深度学习 - RNN训练过程推演

标签：loss partial RNN tanh text 推演 pmatrix 深度 frac

1. 数据准备

字符序列 “hello” 转换为 one-hot 编码表示：

• 输入: [‘h’, ‘e’, ‘l’, ‘l’]
• 输出: [‘e’, ‘l’, ‘l’, ‘o’]

2. 初始化参数

假设我们使用一个单层的 RNN，隐藏层大小为2。初始参数如下：

W x h = ( 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ) , W h h = ( 0.1 0.2 0.3 0.4 ) , W h y = ( 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ) W_{xh} = \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \\ 0.5 & 0.6 \\ 0.7 & 0.8 \end{pmatrix}, \quad W_{hh} = \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \end{pmatrix}, \quad W_{hy} = \begin{pmatrix} 0.1 & 0.2 & 0.3 & 0.4 \\ 0.5 & 0.6 & 0.7 & 0.8 \end{pmatrix} Wxh​= ​0.10.30.50.7​0.20.40.60.8​ ​,Whh​=(0.10.3​0.20.4​),Why​=(0.10.5​0.20.6​0.30.7​0.40.8​)

偏置项初始化为0。

3. 前向传播和反向传播

时间步 1（输入 ‘h’）：

输入向量 x 1 = [ 1 , 0 , 0 , 0 ] x_1 = [1, 0, 0, 0] x1​=[1,0,0,0]

h 1 = tanh ⁡ ( W x h x 1 + W h h h 0 ) = tanh ⁡ ( ( 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ) ( 1 0 0 0 ) + ( 0.1 0.2 0.3 0.4 ) ( 0 0 ) ) = tanh ⁡ ( ( 0.1 0.3 ) ) = ( 0.0997 0.2913 ) h_1 = \tanh(W_{xh} x_1 + W_{hh} h_0) = \tanh \left( \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \\ 0.5 & 0.6 \\ 0.7 & 0.8 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right) = \tanh \left( \begin{pmatrix} 0.1 \\ 0.3 \end{pmatrix} \right) = \begin{pmatrix} 0.0997 \\ 0.2913 \end{pmatrix} h1​=tanh(Wxh​x1​+Whh​h0​)=tanh ​ ​0.10.30.50.7​0.20.40.60.8​ ​ ​1000​ ​+(0.10.3​0.20.4​)(00​) ​=tanh((0.10.3​))=(0.09970.2913​)

y 1 = W h y h 1 = ( 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ) ( 0.0997 0.2913 ) = ( 0.1695 0.3889 0.6083 0.8277 ) y_1 = W_{hy} h_1 = \begin{pmatrix} 0.1 & 0.2 & 0.3 & 0.4 \\ 0.5 & 0.6 & 0.7 & 0.8 \end{pmatrix} \begin{pmatrix} 0.0997 \\ 0.2913 \end{pmatrix} = \begin{pmatrix} 0.1695 \\ 0.3889 \\ 0.6083 \\ 0.8277 \end{pmatrix} y1​=Why​h1​=(0.10.5​0.20.6​0.30.7​0.40.8​)(0.09970.2913​)= ​0.16950.38890.60830.8277​ ​

预测值 y ^ 1 = softmax ( y 1 ) \hat{y}_1 = \text{softmax}(y_1) y^​1​=softmax(y1​)

假设真实输出为 ‘e’，对应 one-hot 编码为 y 1 = [ 0 , 1 , 0 , 0 ] y_1 = [0, 1, 0, 0] y1​=[0,1,0,0]。

交叉熵损失函数：

loss 1 = − ∑ i y 1 i log ⁡ ( y ^ 1 i ) \text{loss}_1 = - \sum_{i} y_{1i} \log(\hat{y}_{1i}) loss1​=−i∑​y1i​log(y^​1i​)

梯度计算：

∂ loss 1 ∂ W h y = ( y ^ 1 − y 1 ) h 1 T \frac{\partial \text{loss}_1}{\partial W_{hy}} = (\hat{y}_1 - y_1) h_1^T ∂Why​∂loss1​​=(y^​1​−y1​)h1T​

∂ loss 1 ∂ W x h = ∂ loss 1 ∂ h 1 ⋅ ∂ h 1 ∂ W x h \frac{\partial \text{loss}_1}{\partial W_{xh}} = \frac{\partial \text{loss}_1}{\partial h_1} \cdot \frac{\partial h_1}{\partial W_{xh}} ∂Wxh​∂loss1​​=∂h1​∂loss1​​⋅∂Wxh​∂h1​​

