imread(‘j.png’, 0) # Create a 5x5 kernel of ones kernel = np.ones((5,5),np.uint8) # Dilate the image dilation
由于未知的降雨模式，单图像去噪相当具有挑战性。现有的方法通常对降雨模型做出特定的假设，这些假设很难涵盖现实世界中的许多不同情况，这使得它们不得不采用复杂的优化...
): ''' inplanes: input planes: output dilation: dilation rate...=middle_block_dilation) self.block5 = Block(728, 728, reps=3, stride=1, dilation=middle_block_dilation...stride=1, dilation=middle_block_dilation) self.block10 = Block(728, 728, reps=3, stride=1, dilation...=middle_block_dilation) self.block11 = Block(728, 728, reps=3, stride=1, dilation=middle_block_dilation...dilation=middle_block_dilation) self.block17 = Block(728, 728, reps=3, stride=1, dilation=middle_block_dilation
self.num_dilation = len(dilation) assert num_heads % self.num_dilation == 0, f"num_heads{num_heads...} must be the times of num_dilation{self.num_dilation}!!"..., C //self.num_dilation, H, W).permute(2, 1, 0, 3, 4, 5) #num_dilation,3,B,C//num_dilation,H,W...x = x.reshape(B, self.num_dilation, C//self.num_dilation, H, W).permute(1, 0, 3, 4, 2 )...# num_dilation, B, H, W, C//num_dilation for i in range(self.num_dilation): x[i] =
_norm_layer = norm_layer self.inplanes = 32 self.dilation = 1 if replace_stride_with_dilation...= [False, False, False, False] if len(replace_stride_with_dilation) !..._norm_layer downsample = None previous_dilation = self.dilation if dilate:...self.dilation *= stride stride = 1 if stride !...=self.dilation, norm_layer=norm_layer)) return nn.Sequential
, groups=groups, bias=False, dilation=dilation) def conv1x1(in_planes, out_planes, stride=1): "..._norm_layer = norm_layer self.inplanes = 64 self.dilation = 1 if replace_stride_with_dilation...= [False, False, False] if len(replace_stride_with_dilation) !..._norm_layer downsample = None previous_dilation = self.dilation if dilate:...self.dilation *= stride stride = 1 if stride !
, groups=groups, bias=False, dilation=dilation) class BasicBlock(nn.Module): expansion = 1..., groups=groups, bias=False, dilation=dilation) def conv1x1(in_planes, out_planes, stride=1): ""..._norm_layer = norm_layer self.inplanes = 64 self.dilation = 1 if replace_stride_with_dilation..._norm_layer downsample = None previous_dilation = self.dilation if dilate:...self.dilation *= stride stride = 1 if stride !
图©和图(b)是类似的，只是dilation rate=4，相当于变成了15×1515\times1515×15的卷积核。...上图中示例是三个dilation rate=2扩张卷积层连续卷积后的结果，蓝色的标识是参与计算的卷积中心，而颜色的深度表征次数。...解决方法 解决这个问题最直接的方法当然就是不使用连续的dilation rate相同的扩展卷积，但是这还不够，因为如果dilation rate是成倍数的，那么问题还是存在。...此外，扩张卷积已经在caffe原始代码中实现了，就是dilation参数。...: 1 } } 上面这种情况就是一个基础的卷积层，它的dilation=1。
def conv2d_same_padding(input, weight, bias=None, stride=1, padding=1, dilation=1, groups=1): # 函数中...= dilation self.transposed = transposed self.output_padding = output_padding...= (0,) * len(self.padding): s += ', padding={padding}' if self.dilation !...= (1,) * len(self.dilation): s += ', dilation={dilation}' if self.output_padding...= _pair(dilation) super(Conv2d, self).
