问题描述
我正在尝试在 pytorch 中实现 3D 体积的类似高斯模糊。我可以很容易地通过与 2D 高斯核进行卷积来对 2D 图像进行 2D 模糊,并且相同的方法似乎适用于具有 3D 高斯核的 3D。但是,它在 3D 中非常慢(尤其是在 sigmas/kernel 尺寸较大的情况下)。我知道这也可以通过与 2D 内核卷积 3 次来完成,这应该更快,但我无法让它工作。我的测试用例如下。
import torch
import torch.nn.functional as F
VOL_SIZE = 21
def make_gaussian_kernel(sigma):
ks = int(sigma * 5)
if ks % 2 == 0:
ks += 1
ts = torch.linspace(-ks // 2,ks // 2 + 1,ks)
gauss = torch.exp((-(ts / sigma)**2 / 2))
kernel = gauss / gauss.sum()
return kernel
def test_3d_gaussian_blur(blur_sigma=2):
# Make a test volume
vol = torch.zeros([VOL_SIZE] * 3)
vol[VOL_SIZE // 2,VOL_SIZE // 2,VOL_SIZE // 2] = 1
# 3D convolution
vol_in = vol.reshape(1,1,*vol.shape)
k = make_gaussian_kernel(blur_sigma)
k3d = torch.einsum('i,j,k->ijk',k,k)
k3d = k3d / k3d.sum()
vol_3d = F.conv3d(vol_in,k3d.reshape(1,*k3d.shape),stride=1,padding=len(k) // 2)
# Separable 2D convolution
vol_in = vol.reshape(1,*vol.shape)
k2d = torch.einsum('i,j->ij',k)
k2d = k2d / k2d.sum()
k2d = k2d.expand(VOL_SIZE,*k2d.shape)
for i in range(3):
vol_in = vol_in.permute(0,3,2)
vol_in = F.conv2d(vol_in,k2d,padding=len(k) // 2,groups=VOL_SIZE)
vol_3d_sep = vol_in
torch.allclose(vol_3d,vol_3d_sep) # --> False
非常感谢任何帮助!
解决方法
理论上您可以使用三个 2d 卷积计算 3d 高斯卷积,但这意味着您必须减小 2d 内核的大小,因为您在每个方向上都有效地进行了卷积两次.
但在计算上更高效(以及您通常想要的)是分离成一维内核。我更改了函数的第二部分来实现这一点。 (我必须说我真的很喜欢你基于排列的方法!)因为你使用的是 3d 卷,所以你不能很好地使用 useEffect(() => {
async callAxios() => {
const result = await axios.get(`${url_one}`)
const user_orders = await axios.get(`${url_two}`)
const user_details = await axios.get(`${url_three}`)
}
callAxios()
},[])
或 conv2d
函数,所以最好的事情是真的即使您只是在计算一维卷积,也只需使用 conv1d
。
请注意,conv3d
使用的阈值 allclose
我们无法通过此方法达到,可能是由于取消错误。
1e-8
附录:如果您真的想滥用 def test_3d_gaussian_blur(blur_sigma=2):
# Make a test volume
vol = torch.randn([VOL_SIZE] * 3) # using something other than zeros
vol[VOL_SIZE // 2,VOL_SIZE // 2,VOL_SIZE // 2] = 1
# 3D convolution
vol_in = vol.reshape(1,1,*vol.shape)
k = make_gaussian_kernel(blur_sigma)
k3d = torch.einsum('i,j,k->ijk',k,k)
k3d = k3d / k3d.sum()
vol_3d = F.conv3d(vol_in,k3d.reshape(1,*k3d.shape),stride=1,padding=len(k) // 2)
# Separable 1D convolution
vol_in = vol[None,None,...]
# k2d = torch.einsum('i,j->ij',k)
# k2d = k2d / k2d.sum() # not necessary if kernel already sums to zero,check:
# print(f'{k2d.sum()=}')
k1d = k[None,:,None]
for i in range(3):
vol_in = vol_in.permute(0,4,2,3)
vol_in = F.conv3d(vol_in,k1d,padding=(len(k) // 2,0))
vol_3d_sep = vol_in
print((vol_3d- vol_3d_sep).abs().max()) # something ~1e-7
print(torch.allclose(vol_3d,vol_3d_sep)) # allclose checks if it is around 1e-8
来处理卷,您可以尝试
conv2d
或者单独使用 # separate 3d kernel into 1d + 2d
vol_in = vol[None,...]
k2d = torch.einsum('i,k)
k2d = k2d.expand(VOL_SIZE,len(k),len(k))
# k2d = k2d / k2d.sum() # not necessary if kernel already sums to zero,check:
# print(f'{k2d.sum()=}')
k1d = k[None,None]
vol_in = F.conv3d(vol_in,0))
vol_in = vol_in[0,...]
# abuse conv2d-groups argument for volume dimension,works only for 1 channel volumes
vol_in = F.conv2d(vol_in,k2d,len(k) // 2),groups=VOL_SIZE)
vol_3d_sep = vol_in
你可以这样做:
conv2d
这些应该仍然比三个连续的二维卷积更快。