∂ loss 1 ∂ W h h = ∂ loss 1 ∂ h 1 ⋅ ∂ h 1 ∂ W h h \frac{\partial \text{loss}_1}{\partial W_{hh}} = \frac{\partial \text{loss}_1}{\partial h_1} \cdot \frac{\partial h_1}{\partial W_{hh}} ∂Whh​∂loss1​​=∂h1​∂loss1​​⋅∂Whh​∂h1​​

参数更新：

W x h = W x h − η ∂ loss 1 ∂ W x h W_{xh} = W_{xh} - \eta \frac{\partial \text{loss}_1}{\partial W_{xh}} Wxh​=Wxh​−η∂Wxh​∂loss1​​

W h h = W h h − η ∂ loss 1 ∂ W h h W_{hh} = W_{hh} - \eta \frac{\partial \text{loss}_1}{\partial W_{hh}} Whh​=Whh​−η∂Whh​∂loss1​​

W h y = W h y − η ∂ loss 1 ∂ W h y W_{hy} = W_{hy} - \eta \frac{\partial \text{loss}_1}{\partial W_{hy}} Why​=Why​−η∂Why​∂loss1​​

时间步 2（输入 ‘e’）：

使用更新后的 W x h W_{xh} Wxh​、 W h h W_{hh} Whh​ 和 W h y W_{hy} Why​ 参数。

输入向量 x 2 = [ 0 , 1 , 0 , 0 ] x_2 = [0, 1, 0, 0] x2​=[0,1,0,0]

h 2 = tanh ⁡ ( W x h x 2 + W h h h 1 ) = tanh ⁡ ( ( 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ) ( 0 1 0 0 ) + ( 0.1 0.2 0.3 0.4 ) ( 0.0997 0.2913 ) ) h_2 = \tanh(W_{xh} x_2 + W_{hh} h_1) = \tanh \left( \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \\ 0.5 & 0.6 \\ 0.7 & 0.8 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & 0.4 \end{pmatrix} \begin{pmatrix} 0.0997 \\ 0.2913 \end{pmatrix} \right) h2​=tanh(Wxh​x2​+Whh​h1​)=tanh ​ ​0.10.30.50.7​0.20.40.60.8​ ​ ​0100​ ​+(0.10.3​0.20.4​)(0.09970.2913​) ​

计算后得：

h 2 = tanh ⁡ ( ( 0.3 0.7 ) + ( 0.1283 0.2147 ) ) = tanh ⁡ ( ( 0.4283 0.9147 ) ) h_2 = \tanh \left( \begin{pmatrix} 0.3 \\ 0.7 \end{pmatrix} + \begin{pmatrix} 0.1283 \\ 0.2147 \end{pmatrix} \right) = \tanh \left( \begin{pmatrix} 0.4283 \\ 0.9147 \end{pmatrix} \right) h2​=tanh((0.30.7​)+(0.12830.2147​))=tanh((0.42830.9147​))

y 2 = W h y h 2 y_2 = W_{hy} h_2 y2​=Why​h2​

预测值 y ^ 2 = softmax ( y 2 ) \hat{y}_2 = \text{softmax}(y_2) y^​2​=softmax(y2​)

假设真实输出为 ‘l’，对应 one-hot 编码为 y 2 = [ 0 , 0 , 1 , 0 ] y_2 = [0, 0, 1, 0] y2​=[0,0,1,0]。

交叉熵损失函数：

loss 2 = − ∑ i y 2 i log ⁡ ( y ^ 2 i ) \text{loss}_2 = - \sum_{i} y_{2i} \log(\hat{y}_{2i}) loss2​=−i∑​y2i​log(y^​2i​)

梯度计算：

∂ loss 2 ∂ W h y = ( y ^ 2 − y 2 ) h 2 T \frac{\partial \text{loss}_2}{\partial W_{hy}} = (\hat{y}_2 - y_2) h_2^T ∂Why​∂loss2​​=(y^​2​−y2​)h2T​

∂ loss 2 ∂ W x h = ∂ loss 2 ∂ h 2 ⋅ ∂ h 2 ∂ W x h \frac{\partial \text{loss}_2}{\partial W_{xh}} = \frac{\partial \text{loss}_2}{\partial h_2} \cdot \frac{\partial h_2}{\partial W_{xh}} ∂Wxh​∂loss2​​=∂h2​∂loss2​​⋅∂Wxh​∂h2​​

∂ loss 2 ∂ W h h = ∂ loss 2 ∂ h 2 ⋅ ∂ h 2 ∂ W h h \frac{\partial \text{loss}_2}{\partial W_{hh}} = \frac{\partial \text{loss}_2}{\partial h_2} \cdot \frac{\partial h_2}{\partial W_{hh}} ∂Whh​∂loss2​​=∂h2​∂loss2​​⋅∂Whh​∂h2​​

参数更新：

W x h = W x h − η ∂ loss 2 ∂ W x h W_{xh} = W_{xh} - \eta \frac{\partial \text{loss}_2}{\partial W_{xh}} Wxh​=Wxh​−η∂Wxh​∂loss2​​

W h h = W h h − η ∂ loss 2 ∂ W h h W_{hh} = W_{hh} - \eta \frac{\partial \text{loss}_2}{\partial W_{hh}} Whh​=Whh​−η∂Whh​∂loss2​​

W h y = W h y − η ∂ loss 2 ∂ W h y W_{hy} = W_{hy} - \eta \frac{\partial \text{loss}_2}{\partial W_{hy}} Why​=Why​−η∂Why​∂loss2​​

时间步 3（输入 ‘l’）：

使用更新后的 W x h W_{xh} Wxh​、 W h h W_{hh} Whh​ 和 W h y W_{hy} Why​ 参数。

输入向量 x 3 = [ 0 , 0 , 1 , 0 ] x_3 = [0, 0, 1, 0] x3​=[0,0,1,0]

h 3 = tanh ⁡ ( W x h x 3 + W h h h 2 ) h_3 = \tanh(W_{xh} x_3 + W_{hh} h_2) h3​=tanh(Wxh​x3​+Whh​h2​)

计算后得：

h 3 = tanh ⁡ ( ( 0.5 1.2 ) + W h h h 2 ) h_3 = \tanh \left( \begin{pmatrix} 0.5 \\ 1.2 \end{pmatrix} + W_{hh} h_2 \right) h3​=tanh((0.51.2​)+Whh​h2​)

y 3 = W h y h 3 y_3 = W_{hy} h_3 y3​=Why​h3​

预测值 y ^ 3 = softmax ( y 3 ) \hat{y}_3 = \text{softmax}(y_3) y^​3​=softmax(y3​)

假设真实输出为 ‘l’，对应 one-hot 编码为 y 3 = [ 0 , 0 , 1 , 0 ] y_3 = [0, 0, 1, 0] y3​=[0,0,1,0]。

交叉熵损失函数：

$$
\text{loss}3 = - \sum{i} y_{3i} \log(\hat{y}_{3

i})
$$

梯度计算：

∂ loss 3 ∂ W h y = ( y ^ 3 − y 3 ) h 3 T \frac{\partial \text{loss}_3}{\partial W_{hy}} = (\hat{y}_3 - y_3) h_3^T ∂Why​∂loss3​​=(y^​3​−y3​)h3T​

∂ loss 3 ∂ W x h = ∂ loss 3 ∂ h 3 ⋅ ∂ h 3 ∂ W x h \frac{\partial \text{loss}_3}{\partial W_{xh}} = \frac{\partial \text{loss}_3}{\partial h_3} \cdot \frac{\partial h_3}{\partial W_{xh}} ∂Wxh​∂loss3​​=∂h3​∂loss3​​⋅∂Wxh​∂h3​​

∂ loss 3 ∂ W h h = ∂ loss 3 ∂ h 3 ⋅ ∂ h 3 ∂ W h h \frac{\partial \text{loss}_3}{\partial W_{hh}} = \frac{\partial \text{loss}_3}{\partial h_3} \cdot \frac{\partial h_3}{\partial W_{hh}} ∂Whh​∂loss3​​=∂h3​∂loss3​​⋅∂Whh​∂h3​​

参数更新：

W x h = W x h − η ∂ loss 3 ∂ W x h W_{xh} = W_{xh} - \eta \frac{\partial \text{loss}_3}{\partial W_{xh}} Wxh​=Wxh​−η∂Wxh​∂loss3​​

