如何在两层多层感知器中更新学习率？

Question

鉴于 XOR 问题：

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

还有一个简单的

• 两层多层感知器 (MLP) 与
• 它们和
之间的 sigmoid 激活
• 均方误差 (MSE) 作为损失函数/优化标准

[代码]：

def sigmoid(x): # Returns values that sums to one.
    return 1 / (1 + np.exp(-x))

def sigmoid_derivative(sx): # For backpropagation.
    # See https://math.stackexchange.com/a/1225116
    return sx * (1 - sx)

# Cost functions.
def mse(predicted, truth):
    return np.sum(np.square(truth - predicted))

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

# Define the shape of the weight vector.
num_data, input_dim = X.shape
# Lets set the dimensions for the intermediate layer.
hidden_dim = 5
# Initialize weights between the input layers and the hidden layer.
W1 = np.random.random((input_dim, hidden_dim))

# Define the shape of the output vector. 
output_dim = len(Y.T)
# Initialize weights between the hidden layers and the output layer.
W2 = np.random.random((hidden_dim, output_dim))

并且给定停止标准作为固定编号。固定学习率为 0.3 的时期数（通过 X 和 Y 的迭代次数）：

# Initialize weigh
num_epochs = 10000
learning_rate = 0.3

当我运行前向-后向传播并更新每个 epoch 中的权重时，我应该如何更新权重？

我尝试简单地将学习率的乘积与反向传播导数的点积与层输出相加，但模型仍然只更新一个方向的权重，导致所有权重降级到接近零。

for epoch_n in range(num_epochs):
    layer0 = X
    # Forward propagation.

    # Inside the perceptron, Step 2. 
    layer1 = sigmoid(np.dot(layer0, W1))
    layer2 = sigmoid(np.dot(layer1, W2))

    # Back propagation (Y -> layer2)

    # How much did we miss in the predictions?
    layer2_error = mse(layer2, Y)

    #print(layer2_error)
    # In what direction is the target value?
    # Were we really close? If so, don't change too much.
    layer2_delta = layer2_error * sigmoid_derivative(layer2)

    # Back propagation (layer2 -> layer1)
    # How much did each layer1 value contribute to the layer2 error (according to the weights)?
    layer1_error = np.dot(layer2_delta, W2.T)
    layer1_delta = layer1_error * sigmoid_derivative(layer1)

    # update weights
    W2 += - learning_rate * np.dot(layer1.T, layer2_delta)
    W1 += - learning_rate * np.dot(layer0.T, layer1_delta)
    #print(np.dot(layer0.T, layer1_delta))
    #print(epoch_n, list((layer2)))

    # Log the loss value as we proceed through the epochs.
    losses.append(layer2_error.mean())

应该如何正确更新权重？

完整代码：

from itertools import chain
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(0)

def sigmoid(x): # Returns values that sums to one.
    return 1 / (1 + np.exp(-x))

def sigmoid_derivative(sx):
    # See https://math.stackexchange.com/a/1225116
    return sx * (1 - sx)

# Cost functions.
def mse(predicted, truth):
    return np.sum(np.square(truth - predicted))

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

# Define the shape of the weight vector.
num_data, input_dim = X.shape
# Lets set the dimensions for the intermediate layer.
hidden_dim = 5
# Initialize weights between the input layers and the hidden layer.
W1 = np.random.random((input_dim, hidden_dim))

# Define the shape of the output vector. 
output_dim = len(Y.T)
# Initialize weights between the hidden layers and the output layer.
W2 = np.random.random((hidden_dim, output_dim))

# Initialize weigh
num_epochs = 10000
learning_rate = 0.3

losses = []

for epoch_n in range(num_epochs):
    layer0 = X
    # Forward propagation.

