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wuhuikai avatar wuhuikai commented on July 23, 2024
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wuhuikai avatar wuhuikai commented on July 23, 2024

需要同时增加pooling/stride conv

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BlossomingL avatar BlossomingL commented on July 23, 2024

不好意思,我不是太明白。。。

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BlossomingL avatar BlossomingL commented on July 23, 2024

因为我的原本网络的dilated conv层前后分别有stride conv和deconv层,您的意思是将哪里的去除。

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wuhuikai avatar wuhuikai commented on July 23, 2024

在分割中,最后两个阶段的操作是去除降采样同时加大dilation,这样可以保证最后得到的feature map是输入图像的1/8。在我们的方法中,最后两个阶段的dilation是1,降采样操作也没有去除,最后得到的feature map是输入图像的1/32,然后经过我们的JPU变为输入图像的1/8.

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wuhuikai avatar wuhuikai commented on July 23, 2024

在本身就有stride conv的情况下,是不需要使用我们的方法的

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2qwerty avatar 2qwerty commented on July 23, 2024

我们的网络,前面使用了类似论文中提出的5层(conv)encoder结构,使用您提出的JPU 替换膨胀卷积层 连接encoder部分,但是速度并没有提升?

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BlossomingL avatar BlossomingL commented on July 23, 2024

不好意思,我还有个问题就是代码里面SeparableConv2d(3*width, width, kernel_size=3, padding=1, dilation=1, bias=False),这里经过了一次dilated conv,论文里面用了四次,那速度不跟原始的有dilated conv的一样(甚至速度会有下降)?

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wuhuikai avatar wuhuikai commented on July 23, 2024

@2qwerty Backbone是用的哪一个网络?Backbone的output stride是多少?

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wuhuikai avatar wuhuikai commented on July 23, 2024

@Linxxxx JPU能加速的原理是Backbone的output stride从8升到32。其主要作用是把1/32的feature map重新变回1/8,同时取得相近或更好的性能。

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BlossomingL avatar BlossomingL commented on July 23, 2024

代码中的feat1 = F.upsample(self.conv1(self.pool1(x)), (h, w), **self._up_kwargs),F.upsample在tensorflow中能用简单的tf.image.resize_images()达到一样效果吗?感谢回答!

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wuhuikai avatar wuhuikai commented on July 23, 2024

F.upsample用于双线性插值上采样,等价于tf.image.resize_images.

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lifeiwen avatar lifeiwen commented on July 23, 2024

请问你是怎么开始训练的?用的什么指令

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