Pytorch的“折叠"效果如何和“展开"工作? [英] How does Pytorch's "Fold" and "Unfold" work?

问题描述

我已经阅读了官方文档.我很难理解此功能的用途以及它的工作方式.有人可以用Layman的方式解释吗?

I've gone through the official doc. I'm having a hard time understanding what this function is used for and how it works. Can someone explain this in Layman terms?

尽管我使用的Pytorch版本与文档匹配，但我提供的示例却出现错误.也许修复错误(我所做的错误)应该教给我一些东西?文档中给出的代码段为:

I get an error for the example they provide, although the Pytorch version I'm using matches the documentation. Perhaps fixing the error, which I did, is supposed to teach me something? The snippet given in the documentation is:

   fold = nn.Fold(output_size=(4, 5), kernel_size=(2, 2))
   input = torch.randn(1, 3 * 2 * 2, 1)
   output = fold(input)
   output.size()

，固定的代码段是:

   fold = nn.Fold(output_size=(4, 5), kernel_size=(2, 2))
   input = torch.randn(1, 3 * 2 * 2, 3 * 2 * 2)
   output = fold(input)
   output.size()

谢谢！

推荐答案

unfold 和 fold 用于促进滑动窗口"操作(如卷积).
假设您要将功能foo应用于要素地图/图像中的每个5x5窗口:

unfold and fold are used to facilitate "sliding window" operation (like convolutions).
Suppose you want to apply a function foo to every 5x5 window in a feature map/image:

from torch.nn import functional as f
windows = f.unfold(x, kernel_size=5)

现在windows具有批处理-(5 * 5 * x.size(1))-num_windows的size，您可以在windows上应用foo:

Now windows has size of batch-(5*5*x.size(1))-num_windows, you can apply foo on windows:

processed = foo(windows)

现在，您需要将"processed"折回"到x的原始大小:

Now you need to "fold" processed back to the original size of x:

out = f.fold(processed, x.shape[-2:], kernel_size=5)

您需要注意padding和kernel_size可能会影响您将processed折回x大小的能力.
此外，fold 求和在重叠的元素上，因此您可能想将fold的输出除以补丁大小.

You need to take care of padding, and kernel_size that may affect your ability to "fold" back processed to the size of x.
Moreover, fold sums over overlapping elements, so you might want to divide the output of fold by patch size.

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