Question

我正在使用Tensorflow实现Canny算法（这需要使用边框作为评估指标，但这不是主题）。其中一个步骤是计算“非最大抑制”，其中包括将3x3区域中的中心元素归零，除非两个特定邻居较小。更多详情here。

如何使用Tensorflow实现此操作？

我实际上正在使用Keras，但Tensorflow解决方案也可以使用，作为参考，我的代码到目前为止看起来像这样：

def canny(img):
    '''Canny border detection. The input should be a grayscale image.'''
    gauss_kernel = np.array([[2,  4,  5,  4, 2],
                             [4,  9, 12,  9, 4],
                             [5, 12, 15, 12, 5],
                             [4,  9, 12,  9, 4],
                             [2,  4,  5,  4, 2]]).reshape(5, 5, 1, 1)
    gauss_kernel = K.variable(1./159 * gauss_kernel)

    Gx = K.variable(np.array([[-1., 0. ,1.],
                              [-2., 0., 2.],
                              [-1., 0., 1.]]).reshape(3, 3, 1, 1))

    Gy = K.variable(np.array([[-1., -2., -1.],
                              [ 0.,  0.,  0.],
                              [ 1.,  2.,  1.]]).reshape(3, 3, 1, 1))
    # Smooth image
    smoothed = K.conv2d(img, gauss_kernel, padding='same')
    # Derivative in x
    Dx = K.conv2d(smoothed, Gx, padding='same')
    # Derivative in y
    Dy = K.conv2d(smoothed, Gy, padding='same')
    # Take gradient strength
    G = K.sqrt(K.square(Dx) + K.square(Dy))

    # TODO: Non-maximum Suppression & Hysteresis Thresholding   

    return G

Answer 1

你可以使用卷积滤波器来隔离两个目标像素，并使它们与中心像素“同心”。

例如，为了与两个目标像素进行比较，我们可以使用此滤波器，形状为(3, 3, 1, 2) - 一个输入通道，两个输出通道。每个通道都将返回一个目标像素。

滤镜在目标像素处应该为1。剩下的就是零：

#taking two diagonal pixels
filter = np.zeros((3,3,1,2))
filter[0,0,0,0] = 1 #first pixel is top/left, passed to the first channel
filter[2,2,0,1] = 1 #second pixel is bottom/right, passed to the second channel     
    #which ones are really bottom or top, left or right depend on your preprocessing, 
    #but they should be consistent with the rest of your modeling 

filter = K.variable(filter)

如果你是上下左右，你可以制作较小的过滤器。不需要3x3（也没问题），但只需要1x3或3x1：

filter1 = np.zeros((1,3,1,2)) #horizontal filter
filter2 = np.zeros((3,1,1,2)) #vertical filter

filter1[0,0,0,0] = 1    #left pixel -  if filter is 3x3: [1,0,0,0]
filter1[0,2,0,1] = 1    #right pixel - if filter is 3x3: [1,2,0,1]
filter1 = K.variable(filter1)

filter2[0,0,0,0] = 1    #top pixel -  if filter is 3x3: [0,1,0,0]
filter2[2,0,0,1] = 1    #bottom pxl - if filter is 3x3: [2,1,0,1]
filter2 = K.variable(filter2)

然后你将这些作为卷积来应用。您将为一个像素获得一个通道，为另一个像素获得另一个通道。然后，您可以比较它们，就好像它们都在同一个地方，只是在不同的渠道中：

targetPixels = K.conv2d(originalImages, kernel=filter, padding='same')

#two channels telling if the center pixel is greater than the pixel in the channel
isGreater = K.greater(originalImages,targetPixels)

#merging the two channels, considering they're 0 for false and 1 for true
isGreater = K.cast(isGreater,K.floatx())
isGreater = isGreater[:,:,:,:1] * isGreater[:,:,:,1:]

#now, the center pixel will remain if isGreater = 1 at that position:
result = originalImages * isGreater

张量相对于元素邻居的条件值

1 个答案: