深度学习笔记--Transformer中position encoding的源码理解与实现_positionalencoding代码

1--源码

import torch
import math
import numpy as np
import torch.nn as nn
 
class Pos_Embed(nn.Module):
    def __init__(self, channels, num_frames, num_joints):
        super().__init__()
        
        # 根据帧序和节点序生成位置向量
        pos_list = [] 
        for tk in range(num_frames):
            for st in range(num_joints):
                pos_list.append(st)
 
        position = torch.from_numpy(np.array(pos_list)).unsqueeze(1).float()  # num_frames*num_joints, 1
 
        pe = torch.zeros(num_frames * num_joints, channels)  # T*N, C
 
        div_term = torch.exp(torch.arange(0, channels, 2).float() * -(math.log(10000.0) / channels))
 
        pe[:, 0::2] = torch.sin(position * div_term)  # 偶数列 # 偶数C维度sin
        pe[:, 1::2] = torch.cos(position * div_term)  # 奇数列 # 奇数C维度cos
        pe = pe.view(num_frames, num_joints, channels).permute(2, 0, 1).unsqueeze(0)  # T N C -> C T N -> 1 C T N
        self.register_buffer('pe', pe)
 
    def forward(self, x):  # nctv # BCTN
        x = self.pe[:, :, :x.size(2)]
        return x
 
if __name__ == "__main__":
    B = 2
    C = 4
    T = 120
    N = 25
    x = torch.rand((B, C, T, N))
 
    Pos_embed_1 = Pos_Embed(C, T, N)
    PE = Pos_embed_1(x)
    # print(PE.shape) # 1 C T N
    x = x + PE
 
    print("All Done !")

2--源码分析与理解

原理理解：Positional Encoding（位置编码）

推荐视频：位置编码

代码解释：

①代码 div_term = torch.exp(torch.arange(0, channels, 2).float() * -(math.log(10000.0) / channels))：

令：channels = C, torch.arange(0, channels, 2).float() = k（则k = 0, 2, ..., C-2）；

-(math.log(10000.0) / channels)  $\large {\color{Red} =\frac{-\log_{e}1000}{C}}$；

则：torch.arange(0, channels, 2).float() * -(math.log(10000.0) / channels)$\large {\color{Red} =\frac{-k\log_{e}10000}{C}}$

torch.exp(torch.arange(0, channels, 2).float() * -(math.log(10000.0) / channels))$\LARGE {\color{Red} =e^{\frac{-k\log_{e}10000}{C}} = e^{\log_{e}\frac{-10000k}{C}} = \frac{-10000k}{C}}$;

②代码：pe[:, 0::2] = torch.sin(position * div_term) 和 pe[:, 1::2] = torch.cos(position * div_term)：

令：position = p，则position * div_term$\large {\color{Red} =p*\frac{-10000k}{C}=\frac{p}{10000^{\frac{k}{c}}}}$;

将k等价为2i，pe[:, 0::2]和pe[:, 1::2]分别取维度C的偶数列和奇数列，就可以得到上图绿框所示的公式。

3--参考

参考1

参考2