I tried to implement a function propagate() that computes the cost function and its gradient knowing that :
Forward Propagation:
- You get X
- You compute $A = \sigma(w^T X + b) = (a^{(0)}, a^{(1)}, ..., a^{(m-1)}, a^{(m)})$
- You calculate the cost function: $J = -\frac{1}{m}\sum_{i=1}^{m}y^{(i)}\log(a^{(i)})+(1-y^{(i)})\log(1-a^{(i)})$
Here are the two formulas I was using:
$$ \frac{\partial J}{\partial w} = \frac{1}{m}X(A-Y)^T\tag{7}$$ $$ \frac{\partial J}{\partial b} = \frac{1}{m} \sum_{i=1}^m (a^{(i)}-y^{(i)})\tag{8}$$
Which I coded as :
def propagate(w, b, X, Y):
"""
Implement the cost function and its gradient for the propagation explained above
Arguments:
w -- weights, a numpy array of size (num_px * num_px * 3, 1)
b -- bias, a scalar
X -- data of size (num_px * num_px * 3, number of examples)
Y -- true "label" vector (containing 0 if non-cat, 1 if cat) of size (1, number of examples)
Return:
cost -- negative log-likelihood cost for logistic regression
dw -- gradient of the loss with respect to w, thus same shape as w
db -- gradient of the loss with respect to b, thus same shape as b
Tips:
- Write your code step by step for the propagation. np.log(), np.dot()
"""
m = X.shape[1]
# FORWARD PROPAGATION (FROM X TO COST)
### START CODE HERE ### (≈ 2 lines of code)
A = sigmoid(np.dot(w.T,X)+b) # compute activation
cost = -1/m * np.sum(np.dot(Y.T,np.log(A)) + np.dot((1-Y).T,np.log(1-A))) # compute cost
### END CODE HERE ###
# BACKWARD PROPAGATION (TO FIND GRAD)
### START CODE HERE ### (≈ 2 lines of code)
dw = 1/m*np.dot(X,(A-Y).T)
db = 1/m*np.sum(A-Y)
### END CODE HERE ###
assert(dw.shape == w.shape)
assert(db.dtype == float)
cost = np.squeeze(cost)
assert(cost.shape == ())
grads = {"dw": dw,
"db": db}
return grads, cost
With the following example :
w, b, X, Y = np.array([[1.],[2.]]), 2., np.array([[1.,2.,-1.],[3.,4.,-3.2]]), np.array([[1,0,1]])
grads, cost = propagate(w, b, X, Y)
My output was :
dw = [[ 0.99845601]
[ 2.39507239]]
db = 0.00145557813678
cost = 10.6046359582
Whereas the expected output was
