我写了这个函数：
def two_layer_model(X, Y, layers_dims, learning_rate = 0.0075, num_iterations = 3000, print_cost=False):
"""
Implements a two-layer neural network: LINEAR->RELU->LINEAR->SIGMOID.
Arguments:
X -- input data, of shape (n_x, number of examples)
Y -- true "label" vector (containing 1 if cat, 0 if non-cat), of shape (1, number of examples)
layers_dims -- dimensions of the layers (n_x, n_h, n_y)
num_iterations -- number of iterations of the optimization loop
learning_rate -- learning rate of the gradient descent update rule
print_cost -- If set to True, this will print the cost every 100 iterations
Returns:
parameters -- a dictionary containing W1, W2, b1, and b2
"""
np.random.seed(1)
grads = {}
costs = [] # to keep track of the cost
m = X.shape[1] # number of examples
#(n_x, n_h, n_y) = layers_dims
layers_dims=(n_x, n_h, n_y)
# Initialize parameters dictionary, by calling one of the functions you'd previously implemented
### START CODE HERE ### (≈ 1 line of code)
parameters = initialize_parameters(n_x, n_h, n_y)
### END CODE HERE ###
# Get W1, b1, W2 and b2 from the dictionary parameters.
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Loop (gradient descent)
for i in range(0, num_iterations):
# Forward propagation: LINEAR -> RELU -> LINEAR -> SIGMOID. Inputs: "X, W1, b1, W2, b2". Output: "A1, cache1, A2, cache2".
### START CODE HERE ### (≈ 2 lines of code)
A1, cache1 = linear_activation_forward(X, W1, b1, activation="relu")
A2, cache2 = linear_activation_forward(A1, W2, b2, activation="sigmoid")
### END CODE HERE ###
# Compute cost
### START CODE HERE ### (≈ 1 line of code)
cost = compute_cost(A2, Y)
### END CODE HERE ###
# Initializing backward propagation
dA2 = - (np.divide(Y, A2) - np.divide(1 - Y, 1 - A2))
# Backward propagation. Inputs: "dA2, cache2, cache1". Outputs: "dA1, dW2, db2; also dA0 (not used), dW1, db1".
### START CODE HERE ### (≈ 2 lines of code)
dA1, dW2, db2 = linear_activation_backward(dA2, cache2, activation="sigmoid")
dA0, dW1, db1 = linear_activation_backward(dA1, cache1, activation="relu")
### END CODE HERE ###
# Set grads['dWl'] to dW1, grads['db1'] to db1, grads['dW2'] to dW2, grads['db2'] to db2
grads['dW1'] = dW1
grads['db1'] = db1
grads['dW2'] = dW2
grads['db2'] = db2
# Update parameters.
### START CODE HERE ### (approx. 1 line of code)
parameters = update_parameters(parameters, grads, learning_rate)
# Retrieve W1, b1, W2, b2 from parameters
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Print the cost every 100 training example
if print_cost and i % 100 == 0:
print("Cost after iteration {}: {}".format(i, np.squeeze(cost)))
if print_cost and i % 100 == 0:
costs.append(cost)
# plot the cost
plt.plot(np.squeeze(costs))
plt.ylabel('cost')
plt.xlabel('iterations (per hundreds)')
plt.title("Learning rate =" + str(learning_rate))
plt.show()
return parameters
然后
parameters, costs = two_layer_model(train_x, train_y, layers_dims = (n_x, n_h, n_y), num_iterations = 2, print_cost=False)
print("Cost after first iteration: " + str(costs[0]))
two_layer_model_test(two_layer_model)
然后我收到了这个错误：
ValueError Traceback（最近一次调用最后一次） 在 ----> 1个参数，cost =two_layer_model(train_x,train_y,layers_dims =(n_x,n_h,n_y),num_iterations = 2,print_cost=False) 2 3 print("第一次迭代后的成本：" + str(costs[0])) 4 5 二层模型测试（二层模型）
ValueError：需要解包的值太多（预计为 2 个）
请帮助我。
在函数 Two_layer_model 中，仅返回
parametersreturn parametersreturn parameters, costs这是来自 mhmd nojim 的正确代码
1- 从 public_tests 导入 * 在 jupyter 的第一个单元格中导入 public_tests
import time
import numpy as np
import h5py
import matplotlib.pyplot as plt
import scipy
from PIL import Image
from scipy import ndimage
from dnn_app_utils_v3 import *
from public_tests import *
%matplotlib inline
plt.rcParams['figure.figsize'] = (5.0, 4.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
%load_ext autoreload
%autoreload 2
np.random.seed(1)
2- 在 deftwo_layer_model 中替换以下函数
### GRADED FUNCTION: two_layer_model
def two_layer_model(X, Y, layers_dims, learning_rate = 0.0075, num_iterations = 3000, print_cost=False):
np.random.seed(1)
grads = {}
costs = [] # to keep track of the cost
m = X.shape[1] # number of examples
(n_x, n_h, n_y) = layers_dims
# Initialize parameters dictionary, by calling one of the functions you'd previously implemented
### START CODE HERE ### (≈ 1 line of code)
parameters = initialize_parameters(n_x, n_h, n_y)
# YOUR CODE ENDS HERE
# Get W1, b1, W2 and b2 from the dictionary parameters.
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Loop (gradient descent)
for i in range(0, num_iterations):
# Forward propagation: LINEAR -> RELU -> LINEAR -> SIGMOID. Inputs: "X, W1, b1, W2, b2". Output: "A1, cache1, A2, cache2".
### START CODE HERE ### (≈ 2 lines of code)
A1, cache1 = linear_activation_forward(X, W1, b1, activation = 'relu')
A2, cache2 = linear_activation_forward(A1, W2, b2, activation = 'sigmoid')
# YOUR CODE ENDS HERE
# Compute cost
#(≈ 1 line of code)
# cost = ...
# YOUR CODE STARTS HERE
cost = compute_cost(A2, Y)
### END CODE HERE ###
# Initializing backward propagation
dA2 = - (np.divide(Y, A2) - np.divide(1 - Y, 1 - A2))
# Backward propagation. Inputs: "dA2, cache2, cache1". Outputs: "dA1, dW2, db2; also dA0 (not used), dW1, db1".
### START CODE HERE ### (≈ 2 lines of code)
dA1, dW2, db2 = linear_activation_backward(dA2, cache2, activation = 'sigmoid')
dA0, dW1, db1 = linear_activation_backward(dA1, cache1, activation = 'relu')
### END CODE HERE ###
# Set grads['dWl'] to dW1, grads['db1'] to db1, grads['dW2'] to dW2, grads['db2'] to db2
grads['dW1'] = dW1
grads['db1'] = db1
grads['dW2'] = dW2
grads['db2'] = db2
# Update parameters.
### START CODE HERE ### (approx. 1 line of code)
parameters = update_parameters(parameters, grads, learning_rate)
# YOUR CODE ENDS HERE
# Retrieve W1, b1, W2, b2 from parameters
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Print the cost every 100 training example
if print_cost and i % 100 == 0 or i == num_iterations - 1:
print("Cost after iteration {}: {}".format(i, np.squeeze(cost)))
if i % 100 == 0 or i == num_iterations:
costs.append(cost)
return parameters , costs
问题解决了吗？我还是有同样的错误；