python实现单隐层神经网络
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2022-06-14 11:26:09
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功能:实现用神经网络对多样本数据进行分类训练
改进方法:应该把每次训练好的参数写入txt或json中,可以通过选择是否重新训练来调节 预测速度和 准确率的关系
demo.py
import numpy as np
import matplotlib.pyplot as plt
from planar_utils import plot_decision_boundary, sigmoid, load_planar_dataset
def layer_sizes(X, Y):
"""
Arguments:
X -- input dataset of shape (input size, number of examples)
Y -- labels of shape (output size, number of examples)
Returns:
n_x -- the size of the input layer
n_h -- the size of the hidden layer
n_y -- the size of the output layer
"""
### START CODE HERE ### (≈ 3 lines of code)
n_x = X.shape[0] # size of input layer
n_y = Y.shape[0] # size of output layer
### END CODE HERE ###
return (n_x, n_y)
np.random.seed(1) # set a seed so that the results are consistent
def initialize_parameters(n_x, n_h, n_y):
"""
Argument:
n_x -- size of the input layer
n_h -- size of the hidden layer
n_y -- size of the output layer
Returns:
params -- python dictionary containing your parameters:
W1 -- weight matrix of shape (n_h, n_x)
b1 -- bias vector of shape (n_h, 1)
W2 -- weight matrix of shape (n_y, n_h)
b2 -- bias vector of shape (n_y, 1)
"""
np.random.seed(2) # we set up a seed so that your output matches ours although the initialization is random.
### START CODE HERE ### (≈ 4 lines of code)
W1 = np.random.randn(n_h, n_x)*0.01
b1 = np.zeros((n_h, 1))
W2 = np.random.randn(n_y, n_h)*0.01
b2 = np.zeros((n_y, 1))
### END CODE HERE ###
assert (W1.shape == (n_h, n_x))
assert (b1.shape == (n_h, 1))
assert (W2.shape == (n_y, n_h))
assert (b2.shape == (n_y, 1))
parameters = {"W1": W1,
"b1": b1,
"W2": W2,
"b2": b2}
return parameters
def forward_propagation(X, parameters):
"""
Argument:
X -- input data of size (n_x, m)
parameters -- python dictionary containing your parameters (output of initialization function)
Returns:
A2 -- The sigmoid output of the second activation
cache -- a dictionary containing "Z1", "A1", "Z2" and "A2"
"""
# Retrieve each parameter from the dictionary "parameters"
### START CODE HERE ### (≈ 4 lines of code)
W1 = parameters['W1']
b1 = parameters['b1']
W2 = parameters['W2']
b2 = parameters['b2']
### END CODE HERE ###
# Implement Forward Propagation to calculate A2 (probabilities)
### START CODE HERE ### (≈ 4 lines of code)
Z1 = np.dot(W1, X) + b1
A1 = np.tanh(Z1)
Z2 = np.dot(W2, A1) + b2
A2 = sigmoid(Z2)
### END CODE HERE ###
assert(A2.shape == (1, X.shape[1]))
cache = {"Z1": Z1,
"A1": A1,
"Z2": Z2,
"A2": A2}
return A2, cache
def compute_cost(A2, Y, parameters):
"""
Computes the cross-entropy cost given in equation (13)
Arguments:
A2 -- The sigmoid output of the second activation, of shape (1, number of examples)
Y -- "true" labels vector of shape (1, number of examples)
parameters -- python dictionary containing your parameters W1, b1, W2 and b2
Returns:
cost -- cross-entropy cost given equation (13)
"""
m = float(Y.shape[1]) # number of example
# Compute the cross-entropy cost
### START CODE HERE ### (≈ 2 lines of code)
logprobs = np.multiply(np.log(A2),Y) + np.multiply((1-Y), (np.log(1-A2)))
cost = -1/m * np.sum(logprobs)
### END CODE HERE ###
cost = np.squeeze(cost) # makes sure cost is the dimension we expect.