W h h = W h h − η ∂ loss 3 ∂ W h h W_{hh} = W_{hh} - \eta \frac{\partial \text{loss}_3}{\partial W_{hh}} Whh​=Whh​−η∂Whh​∂loss3​​

W h y = W h y − η ∂ loss 3 ∂ W h y W_{hy} = W_{hy} - \eta \frac{\partial \text{loss}_3}{\partial W_{hy}} Why​=Why​−η∂Why​∂loss3​​

时间步 4（输入 ‘l’）：

使用更新后的 W x h W_{xh} Wxh​、 W h h W_{hh} Whh​ 和 W h y W_{hy} Why​ 参数。

输入向量 x 4 = [ 0 , 0 , 1 , 0 ] x_4 = [0, 0, 1, 0] x4​=[0,0,1,0]

h 4 = tanh ⁡ ( W x h x 4 + W h h h 3 ) h_4 = \tanh(W_{xh} x_4 + W_{hh} h_3) h4​=tanh(Wxh​x4​+Whh​h3​)

计算后得：

h 4 = tanh ⁡ ( ( 0.5 1.2 ) + W h h h 3 ) h_4 = \tanh \left( \begin{pmatrix} 0.5 \\ 1.2 \end{pmatrix} + W_{hh} h_3 \right) h4​=tanh((0.51.2​)+Whh​h3​)

y 4 = W h y h 4 y_4 = W_{hy} h_4 y4​=Why​h4​

预测值 y ^ 4 = softmax ( y 4 ) \hat{y}_4 = \text{softmax}(y_4) y^​4​=softmax(y4​)

假设真实输出为 ‘o’，对应 one-hot 编码为 y 4 = [ 0 , 0 , 0 , 1 ] y_4 = [0, 0, 0, 1] y4​=[0,0,0,1]。

交叉熵损失函数：

loss 4 = − ∑ i y 4 i log ⁡ ( y ^ 4 i ) \text{loss}_4 = - \sum_{i} y_{4i} \log(\hat{y}_{4i}) loss4​=−i∑​y4i​log(y^​4i​)

梯度计算：

∂ loss 4 ∂ W h y = ( y ^ 4 − y 4 ) h 4 T \frac{\partial \text{loss}_4}{\partial W_{hy}} = (\hat{y}_4 - y_4) h_4^T ∂Why​∂loss4​​=(y^​4​−y4​)h4T​

∂ loss 4 ∂ W x h = ∂ loss 4 ∂ h 4 ⋅ ∂ h 4 ∂ W x h \frac{\partial \text{loss}_4}{\partial W_{xh}} = \frac{\partial \text{loss}_4}{\partial h_4} \cdot \frac{\partial h_4}{\partial W_{xh}} ∂Wxh​∂loss4​​=∂h4​∂loss4​​⋅∂Wxh​∂h4​​

∂ loss 4 ∂ W h h = ∂ loss 4 ∂ h 4 ⋅ ∂ h 4 ∂ W h h \frac{\partial \text{loss}_4}{\partial W_{hh}} = \frac{\partial \text{loss}_4}{\partial h_4} \cdot \frac{\partial h_4}{\partial W_{hh}} ∂Whh​∂loss4​​=∂h4​∂loss4​​⋅∂Whh​∂h4​​

参数更新：

W x h = W x h − η ∂ loss 4 ∂ W x h W_{xh} = W_{xh} - \eta \frac{\partial \text{loss}_4}{\partial W_{xh}} Wxh​=Wxh​−η∂Wxh​∂loss4​​

W h h = W h h − η ∂ loss 4 ∂ W h h W_{hh} = W_{hh} - \eta \frac{\partial \text{loss}_4}{\partial W_{hh}} Whh​=Whh​−η∂Whh​∂loss4​​

W h y = W h y − η ∂ loss 4 ∂ W h y W_{hy} = W_{hy} - \eta \frac{\partial \text{loss}_4}{\partial W_{hy}} Why​=Why​−η∂Why​∂loss4​​

4.代码实现

下面是一个使用 PyTorch 实现简单 RNN（循环神经网络）的示例代码，该代码将字符序列作为输入并预测下一个字符。我们将使用一个小的字符集进行演示。

安装 PyTorch

在开始之前，请确保您已安装 PyTorch。您可以使用以下命令进行安装：

pip install torch

RNN 实现示例

我们将实现一个字符级 RNN，用于从序列 “hello” 中预测下一个字符。字符集为 {‘h’, ‘e’, ‘l’, ‘o’}。

import torch
import torch.nn as nn
import torch.optim as optim
import numpy as np

