Thanks for your work!!

Have you tried using fcanet to train classification tasks on cifar10 or cifar100?. If you have tried, what is the frequency components setting?

在这个类中MultiSpectralAttentionLayer有以下部分。 if h != self.dct_h or w != self.dct_w: x_pooled = torch.nn.functional.adaptive_avg_pool2d(x, (self.dct_h, self.dct_w)) # If you have concerns about one-line-change, don't worry. :) # In the ImageNet models, this line will never be triggered. # This is for compatibility in instance segmentation and object detection.

如果我的任务是目标检测，我该怎么设置self.dct_h和self.dct_w？

Hi, I noticed that in your paper you computed FCAnet model FLOPs.

I wonder how do you compute the FLOPs of 2d dct? Could you provide your formula or code?

Thanks!

As in your code, the FcaBottleneck expansion is 4 and FcaBasicBlock is 1, FcaBottleneck has one more layer of convolution than FcaBasicBlock, so how should I choose which module to use ?

你好，我是一名深度学习初学者，我添加了两个FCA模块使原模型的mIOU提升了2.3，效果很好； 然而对于通道分组，我有一些其他的看法； 如果分组的通道中表示不同的信息，每个分组再使用不同的频率分量，这似乎会造成更多的信息丢失吧，因为DCT可以看作是一种加权和，可以从论文中看到除了GAP是对每个通道上像素的一视同仁，其他的都是对空间上某一个或几个部分注意的更多，这显然是存在偏颇的，这似乎也能解释为什么单个频率分量实验中GAP的效果最好；在这种情况下，对通道进行分组，或许会造成更多的信息损失？ 我仔细思考了下，我认为FCAwork的原因主要是存在通道冗余以及DCT加权形成的一种“互补” 因为存在通道冗余，进行通道分组时可能某些分组中的信息相近，并且这些分组的权重是“互补”的，比如一个权重矩阵更注重左半边，一个更注重右半边这样。似乎模块学习这种‘稀疏’的关系效果会更好。 可以认为FAC比SE更充分的使用了冗余的通道。 考虑了两个实验来证明， 不对减小输入的通道数，将FCA与原模型或是SE进行对比，当通道减少到一定程度时，信息没有那么冗余，这时应该会有大量的信息丢失，精度相较于原模型更低； 关于频率分量的选择，选取某些“对称”“互补”的权重矩阵，而不是通过单个频率分量的性能的来选择，并且去除那些"混乱”的权重矩阵，因为单个频率分量证明这种混乱的权重并没有简单分块的效果好 另外可以在大通道数使用大的分组，在小通道数使用小的分组，来检验是否会获得更好的性能

不能完全表达我的意思，如有错误，恳请指出！

您好，非常欣赏您的idea，所以尝试跑一下您的分类模型。 我下载了ImageNet2012数据集之后，尝试启动您的模型，遇到了以下问题，想请教一下是否我的哪些设置出错了？

错误信息如下： Traceback (most recent call last): File "main.py", line 643, in main() File "main.py", line 389, in main avg_train_time = train(train_loader, model, criterion, optimizer, epoch, logger, scheduler) File "main.py", line 471, in train prec1, prec5 = accuracy(output.data, target, topk=(1, 5)) File "main.py", line 631, in accuracy correct_k = correct[:k].view(-1).float().sum(0, keepdim=True) RuntimeError: view size is not compatible with input tensor's size and stride (at least one dimension spans across two contiguous subspaces). Use .reshape(...) instead.

Hi, I want to know how did you select the frequency components like Figure6? I want to select 1, 3, 6, 10 frequencies like zigzag DCT.

And, I want to know the meaning of the numbers in the layer.py.

num_freq = int(method[3:])
if 'top' in method:
    all_top_indices_x = [0,0,6,0,0,1,1,4,5,1,3,0,0,0,3,2,4,6,3,5,5,2,6,5,5,3,3,4,2,2,6,1]
    all_top_indices_y = [0,1,0,5,2,0,2,0,0,6,0,4,6,3,5,2,6,3,3,3,5,1,1,2,4,2,1,1,3,0,5,3]
    mapper_x = all_top_indices_x[:num_freq]
    mapper_y = all_top_indices_y[:num_freq]
elif 'low' in method:
    all_low_indices_x = [0,0,1,1,0,2,2,1,2,0,3,4,0,1,3,0,1,2,3,4,5,0,1,2,3,4,5,6,1,2,3,4]
    all_low_indices_y = [0,1,0,1,2,0,1,2,2,3,0,0,4,3,1,5,4,3,2,1,0,6,5,4,3,2,1,0,6,5,4,3]
    mapper_x = all_low_indices_x[:num_freq]
    mapper_y = all_low_indices_y[:num_freq]
elif 'bot' in method:
    all_bot_indices_x = [6,1,3,3,2,4,1,2,4,4,5,1,4,6,2,5,6,1,6,2,2,4,3,3,5,5,6,2,5,5,3,6]
    all_bot_indices_y = [6,4,4,6,6,3,1,4,4,5,6,5,2,2,5,1,4,3,5,0,3,1,1,2,4,2,1,1,5,3,3,3]
    mapper_x = all_bot_indices_x[:num_freq]
    mapper_y = all_bot_indices_y[:num_freq]
else:
    raise NotImplementedError
return mapper_x, mapper_y

在layer.py中有： class MultiSpectralAttentionLayer(torch.nn.Module):中有 self.dct_layer = MultiSpectralDCTLayer(dct_h, dct_w, mapper_x, mapper_y, channel) 可见dct_h在前， dct_w在后 就是h在前，w在后 而在class MultiSpectralDCTLayer(nn.Module):中 def init(self, width, height, mapper_x, mapper_y, channel): 可见 width在前，height在后，就是w在前，h在后 请问这有什么说处么？我晕了

How can I set dct_h and dct_w if i want to add FCA layer into another model. My feature maps for the layer I want to inset Fca layer are 160x160, 80x80, 40x40, 20x20

Please advise.

elif 'bot' in method:
    all_bot_indices_x = [6,1,3,3,2,4,1,2,4,4,5,1,4,6,2,5,6,1,6,2,2,4,3,3,5,5,6,2,5,5,3,6]
    all_bot_indices_y = [6,4,4,6,6,3,1,4,4,5,6,5,2,2,5,1,4,3,5,0,3,1,1,2,4,2,1,1,5,3,3,3]