https://arxiv.org/abs/1709.00179 针对卫星图像中的小目标分割问题，本文从 dilated convolution 的有效使用给出了解决方法，主要是 先 increasing dilation...factor，再 decreasing dilation factor 卫星图像分割和普通图像分割是有明显的差异性的，如下图所示： ?...This means that whereas increasing dilation factors is important in terms of resolution and context,...当我们增加 dilation factor 时， 没有联系的神经元空隙会变大。 造成的后果就是分割结果有锯齿现象。...Local feature extraction module: 解决的方法就是 先 increasing dilation factor，再 decreasing dilation factor
= cv2.dilate(image,kernel,iterations = 1) cv2.imwrite('dilation.jpg', dilation) kernel = np.ones...((2,2), dtype = "uint8")/9 dilation = cv2.dilate(image,kernel,iterations = 1) cv2.imwrite('dilation.jpg...= 3) cv2.imwrite('dilation.jpg', dilation) kernel = np.ones((2,2), dtype = "uint8")/9 dilation...= cv2.dilate(image,kernel,iterations = 5) cv2.imwrite('dilation.jpg', dilation) kernel = np.ones(...', dilation) 侵蚀 侵蚀函数正好与膨胀功函数相反。
= 1 if use_dilation_conv5: stride = 1 dilation = 2 ResBody(net, from_layer, '5a...', out2a=512, out2b=512, out2c=2048, stride=stride, use_branch1=True, dilation=dilation, **bn_param)...=False, dilation=dilation, **bn_param) if use_pool5: net.pool5 = L.Pooling(net.res5c, pool...= 1 if use_dilation_conv5: stride = 1 dilation = 2 ResBody(net, from_layer, '5a...', out2a=512, out2b=512, out2c=2048, stride=stride, use_branch1=True, dilation=dilation, **bn_param)
在 TensorFlow 中，提供 tf.nn.dilation2d 和 tf.nn.erosion2d 这两种形态学网络层，分别对应着形态学操作上的膨胀和腐蚀操作。...tf.nn.erosion2d( value, kernel, strides, rates, padding, name=None ) tf.nn.dilation2d...filter, strides, rates, padding, name=None ) 拿膨胀来说明，在文档中提到： Computes the grayscale dilation...参考 [1]. tf.nn.dilation2d [2]. Tensorflow dilation behave differently than morphological dilation
=dilation #空洞率 assert stride in [1,2,3] # inter_channels=out_channels...,dilation,groups) self.conv3=nn.Sequential( nn.Conv2d(inter_channels,...= dilation inter_channels = out_channels // 2 #channel split if(stride>1 or...dilation>1):#带下采样的模块，左边的路径的特征图也需要进行相应的下采样，同时也不使用channel split self.branch1=nn.Sequential...( DWConv(in_channels,in_channels,3,stride, dilation, dilation),
然而光采用大 dilation rate 的信息或许只对一些大物体分割有效果，而对小物体来说可能则有弊无利了。...是 i 层的 dilation rate 而 ? 是指在 i 层的最大dilation rate，那么假设总共有n层的话，默认 ? 。...，这样我们至少可以用 dilation rate 1 即 standard convolution 的方式来覆盖掉所有洞。...一个简单的例子: dilation rate [1, 2, 5] with 3 x 3 kernel (可行的方案) ?...而这样的锯齿状本身的性质就比较好的来同时满足小物体大物体的分割要求(小 dilation rate 来关心近距离信息，大 dilation rate 来关心远距离信息)。
如图，这个就是dilation=2的时候的情况，与之前的区别有两个： 看红色区域：可以看到卷积核大小依然是2，但是卷积核之间变得空洞了，隔过去了一个数据；如果dilation=3的话，那么可以想而知，这个卷积核中间会空的更大...可以看到，第一次卷积使用dilation=1的卷积，然后第二次使用dilation=2的卷积，这样通过两次卷积就可以实现视野域是4. 那么假设视野域要是8呢？那就再加一个dilation=4的卷积。...dilation的值是2的次方，然后视野域也是2的次方的增长，那么就算是要1000视野域，那十层大概就行了。...然后TCN中并不是每一次卷积都会扩大一倍的dilation，而是每两次扩大一倍的dilation 总之TCN中的基本组件：TemporalBlock()是两个dilation相同的卷积层，卷积+修改数据尺寸...=dilation_size, padding=(kernel_size-1) * dilation_size, dropout
padding=(1, 1)) (3): ReLU(inplace=True) (4): MaxPool2d(kernel_size=2, stride=2, padding=0, dilation...padding=(1, 1)) (8): ReLU(inplace=True) (9): MaxPool2d(kernel_size=2, stride=2, padding=0, dilation...padding=(1, 1)) (15): ReLU(inplace=True) (16): MaxPool2d(kernel_size=2, stride=2, padding=0, dilation...padding=(1, 1)) (22): ReLU(inplace=True) (23): MaxPool2d(kernel_size=2, stride=2, padding=0, dilation...padding=(1, 1)) (3): ReLU(inplace=True) (4): MaxPool2d(kernel_size=2, stride=2, padding=0, dilation
= 1) # 再次膨胀 Dilation2 = cv2.dilate(Erosion, element2,iterations = 3) cv2.imshow('Dilation2 ', Dilation2...Image', 'Erosion Image', 'Dilation2 Image'] images = [lenna_img, GrayImage, Gaussian, Median..., Sobel, Binary, Dilation, Erosion, Dilation2] for i in xrange(9): plt.subplot(3,3,i...= 1) # 再次膨胀，让轮廓明显一些 Dilation2 = cv2.dilate(Erosion, element2,iterations = 3) cv2.imshow('Dilation2 '..., Sobel, Binary, Dilation, Erosion, closed] for i in xrange(9): plt.subplot(3,3,i+1)
第一层为普通卷积，stride = 2，紧跟着 stride = 2 的 max-pooling，然后一个普通的 bottleneck ，一个 stride = 2 的 bottleneck，然后 dilation...=2、dilation =4 的bottleneck。...=dilation * multi_grids[i], downsample=(True if i == 0 else False),...=dilation * multi_grids[i], downsample=(True if i == 0 else False),...__init__() # Stride and dilation if output_stride == 8: s = [1, 2, 1, 1]
, bias=False, dilation=dilation) self.bn1 = BatchNorm(planes)..., bias=False, dilation=dilation) self.bn2 = BatchNorm(planes)..., bias=False, dilation=dilation) self.bn2 = BatchNorm(bottle_planes..., bias=False, dilation=dilation, groups=cardinality) self.bn2...dilation=dilation) else: self.tree1 = Tree(levels - 1, block, in_channels, out_channels
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