# 定义字符集和字符到索引的映射
chars = ['h', 'e', 'l', 'o']
char_to_idx = {ch: idx for idx, ch in enumerate(chars)}
idx_to_char = {idx: ch for idx, ch in enumerate(chars)}

# 超参数
input_size = len(chars)
hidden_size = 10
output_size = len(chars)
num_layers = 1
learning_rate = 0.01
num_epochs = 100

# 准备数据
def char_to_tensor(char):
    tensor = torch.zeros(input_size)
    tensor[char_to_idx[char]] = 1.0
    return tensor

def string_to_tensor(string):
    tensor = torch.zeros(len(string), input_size)
    for idx, char in enumerate(string):
        tensor[idx][char_to_idx[char]] = 1.0
    return tensor

input_seq = "hell"
target_seq = "ello"

input_tensor = string_to_tensor(input_seq)
target_tensor = torch.tensor([char_to_idx[ch] for ch in target_seq])

# 定义 RNN 模型
class RNN(nn.Module):
    def __init__(self, input_size, hidden_size, output_size):
        super(RNN, self).__init__()
        self.hidden_size = hidden_size
        self.rnn = nn.RNN(input_size, hidden_size, num_layers, batch_first=True)
        self.fc = nn.Linear(hidden_size, output_size)

    def forward(self, x, hidden):
        out, hidden = self.rnn(x, hidden)
        out = self.fc(out[:, -1, :])
        return out, hidden

    def init_hidden(self):
        return torch.zeros(num_layers, 1, hidden_size)

# 初始化模型、损失函数和优化器
model = RNN(input_size, hidden_size, output_size)
criterion = nn.CrossEntropyLoss()
optimizer = optim.Adam(model.parameters(), lr=learning_rate)

# 训练模型
for epoch in range(num_epochs):
    hidden = model.init_hidden()
    model.zero_grad()
    
    input_seq = input_tensor.unsqueeze(0)
    output, hidden = model(input_seq, hidden)
    
    loss = criterion(output, target_tensor.unsqueeze(0))
    loss.backward()
    optimizer.step()
    
    if (epoch + 1) % 10 == 0:
        print(f'Epoch [{epoch + 1}/{num_epochs}], Loss: {loss.item():.4f}')

# 测试模型
def predict(model, char, hidden=None):
    if hidden is None:
        hidden = model.init_hidden()
    input_tensor = char_to_tensor(char).unsqueeze(0).unsqueeze(0)
    output, hidden = model(input_tensor, hidden)
    _, predicted_idx = torch.max(output, 1)
    return idx_to_char[predicted_idx.item()], hidden

hidden = model.init_hidden()
input_char = 'h'
predicted_seq = input_char
for _ in range(len(input_seq)):
    next_char, hidden = predict(model, input_char, hidden)
    predicted_seq += next_char
    input_char = next_char

print(f'Predicted sequence: {predicted_seq}')

代码说明

1. 数据准备：

   • 我们定义了一个简单的字符集 {‘h’, ‘e’, ‘l’, ‘o’}，并创建了字符到索引和索引到字符的映射。
   • char_to_tensor 函数将字符转换为 one-hot 向量。
   • string_to_tensor 函数将字符串转换为一系列 one-hot 向量。

2. 定义 RNN 模型：

   • RNN 类继承自 nn.Module，包含一个 RNN 层和一个全连接层。
   • forward 方法执行前向传播。
   • init_hidden 方法初始化隐藏状态。

3. 训练模型：

   • 我们使用交叉熵损失函数和 Adam 优化器。
   • 在每个训练周期，我们进行前向传播、计算损失、反向传播和参数更新。

4. 测试模型：

   • predict 函数根据给定的输入字符生成下一个字符。
   • 我们使用训练好的模型从字符 ‘h’ 开始生成一个字符序列。

运行该代码后，您将看到模型预测的字符序列，它会逐渐学会从输入序列中预测下一个字符。

标签：loss,partial,RNN,tanh,text,推演,pmatrix,深度,frac
From： https://blog.csdn.net/weixin_47552266/article/details/139659451