# E.g., turns [[17]] into 17
assert(isinstance(cost, float))
return cost
def backward_propagation(parameters, cache, X, Y):
"""
Implement the backward propagation using the instructions above.
Arguments:
parameters -- python dictionary containing our parameters
cache -- a dictionary containing "Z1", "A1", "Z2" and "A2".
X -- input data of shape (2, number of examples)
Y -- "true" labels vector of shape (1, number of examples)
Returns:
grads -- python dictionary containing your gradients with respect to different parameters
"""
m = float(X.shape[1])
# First, retrieve W1 and W2 from the dictionary "parameters".
### START CODE HERE ### (≈ 2 lines of code)
W1 = parameters['W1']
W2 = parameters['W2']
### END CODE HERE ###
# Retrieve also A1 and A2 from dictionary "cache".
### START CODE HERE ### (≈ 2 lines of code)
A1 = cache['A1']
A2 = cache['A2']
### END CODE HERE ###
# Backward propagation: calculate dW1, db1, dW2, db2.
### START CODE HERE ### (≈ 6 lines of code, corresponding to 6 equations on slide above)
dZ2= A2 - Y
dW2 =1/m * np.dot(dZ2, A1.T)
db2 =1/m * np.sum(dZ2, axis=1, keepdims=True)
dZ1 = np.dot(W2.T, dZ2) * (1 - np.power(A1, 2))
dW1 = 1/m * np.dot(dZ1, X.T)
db1 =1/m * np.sum(dZ1, axis=1, keepdims=True)
### END CODE HERE ###
grads = {"dW1": dW1,
"db1": db1,
"dW2": dW2,
"db2": db2}
return grads
def update_parameters(parameters, grads, learning_rate = 1.2):
"""
Updates parameters using the gradient descent update rule given above
Arguments:
parameters -- python dictionary containing your parameters
grads -- python dictionary containing your gradients
Returns:
parameters -- python dictionary containing your updated parameters
"""
# Retrieve each parameter from the dictionary "parameters"
### START CODE HERE ### (≈ 4 lines of code)
W1 = parameters['W1']
b1 = parameters['b1']
W2 = parameters['W2']
b2 = parameters['b2']
### END CODE HERE ###
# Retrieve each gradient from the dictionary "grads"
### START CODE HERE ### (≈ 4 lines of code)
dW1 = grads["dW1"]
db1 = grads["db1"]
dW2 = grads["dW2"]
db2 = grads["db2"]
## END CODE HERE ###
# Update rule for each parameter
### START CODE HERE ### (≈ 4 lines of code)
W1 = W1 - learning_rate * dW1
b1 = b1 - learning_rate * db1
W2 = W2 - learning_rate * dW2
b2 = b2 - learning_rate * db2
### END CODE HERE ###
parameters = {"W1": W1,
"b1": b1,
"W2": W2,
"b2": b2}
return parameters
def nn_model(X, Y, n_h, num_iterations = 10000, print_cost=False):
"""
Arguments:
X -- dataset of shape (2, number of examples)
Y -- labels of shape (1, number of examples)
n_h -- size of the hidden layer
num_iterations -- Number of iterations in gradient descent loop
print_cost -- if True, print the cost every 1000 iterations
Returns:
parameters -- parameters learnt by the model. They can then be used to predict.
"""
np.random.seed(3)
# Initialize parameters, then retrieve W1, b1, W2, b2. Inputs: "n_x, n_h, n_y". Outputs = "W1, b1, W2, b2, parameters".
### START CODE HERE ### (≈ 5 lines of code)
n_x, n_y = layer_sizes(X, Y)
parameters = initialize_parameters(n_x, n_h, n_y)
### END CODE HERE ###
# Loop (gradient descent)
for i in range(0, num_iterations):
