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.
In[13]:
# 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 ###
returnJ
In[14]:
x,theta=2,4
J=forward_propagation(x,theta)
print("J = "+str(J))
J = 8
Expected Output:
** J **8
Exercise: Now, implement the backward propagation step (derivative computation) of Figure 1. That is, compute the derivative of 𝐽(𝜃)=𝜃𝑥J(θ)=θx with respect to 𝜃θ. To save you from doing the calculus, you should get 𝑑𝑡ℎ𝑒𝑡𝑎=∂𝐽∂𝜃=𝑥dtheta=∂J∂θ=x.
In[15]:
# GRADED FUNCTION: backward_propagation
defbackward_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 ###
returndtheta
In[16]:
x,theta=2,4
dtheta=backward_propagation(x,theta)
print("dtheta = "+str(dtheta))
dtheta = 2
Expected 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:
𝜃+=𝜃+𝜀θ+=θ+ε
𝜃−=𝜃−𝜀θ−=θ−ε
𝐽+=𝐽(𝜃+)J+=J(θ+)
𝐽−=𝐽(𝜃−)J−=J(θ−)
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:
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=∣∣𝑔𝑟𝑎𝑑−𝑔𝑟𝑎𝑑𝑎𝑝𝑝𝑟𝑜𝑥∣∣2∣∣𝑔𝑟𝑎𝑑∣∣2+∣∣𝑔𝑟𝑎𝑑𝑎𝑝𝑝𝑟𝑜𝑥∣∣2(2)(2)difference=∣∣grad−gradapprox∣∣2∣∣grad∣∣2+∣∣gradapprox∣∣2
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.
In[18]:
# GRADED FUNCTION: gradient_check
defgradient_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# Step 1
thetaminus=theta-epsilon# Step 2
J_plus=thetaplus*x# Step 3
J_minus=thetaminus*x# Step 4
gradapprox=(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)
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 ###
ifdifference<1e-7:
print("The gradient is correct!")
else:
print("The gradient is wrong!")
returndifference
In[19]:
x,theta=2,4
difference=gradient_check(x,theta)
print("difference = "+str(difference))
The gradient is correct!
difference = 2.919335883291695e-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.
**Figure 2** : **deep neural network**
*LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID*
Let's look at your implementations for forward propagation and backward propagation.
In[52]:
defforward_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)
returncost,cache
Now, run backward propagation.
In[53]:
defbackward_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) ##这里没有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=1./m*np.sum(dZ1,axis=1,keepdims=True) ##这里把4改成了1
gradients={"dZ3":dZ3,"dW3":dW3,"db3":db3,
"dA2":dA2,"dZ2":dZ2,"dW2":dW2,"db2":db2,
"dA1":dA1,"dZ1":dZ1,"dW1":dW1,"db1":db1}
returngradients
You 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.
**Figure 2** : **dictionary_to_vector() and vector_to_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]:
Set 𝜃+θ+ to np.copy(parameters_values)
Set 𝜃+𝑖θi+ to 𝜃+𝑖+𝜀θi++ε
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:
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=‖𝑔𝑟𝑎𝑑−𝑔𝑟𝑎𝑑𝑎𝑝𝑝𝑟𝑜𝑥‖2‖𝑔𝑟𝑎𝑑‖2+‖𝑔𝑟𝑎𝑑𝑎𝑝𝑝𝑟𝑜𝑥‖2(3)(3)difference=‖grad−gradapprox‖2‖grad‖2+‖gradapprox‖2
In[69]:
# GRADED FUNCTION: gradient_check_n
defgradient_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
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))
# Compute gradapprox
foriinrange(num_parameters):
# 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)# Step 1
thetaplus[i][0]=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]=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# Step 3'
### END CODE HERE ###
ifdifference>1e-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")
returndifference
In[70]:
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)
There is a mistake in the backward propagation! difference = 1.1890417878779317e-07
##最后的结果是1.1890417878779317e-07，比1e-07还要大，所以会提示Mistake，找来找去没有其他错误要改了，看了其他网友的结果也是一样，看来是作业题的问题
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.
网友评论