- ?J?θ?J?θ is what you want to make sure you’re computing correctly.
- You can compute J(θ+ε)J(θ+ε) and J(θ?ε)J(θ?ε) (in the case that θθ is a real number), since you’re confident your implementation for JJ is correct.
Lets use equation (1) and a small value for εε to convince your CEO that your code for computing ?J?θ?J?θ is correct!
2) 1-dimensional gradient checking
Consider a 1D linear function J(θ)=θxJ(θ)=θx. The model contains only a single real-valued parameter θθ, and takes xx as input.
You will implement code to compute J(.)J(.) and its derivative ?J?θ?J?θ. You will then use gradient checking to make sure your derivative computation for JJ is correct.
The diagram above shows the key computation steps: First start with xx, then evaluate the function J(x)J(x) (“forward propagation”). Then compute the derivative ?J?θ?J?θ (“backward propagation”).
Exercise: implement “forward propagation” and “backward propagation” for this simple function. I.e., compute both J(.)J(.) (“forward propagation”) and its derivative with respect to θθ (“backward propagation”), in two separate functions.
# GRADED FUNCTION: forward_propagationdef forward_propagation(x, theta):"""Implement the linear forward propagation (compute J) presented in Figure 1 (J(theta) = theta * x)Arguments:x -- a real-valued inputtheta -- our parameter, a real number as wellReturns:J -- the value of function J, computed using the formula J(theta) = theta * x"""### START CODE HERE ### (approx. 1 line)J = x * theta### END CODE HERE ###return J x, theta = 2, 4 J = forward_propagation(x, theta) print ("J = " + str(J)) J = 8Expected Output:
# GRADED FUNCTION: backward_propagationdef backward_propagation(x, theta):"""Computes the derivative of J with respect to theta (see Figure 1).Arguments:x -- a real-valued inputtheta -- our parameter, a real number as wellReturns:dtheta -- the gradient of the cost with respect to theta"""### START CODE HERE ### (approx. 1 line)dtheta = x### END CODE HERE ###return dtheta x, theta = 2, 4 dtheta = backward_propagation(x, theta) print ("dtheta = " + str(dtheta)) dtheta = 2Expected Output:
| dtheta | 2 |
Exercise: To show that the backward_propagation() function is correctly computing the gradient ?J?θ?J?θ, let’s implement gradient checking.
Instructions:
- First compute “gradapprox” using the formula above (1) and a small value of εε. Here are the Steps to follow:
1. θ+=θ+εθ+=θ+ε
2. θ?=θ?εθ?=θ?ε
3. J+=J(θ+)J+=J(θ+)
4. J?=J(θ?)J?=J(θ?)
5. gradapprox=J+?J?2εgradapprox=J+?J?2ε
- Then compute the gradient using backward propagation, and store the result in a variable “grad”
- Finally, compute the relative difference between “gradapprox” and the “grad” using the following formula:
You will need 3 Steps to compute this formula:
- 1’. compute the numerator using np.linalg.norm(…)
- 2’. compute the denominator. You will need to call np.linalg.norm(…) twice.
- 3’. divide them.
- If this difference is small (say less than 10?710?7), you can be quite confident that you have computed your gradient correctly. Otherwise, there may be a mistake in the gradient computation. # GRADED FUNCTION: gradient_checkdef gradient_check(x, theta, epsilon = 1e-7):"""Implement the backward propagation presented in Figure 1.Arguments:x -- a real-valued inputtheta -- our parameter, a real number as wellepsilon -- 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 # Step 1thetaminus = theta - epsilon # Step 2J_plus = forward_propagation(x, thetaplus) # Step 3J_minus = forward_propagation(x, thetaminus) # Step 4gradapprox = (J_plus - J_minus) / (2 * epsilon) # Step 5### 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) # print(grad) # print(gradapprox)numerator = np.linalg.norm(grad-gradapprox) # Step 1'denominator = np.linalg.norm(grad) + np.linalg.norm(gradapprox) # Step 2'difference = numerator / denominator # Step 3'### 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 = 2.91933588329e-10
Expected Output:
The gradient is correct!
| difference | 2.9193358103083e-10 |
Congrats, the difference is smaller than the 10?710?7 threshold. So you can have high confidence that you’ve correctly computed the gradient in backward_propagation().