### START CODE HERE ### (≈ 4 lines of code)
# Forward propagation. Inputs: "X, parameters". Outputs: "A2, cache".
A2,cache = forward_propagation(X, parameters)
# Cost function. Inputs: "A2, Y, parameters". Outputs: "cost".
cost = compute_cost(A2, Y, parameters)
# Backpropagation. Inputs: "parameters, cache, X, Y". Outputs: "grads".
grads = backward_propagation(parameters, cache, X, Y)
# Gradient descent parameter update. Inputs: "parameters, grads". Outputs: "parameters".
parameters = update_parameters(parameters, grads)
### END CODE HERE ###
# Print the cost every 1000 iterations
if print_cost and i % 1000 == 0:
print ("Cost after iteration %i: %f" %(i, cost))
return parameters
def predict(parameters, X):
"""
Using the learned parameters, predicts a class for each example in X
Arguments:
parameters -- python dictionary containing your parameters
X -- input data of size (n_x, m)
Returns
predictions -- vector of predictions of our model (red: 0 / blue: 1)
"""
# Computes probabilities using forward propagation, and classifies to 0/1 using 0.5 as the threshold.
### START CODE HERE ### (≈ 2 lines of code)
A2, cache = forward_propagation(X, parameters)
predictions = np.array( [1 if x >0.5 else 0 for x in A2.reshape(-1,1)] ).reshape(A2.shape) # 这一行代码的作用详见下面代码示例
### END CODE HERE ###
return predictions
X, Y = load_planar_dataset()
# print(X.shape) #(2, 400) 在load中进行了转置,其中2代表单个样本特征数,400 代表样本数
# print(Y.shape) #(1, 400) 在load中进行了转置,本例中1代表输出层个数,400 代表样本数
plt.scatter(X[0, :], X[1, :], c=Y, s=40); #c代表色彩或颜色序列 s代表标量或形如shape[n,]数组
# Build a model with a n_h-dimensional hidden layer
parameters = nn_model(X, Y, n_h = 4, num_iterations = 10000, print_cost=True)
# Plot the decision boundary
plot_decision_boundary(lambda x: predict(parameters, x.T), X, Y)
pre = predict(parameters, X)
print ('training accuracy: %.2f' % (np.mean(pre == Y[0])))
plt.title("Decision Boundary for hidden layer size " + str(4))
plt.show()
planar_utils.py
'''
Created on 2018年7月18日
@author: hcl
'''
import matplotlib.pyplot as plt
import numpy as np
def plot_decision_boundary(model, X, y):
# Set min and max values and give it some padding
x_min, x_max = X[0, :].min() - 1, X[0, :].max() + 1
y_min, y_max = X[1, :].min() - 1, X[1, :].max() + 1
h = 0.01
# Generate a grid of points with distance h between them
xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))
# Predict the function value for the whole grid
Z = model(np.c_[xx.ravel(), yy.ravel()])
Z = Z.reshape(xx.shape)
# Plot the contour and training examples
plt.contourf(xx, yy, Z)
plt.ylabel('x2')
plt.xlabel('x1')
plt.scatter(X[0, :], X[1, :], c=y)
def sigmoid(x):
"""
Compute the sigmoid of x
Arguments:
x -- A scalar or numpy array of any size.
