pytorch交叉熵?fù)p失函數(shù)的weight參數(shù)的使用

更新時(shí)間：2021年05月24日 09:59:08 作者：Nick Blog

這篇文章主要介紹了pytorch交叉熵?fù)p失函數(shù)的weight參數(shù)的使用，具有很好的參考價(jià)值，希望對(duì)大家有所幫助。如有錯(cuò)誤或未考慮完全的地方，望不吝賜教

首先

必須將權(quán)重也轉(zhuǎn)為Tensor的cuda格式；

然后

將該class_weight作為交叉熵函數(shù)對(duì)應(yīng)參數(shù)的輸入值。

class_weight = torch.FloatTensor([0.13859937, 0.5821059, 0.63871904, 2.30220396, 7.1588294, 0]).cuda()

補(bǔ)充：關(guān)于pytorch的CrossEntropyLoss的weight參數(shù)

首先這個(gè)weight參數(shù)比想象中的要考慮的多

你可以試試下面代碼

import torch
import torch.nn as nn
inputs = torch.FloatTensor([0,1,0,0,0,1])
outputs = torch.LongTensor([0,1])
inputs = inputs.view((1,3,2))
outputs = outputs.view((1,2))
weight_CE = torch.FloatTensor([1,1,1])
ce = nn.CrossEntropyLoss(ignore_index=255,weight=weight_CE)
loss = ce(inputs,outputs)
print(loss)

tensor(1.4803)

這里的手動(dòng)計(jì)算是：

loss1 = 0 + ln(e0 + e0 + e0) = 1.098

loss2 = 0 + ln(e1 + e0 + e1) = 1.86

求平均 = (loss1 *1 + loss2 *1)/ 2 = 1.4803

加權(quán)呢？

import torch
import torch.nn as nn
inputs = torch.FloatTensor([0,1,0,0,0,1])
outputs = torch.LongTensor([0,1])
inputs = inputs.view((1,3,2))
outputs = outputs.view((1,2))
weight_CE = torch.FloatTensor([1,2,3])
ce = nn.CrossEntropyLoss(ignore_index=255,weight=weight_CE)
loss = ce(inputs,outputs)
print(loss)

tensor(1.6075)

手算發(fā)現(xiàn)，并不是單純的那權(quán)重相乘：

loss1 = 0 + ln(e0 + e0 + e0) = 1.098

loss2 = 0 + ln(e1 + e0 + e1) = 1.86

求平均 = (loss1 * 1 + loss2 * 2)/ 2 = 2.4113

而是

loss1 = 0 + ln(e0 + e0 + e0) = 1.098

loss2 = 0 + ln(e1 + e0 + e1) = 1.86

求平均 = (loss1 *1 + loss2 *2) / 3 = 1.6075

發(fā)現(xiàn)了么，加權(quán)后，除以的是權(quán)重的和，不是數(shù)目的和。

我們?cè)衮?yàn)證一遍：

import torch
import torch.nn as nn
inputs = torch.FloatTensor([0,1,2,0,0,0,0,0,0,1,0,0.5])
outputs = torch.LongTensor([0,1,2,2])
inputs = inputs.view((1,3,4))
outputs = outputs.view((1,4))
weight_CE = torch.FloatTensor([1,2,3])
ce = nn.CrossEntropyLoss(weight=weight_CE)
# ce = nn.CrossEntropyLoss(ignore_index=255)
loss = ce(inputs,outputs)
print(loss)

tensor(1.5472)

手算：

loss1 = 0 + ln(e0 + e0 + e0) = 1.098

loss2 = 0 + ln(e1 + e0 + e1) = 1.86

loss3 = 0 + ln(e2 + e0 + e0) = 2.2395

loss4 = -0.5 + ln(e0.5 + e0 + e0) = 0.7943

求平均 = (loss1 * 1 + loss2 * 2+loss3 * 3+loss4 * 3) / 9 = 1.5472

可能有人對(duì)loss的CE計(jì)算過程有疑問，我這里細(xì)致寫寫交叉熵的計(jì)算過程，就拿最后一個(gè)例子的loss4的計(jì)算說明