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csstudent1418
Asked: 2023-08-26 03:15:01 +0800 CST

PyTorch MSE 损失与直接计算相差 2 倍

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为什么result的结果torch.nn.functional.mse_loss(x1,x2)与直接计算MSE不同？

我要重现的测试代码：

import torch
import numpy as np

# Think of x1 as predicted 2D coordinates and x2 of ground truth
x1 = torch.rand(10,2)
x2 = torch.rand(10,2)

mse_torch = torch.nn.functional.mse_loss(x1,x2)
print(mse_torch) # 0.1557

mse_direct = torch.nn.functional.pairwise_distance(x1,x2).square().mean()
print(mse_direct) # 0.3314

mse_manual = 0
for i in range(len(x1)) :
    mse_manual += np.square(np.linalg.norm(x1[i]-x2[i])) / len(x1)
print(mse_manual) # 0.3314

正如我们所看到的，torch 的结果mse_loss是0.1557，与手动 MSE 计算的结果不同0.3314。

事实上，结果mse_loss恰好等于直接结果乘以点的维度（此处为 2）。

那是怎么回事？

pytorch

1 个回答

  1. Best Answer

    不同之处在于，torch.nn.function.mse_loss(x1,x2) 在计算平方误差时不对坐标应用求和运算。但是，torch.nn.function.pairwise_distance 和 np.linalg.norm 对坐标应用求和运算。您可以通过以下方式重现计算出的 mse 值：

    import torch
import numpy as np

x1 = torch.rand(10,2)
x2 = torch.rand(10,2)

mse_torch = torch.nn.functional.mse_loss(x1,x2)
print(mse_torch) # 0.1557

mse_manual = 0
x3 = torch.zeros(10,2)
for i in range(len(x1)) :
   x3[i,:1] +=(torch.nn.functional.pairwise_distance(x1[i,:1],x2[i,:1],eps=0.0)**2)/len(x1)
   x3[i,1:] += (torch.nn.functional.pairwise_distance(x1[i,1:],x2[i,1:],eps=0.0)**2)/len(x1)
   mse_manual += x3[i]
print(mse_manual.mean()) # 0.1557

mse_manual = 0
for i in range(len(x1)) :
   mse_manual += np.square(x1[i]-x2[i]) / len(x1)
print(mse_manual.mean()) # 0.1557

    或者，如果您想使用修改后的 mse 损失重现成对距离函数，您可以通过以下方式实现：

    import torch
import numpy as np
# Think of x1 as predicted 2D coordinates and x2 of ground truth
x1 = torch.rand(10,2)
x2 = torch.rand(10,2)

mse_torch = torch.nn.functional.mse_loss(x1,x2, reduction='none')
print(mse_torch.sum(-1).mean()) # 0.3314

mse_direct = 
torch.nn.functional.pairwise_distance(x1,x2).square().mean()
print(mse_direct) # 0.3314

mse_manual = 0
for i in range(len(x1)) :
    mse_manual += np.square(np.linalg.norm(x1[i]-x2[i])) / len(x1)
print(mse_manual) # 0.3314
    • 1

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