Now, in the more general case, your cost function JJ has more than a single 1D input. When you are training a neural network, θθ actually consists of multiple matrices W[l]W[l] and biases b[l]b[l]! It is important to know how to do a gradient check with higher-dimensional inputs. Let’s do it!
3) N-dimensional gradient checking
The following figure describes the forward and backward propagation of your fraud detection model.
LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID
Let’s look at your implementations for 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 examplesY -- 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 parametersm = 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 -> SIGMOIDZ1 = np.dot(W1, X) + b1A1 = relu(Z1)Z2 = np.dot(W2, A1) + b2A2 = relu(Z2)Z3 = np.dot(W3, A2) + b3A3 = sigmoid(Z3)# Costlogprobs = 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, cacheNow, 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) = cachedZ3 = A3 - YdW3 = 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))# 這里是故意使用一個錯誤的形式來驗證gradient_check是否正常工作dW2 = 1./m * np.dot(dZ2, A1.T) * 2# 正確的形式，最后再修改的 # dW2 = 1./m * np.dot(dZ2, A1.T)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)# 這里是故意使用一個錯誤的形式來驗證gradient_check是否正常工作db1 = 4./m * np.sum(dZ1, axis=1, keepdims = True)# 正確的形式，最后再修改的 # db1 = 1./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 gradientsYou obtained some results on the fraud detection test set but you are not 100% sure of your model. Nobody’s perfect! Let’s implement gradient checking to verify if your gradients are correct.
How does gradient checking work?.
As in 1) and 2), you want to compare “gradapprox” to the gradient computed by backpropagation. The formula is still:
?J?θ=limε→0J(θ+ε)?J(θ?ε)2ε(1)(1)?J?θ=limε→0J(θ+ε)?J(θ?ε)2ε
However, θθ is not a scalar anymore. It is a dictionary called “parameters”. We implemented a function “dictionary_to_vector()” for you. It converts the “parameters” dictionary into a vector called “values”, obtained by reshaping all parameters (W1, b1, W2, b2, W3, b3) into vectors and concatenating them.
The inverse function is “vector_to_dictionary” which outputs back the “parameters” dictionary.
You will need these functions in gradient_check_n()
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: Here is pseudo-code that will help you implement the gradient check.
For each i in num_parameters:
- To compute J_plus[i]:
1. Set θ+θ+ to np.copy(parameters_values)
2. Set θ+iθi+ to θ+i+εθi++ε
3. Calculate J+iJi+ using to forward_propagation_n(x, y, vector_to_dictionary(θ+θ+ )).
- To compute J_minus[i]: do the same thing with θ?θ?
- Compute gradapprox[i]=J+i?J?i2εgradapprox[i]=Ji+?Ji?2ε
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:
Expected output:
| There is a mistake in the backward propagation! | difference = 0.285093156781 |
It seems that there were errors in the backward_propagation_n code we gave you! Good that you’ve implemented the gradient check. Go back to backward_propagation and try to find/correct the errors (Hint: check dW2 and db1). Rerun the gradient check when you think you’ve fixed it. Remember you’ll need to re-execute the cell defining backward_propagation_n() if you modify the code.
Can you get gradient check to declare your derivative computation correct? Even though this part of the assignment isn’t graded, we strongly urge you to try to find the bug and re-run gradient check until you’re convinced backprop is now correctly implemented.
Note
- Gradient Checking is slow! Approximating the gradient with ?J?θ≈J(θ+ε)?J(θ?ε)2ε?J?θ≈J(θ+ε)?J(θ?ε)2ε is computationally costly. For this reason, we don’t run gradient checking at every iteration during training. Just a few times to check if the gradient is correct.
- Gradient Checking, at least as we’ve presented it, doesn’t work with dropout. You would usually run the gradient check algorithm without dropout to make sure your backprop is correct, then add dropout.
Congrats, you can be confident that your deep learning model for fraud detection is working correctly! You can even use this to convince your CEO. :)
What you should remember from this notebook:
- Gradient checking verifies closeness between the gradients from backpropagation and the numerical approximation of the gradient (computed using forward propagation).
- Gradient checking is slow, so we don’t run it in every iteration of training. You would usually run it only to make sure your code is correct, then turn it off and use backprop for the actual learning process.
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