PyTorch的SoftMax交叉熵损失和梯度用法

时间：2020-11-08 python教程查看: 1022

在PyTorch中可以方便的验证SoftMax交叉熵损失和对输入梯度的计算

关于softmax_cross_entropy求导的过程,可以参考HERE

示例：

# -*- coding: utf-8 -*-
import torch
import torch.autograd as autograd
from torch.autograd import Variable
import torch.nn.functional as F
import torch.nn as nn
import numpy as np

# 对data求梯度, 用于反向传播
data = Variable(torch.FloatTensor([[1.0, 2.0, 3.0], [1.0, 2.0, 3.0], [1.0, 2.0, 3.0]]), requires_grad=True)

# 多分类标签 one-hot格式
label = Variable(torch.zeros((3, 3)))
label[0, 2] = 1
label[1, 1] = 1
label[2, 0] = 1
print(label)

# for batch loss = mean( -sum(Pj*logSj) )
# for one : loss = -sum(Pj*logSj)
loss = torch.mean(-torch.sum(label * torch.log(F.softmax(data, dim=1)), dim=1))

loss.backward()
print(loss, data.grad)

输出：

tensor([[ 0., 0., 1.],
    [ 0., 1., 0.],
    [ 1., 0., 0.]])
# loss:损失 和 input's grad:输入的梯度
tensor(1.4076) tensor([[ 0.0300, 0.0816, -0.1116],
    [ 0.0300, -0.2518, 0.2217],
    [-0.3033, 0.0816, 0.2217]])

注意：

对于单输入的loss 和 grad

data = Variable(torch.FloatTensor([[1.0, 2.0, 3.0]]), requires_grad=True)


label = Variable(torch.zeros((1, 3)))
#分别令不同索引位置label为1
label[0, 0] = 1
# label[0, 1] = 1
# label[0, 2] = 1
print(label)

# for batch loss = mean( -sum(Pj*logSj) )
# for one : loss = -sum(Pj*logSj)
loss = torch.mean(-torch.sum(label * torch.log(F.softmax(data, dim=1)), dim=1))

loss.backward()
print(loss, data.grad)

其输出：

# 第一组:
lable: tensor([[ 1., 0., 0.]])
loss: tensor(2.4076) 
grad: tensor([[-0.9100, 0.2447, 0.6652]])

# 第二组:
lable: tensor([[ 0., 1., 0.]])
loss: tensor(1.4076) 
grad: tensor([[ 0.0900, -0.7553, 0.6652]])

# 第三组:
lable: tensor([[ 0., 0., 1.]])
loss: tensor(0.4076) 
grad: tensor([[ 0.0900, 0.2447, -0.3348]])

"""
解释:
对于输入数据 tensor([[ 1., 2., 3.]]) softmax之后的结果如下
tensor([[ 0.0900, 0.2447, 0.6652]])
交叉熵求解梯度推导公式可知 s[0, 0]-1, s[0, 1]-1, s[0, 2]-1 是上面三组label对应的输入数据梯度
"""

pytorch提供的softmax, 和log_softmax 关系

# 官方提供的softmax实现
In[2]: import torch
 ...: import torch.autograd as autograd
 ...: from torch.autograd import Variable
 ...: import torch.nn.functional as F
 ...: import torch.nn as nn
 ...: import numpy as np
In[3]: data = Variable(torch.FloatTensor([[1.0, 2.0, 3.0]]), requires_grad=True)
In[4]: data
Out[4]: tensor([[ 1., 2., 3.]])
In[5]: e = torch.exp(data)
In[6]: e
Out[6]: tensor([[ 2.7183,  7.3891, 20.0855]])
In[7]: s = torch.sum(e, dim=1)
In[8]: s
Out[8]: tensor([ 30.1929])
In[9]: softmax = e/s
In[10]: softmax
Out[10]: tensor([[ 0.0900, 0.2447, 0.6652]])
In[11]: # 等同于 pytorch 提供的 softmax 
In[12]: org_softmax = F.softmax(data, dim=1)
In[13]: org_softmax
Out[13]: tensor([[ 0.0900, 0.2447, 0.6652]])
In[14]: org_softmax == softmax # 计算结果相同
Out[14]: tensor([[ 1, 1, 1]], dtype=torch.uint8)

# 与log_softmax关系
# log_softmax = log(softmax)
In[15]: _log_softmax = torch.log(org_softmax) 
In[16]: _log_softmax
Out[16]: tensor([[-2.4076, -1.4076, -0.4076]])
In[17]: log_softmax = F.log_softmax(data, dim=1)
In[18]: log_softmax
Out[18]: tensor([[-2.4076, -1.4076, -0.4076]])

官方提供的softmax交叉熵求解结果

# -*- coding: utf-8 -*-
import torch
import torch.autograd as autograd
from torch.autograd import Variable
import torch.nn.functional as F
import torch.nn as nn
import numpy as np

data = Variable(torch.FloatTensor([[1.0, 2.0, 3.0], [1.0, 2.0, 3.0], [1.0, 2.0, 3.0]]), requires_grad=True)
log_softmax = F.log_softmax(data, dim=1)

label = Variable(torch.zeros((3, 3)))
label[0, 2] = 1
label[1, 1] = 1
label[2, 0] = 1
print("lable: ", label)

# 交叉熵的计算方式之一
loss_fn = torch.nn.NLLLoss() # reduce=True loss.sum/batch & grad/batch
# NLLLoss输入是log_softmax, target是非one-hot格式的label
loss = loss_fn(log_softmax, torch.argmax(label, dim=1))
loss.backward()
print("loss: ", loss, "\ngrad: ", data.grad)

"""
# 交叉熵计算方式二
loss_fn = torch.nn.CrossEntropyLoss() # the target label is NOT an one-hotted
#CrossEntropyLoss适用于分类问题的损失函数
#input:没有softmax过的nn.output, target是非one-hot格式label
loss = loss_fn(data, torch.argmax(label, dim=1))
loss.backward()
print("loss: ", loss, "\ngrad: ", data.grad)
"""
"""

输出

lable: tensor([[ 0., 0., 1.],
    [ 0., 1., 0.],
    [ 1., 0., 0.]])
loss: tensor(1.4076) 
grad: tensor([[ 0.0300, 0.0816, -0.1116],
    [ 0.0300, -0.2518, 0.2217],
    [-0.3033, 0.0816, 0.2217]])

通过和示例的输出对比, 发现两者是一样的

以上这篇PyTorch的SoftMax交叉熵损失和梯度用法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持python博客。

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