    # Inside the perceptron, Step 2. 
    layer1 = sigmoid(np.dot(layer0, W1))
    layer2 = sigmoid(np.dot(layer1, W2))

    # Back propagation (Y -> layer2)

    # How much did we miss in the predictions?
    layer2_error = mse(layer2, Y)

    #print(layer2_error)
    # In what direction is the target value?
    # Were we really close? If so, don't change too much.
    layer2_delta = layer2_error * sigmoid_derivative(layer2)

    # Back propagation (layer2 -> layer1)
    # How much did each layer1 value contribute to the layer2 error (according to the weights)?
    layer1_error = np.dot(layer2_delta, W2.T)
    layer1_delta = layer1_error * sigmoid_derivative(layer1)

    # update weights
    W2 += - learning_rate * np.dot(layer1.T, layer2_delta)
    W1 += - learning_rate * np.dot(layer0.T, layer1_delta)
    #print(np.dot(layer0.T, layer1_delta))
    #print(epoch_n, list((layer2)))

    # Log the loss value as we proceed through the epochs.
    losses.append(layer2_error.mean())

# Visualize the losses
plt.plot(losses)
plt.show()

我是否在反向传播中遗漏了什么？

可能我错过了从成本到第二层的导数？

---

已编辑

我意识到我错过了从成本到第二层的偏导数，添加后：

# Cost functions.
def mse(predicted, truth):
    return 0.5 * np.sum(np.square(predicted - truth)).mean()

def mse_derivative(predicted, truth):
    return predicted - truth

使用更新的跨时期反向传播循环：

for epoch_n in range(num_epochs):
    layer0 = X
    # Forward propagation.

    # Inside the perceptron, Step 2. 
    layer1 = sigmoid(np.dot(layer0, W1))
    layer2 = sigmoid(np.dot(layer1, W2))

    # Back propagation (Y -> layer2)

    # How much did we miss in the predictions?
    cost_error = mse(layer2, Y)
    cost_delta = mse_derivative(layer2, Y)

    #print(layer2_error)
    # In what direction is the target value?
    # Were we really close? If so, don't change too much.
    layer2_error = np.dot(cost_delta, cost_error)
    layer2_delta = layer2_error *  sigmoid_derivative(layer2)

    # Back propagation (layer2 -> layer1)
    # How much did each layer1 value contribute to the layer2 error (according to the weights)?
    layer1_error = np.dot(layer2_delta, W2.T)
    layer1_delta = layer1_error * sigmoid_derivative(layer1)

    # update weights
    W2 += - learning_rate * np.dot(layer1.T, layer2_delta)
    W1 += - learning_rate * np.dot(layer0.T, layer1_delta)

它似乎训练和学习了异或……

但现在问题来了，layer2_error 和 layer2_delta 计算是否正确，即 以下部分代码是否正确？

# How much did we miss in the predictions?
cost_error = mse(layer2, Y)
cost_delta = mse_derivative(layer2, Y)

#print(layer2_error)
# In what direction is the target value?
# Were we really close? If so, don't change too much.
layer2_error = np.dot(cost_delta, cost_error)
layer2_delta = layer2_error *  sigmoid_derivative(layer2)

在cost_delta 和cost_error 上为layer2_error 做点积是否正确？还是layer2_error 就等于cost_delta？

即

# How much did we miss in the predictions?
cost_error = mse(layer2, Y)
cost_delta = mse_derivative(layer2, Y)

#print(layer2_error)
# In what direction is the target value?
# Were we really close? If so, don't change too much.
layer2_error = cost_delta
layer2_delta = layer2_error *  sigmoid_derivative(layer2)

Answer 1

是的，当我们更新权重时，将残差 (cost_error) 与增量值相乘是正确的。

但是，是否做点积并不重要，因为它cost_error 是一个标量。所以，一个简单的乘法就足够了。但是，我们必须乘以代价函数的梯度，因为这是我们开始反向传播的地方（即它是反向传播的入口点）。

另外，下面的函数可以简化：

def mse(predicted, truth):
    return 0.5 * np.sum(np.square(predicted - truth)).mean()

作为

def mse(predicted, truth):
    return 0.5 * np.mean(np.square(predicted - truth))