Comments (3)
edsr_x4_baseline.pth使用的是EDSR原论文的方式,即不使用自蒸馏不使用对比学习仅使用L1作为损失训练得到的模型,不支持0.25的推理。论文中给出的0.25baseline实验结果是我们使用n_feats=256*0.25=64时训练的模型的结果,仅用于性能比较,由于与CSD方法本身没有太大关联因此没有给出训练好的模型。
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也就是说0.25baseline的训练与teacher完全没关系,是SR与GT通过l1 loss 直接训练得到的吗?
还有一个问题是CSD是从头开始训练的还是EDSR预训练edsr_x4_baseline.pth作为teacher?
感谢
from csd.
是的。
CSD的完整训练过程是,训练channel数为256的edsr_x4_baseline,作为teacher训练其中小分支的参数。必须要提供预训练的teacher,是因为在对比损失当中使用了T的输出作为正样本来约束小分支,因此此时T的输出需要能够提供可用信息,效果不能太差。
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Related Issues (17)
- Can you provide the full text version of the paper?? I cant find it anywhere on the internet
- Can you provide the full text version of the paper?? I cant find it anywhere on the internet HOT 1
- Is it only works on existing models, or eventually it outputs a pertained model based on a reference one? HOT 1
- code HOT 1
- the code is different to the paper HOT 2
- CL negtive sample question HOT 1
- About EDSR+ teacher performance HOT 2
- baseline比较并不公平 如何验证蒸馏方法的有效性 HOT 5
- 你好
- 关于动态分配两种支路的问题 HOT 3
- Some question about speed up HOT 3
- vgg19_ImageNet.ckpt
- 请问论文中的局部子分支的参数量是如何计算得到的?
- About Contrastive Loss
- ImportError: cannot import name 'SIGPIPE' from 'signal'
- 为什么不直接采用hr作为负样本?
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