Return:
s -- sigmoid(x)
"""
s = 1/(1+np.exp(-x))
return s
def load_planar_dataset():
np.random.seed(1)
m = 400 # number of examples
N = int(m/2) # number of points per class
D = 2 # dimensionality
X = np.zeros((m,D)) # data matrix where each row is a single example
Y = np.zeros((m,1), dtype='uint8') # labels vector (0 for red, 1 for blue)
a = 4 # maximum ray of the flower
for j in range(2):
ix = range(N*j,N*(j+1))
t = np.linspace(j*3.12,(j+1)*3.12,N) + np.random.randn(N)*0.2 # theta
r = a*np.sin(4*t) + np.random.randn(N)*0.2 # radius
X[ix] = np.c_[r*np.sin(t), r*np.cos(t)]
Y[ix] = j
X = X.T
Y = Y.T
return X, Y
输出:
Cost after iteration 0: 0.693048
Cost after iteration 1000: 0.288083
Cost after iteration 2000: 0.254385
Cost after iteration 3000: 0.233864
Cost after iteration 4000: 0.226792
Cost after iteration 5000: 0.222644
Cost after iteration 6000: 0.219731
Cost after iteration 7000: 0.217504
Cost after iteration 8000: 0.219504
Cost after iteration 9000: 0.218571
training accuracy: 0.91
1:通过修改对应的X Y数据集 可以训练不同的数据
X = np.array([ [0,0,1],
[1,1,1],
[1,0,1],
[0,1,1] ]).T
Y = np.array([[0,1,1,0]])
输出;
Cost after iteration 0: 0.693173
Cost after iteration 1000: 0.000294
Cost after iteration 2000: 0.000138
Cost after iteration 3000: 0.000090
Cost after iteration 4000: 0.000066
Cost after iteration 5000: 0.000052
Cost after iteration 6000: 0.000043
Cost after iteration 7000: 0.000037
Cost after iteration 8000: 0.000032
Cost after iteration 9000: 0.000028
2:在planar_utils.py中添加
并在开头导入:import sklearn.datasets
def load_extra_datasets():
N = 200
noisy_circles = sklearn.datasets.make_circles(n_samples=N, factor=.5, noise=.3)
noisy_moons = sklearn.datasets.make_moons(n_samples=N, noise=.2)
blobs = sklearn.datasets.make_blobs(n_samples=N, random_state=5, n_features=2, centers=6)
gaussian_quantiles = sklearn.datasets.make_gaussian_quantiles(mean=None, cov=0.5, n_samples=N, n_features=2, n_classes=2, shuffle=True, random_state=None)
no_structure = np.random.rand(N, 2), np.random.rand(N, 2)
return noisy_circles, noisy_moons, blobs, gaussian_quantiles, no_structure
在demo.py中 导入load_extra_datasets
并将X Y数据来源 改为:
noisy_circles, noisy_moons, blobs, gaussian_quantiles, no_structure = load_extra_datasets()
datasets = {'noisy_circles':noisy_circles,
'noisy_moons':noisy_moons,
'blobs':blobs,
'gaussian_quantiles':gaussian_quantiles,
}
dataset = 'noisy_moons'
X,Y = datasets[dataset]
X,Y = X.T, Y.reshape(1,Y.shape[0])
if dataset == 'blobs':
Y = Y % 2
plt.scatter(X[0, :], X[1, :], c=Y, s=40)
训练输出:
Cost after iteration 0: 0.692994
Cost after iteration 1000: 0.299064
Cost after iteration 2000: 0.131752
Cost after iteration 3000: 0.124443
Cost after iteration 4000: 0.121406
Cost after iteration 5000: 0.118751
Cost after iteration 6000: 0.116311
Cost after iteration 7000: 0.114297
Cost after iteration 8000: 0.112747
Cost after iteration 9000: 0.111537
training accuracy: 0.95
3、查看不同隐藏层节点数对预测数据精确度的影响
在获取X Y数据集以后:
plt.figure(figsize=(16, 32))
hidden_layer_sizes = [1, 2, 3, 4, 5, 20, 50]
for i, n_h in enumerate(hidden_layer_sizes):
plt.subplot(5, 2, i+1)
plt.title('Hidden Layer of size %d' % n_h)
parameters = nn_model(X, Y, n_h, num_iterations = 5000)
plot_decision_boundary(lambda x: predict(parameters, x.T), X, Y)
predictions = predict(parameters, X)
accuracy = float((np.dot(Y,predictions.T) + np.dot(1-Y,1-predictions.T))/float(Y.size)*100)
print ("Accuracy for {} hidden units: {} %".format(n_h, accuracy))
plt.show()
输出:
Accuracy for 1 hidden units: 67.5 %
Accuracy for 2 hidden units: 67.25 %
Accuracy for 3 hidden units: 90.75 %
Accuracy for 4 hidden units: 90.5 %
Accuracy for 5 hidden units: 91.25 %
Accuracy for 20 hidden units: 90.0 %
Accuracy for 50 hidden units: 90.75 %
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