有人可以帮忙吗?我正在从deeplearning.ai做深度学习我在课程1的第二周我的传播功能如下正向传播:
您得到X您计算A =σ(wTX + b)=(a(1),a(2),...,a(m-1),a(m))A =σ(wTX + b)=(a( 1),(2),...,第(m-1),(M))您可以计算成本函数:J = −1m∑mi = 1y(i)log(a(i))+(1-y(i))log(1-a(i))J = −1m∑i = 1my (ⅰ)log(A(I))+(1-γ(i))的log(1-A(I))
# GRADED FUNCTION: propagate
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(Y*np.log(A)+(1-Y)*np.log(1-A), axis=1,keepdims=True) # compute cost
### END CODE HERE ###
# BACKWARD PROPAGATION (TO FIND GRAD)
### START CODE HERE ### (≈ 2 lines of code)
dw = 1/m*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
和
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)
print ("dw = " + str(grads["dw"]))
print ("db = " + str(grads["db"]))
print ("cost = " + str(cost))
但是我遇到以下错误
TypeError Traceback (most recent call last)
----> 3 grads, cost = propagate(w, b, X, Y)
---> 26 A = sigmoid(np.dot((w.T,X)+b)) # compute activation
TypeError: can only concatenate tuple (not "float") to tuple
如何解决?我的S型函数工作正常。]
您的错误在表达式np.dot((w.T,X)+b)
中。在此表达式中,将函数np.dot
应用于one参数(w.T,X)+b
。这又由元组(w.T, X)
和您尝试加在一起的浮点数b
组成(这是不可能的。)
问题在于您的括号。您想使用two
参数w.T
和X
调用该函数,然后将b
添加到结果中:np.dot(w.T,X)+b
。