Question

我正在阅读这篇论文"Self-Invertible 2D Log-Gabor Wavelets"，它定义了2D log gabor过滤器：

本文还指出，滤波器仅覆盖频率空间的一侧，并在此图像中显示

在我尝试实施过滤器时，我得到的结果与论文中的内容不符。让我从实施开始，然后我将陈述问题。

实施

我创建了一个包含滤波器的二维数组，并对每个索引进行了变换，使得频域的原点位于数组的中心，正x轴向右，正y轴向上。< / p>

number_scales = 5         # scale resolution
number_orientations = 9   # orientation resolution
N = constantDim           # image dimensions

def getLogGaborKernal(scale, angle, logfun=math.log2, norm = True):
    # setup up filter configuration
    center_scale = logfun(N) - scale          
    center_angle = ((np.pi/number_orientations) * angle) if (scale % 2) \
                else ((np.pi/number_orientations) * (angle+0.5))
    scale_bandwidth =  0.996 * math.sqrt(2/3)
    angle_bandwidth =  0.996 * (1/math.sqrt(2)) * (np.pi/number_orientations)

    # 2d array that will hold the filter
    kernel = np.zeros((N, N))
    # get the center of the 2d array so we can shift origin
    middle = math.ceil((N/2)+0.1)-1

    # calculate the filter
    for x in range(0,constantDim):
        for y in range(0,constantDim):
            # get the transformed x and y where origin is at center
            # and positive x-axis goes right while positive y-axis goes up
            x_t, y_t = (x-middle),-(y-middle)
            # calculate the filter value at given index
            kernel[y,x] = logGaborValue(x_t,y_t,center_scale,center_angle,
        scale_bandwidth, angle_bandwidth,logfun)

    # normalize the filter energy
    if norm:
        Kernel = kernel / np.sum(kernel**2)
    return kernel

要计算每个索引处的过滤器值，我们会转到对数极坐标空间的另一个变换

def logGaborValue(x,y,center_scale,center_angle,scale_bandwidth,
              angle_bandwidth, logfun):
    # transform to polar coordinates
    raw, theta = getPolar(x,y)
    # if we are at the center, return 0 as in the log space
    # zero is not defined
    if raw == 0:
        return 0

    # go to log polar coordinates
    raw = logfun(raw)

    # calculate (theta-center_theta), we calculate cos(theta-center_theta) 
    # and sin(theta-center_theta) then use atan to get the required value,
    # this way we can eliminate the angular distance wrap around problem
    costheta, sintheta = math.cos(theta), math.sin(theta)
    ds = sintheta * math.cos(center_angle) - costheta * math.sin(center_angle)    
    dc = costheta * math.cos(center_angle) + sintheta * math.sin(center_angle)  
    dtheta = math.atan2(ds,dc)

    # final value, multiply the radial component by the angular one
    return math.exp(-0.5 * ((raw-center_scale) / scale_bandwidth)**2) * \
            math.exp(-0.5 * (dtheta/angle_bandwidth)**2)

问题：

角度：论文指出，将角度从1-> 8索引会产生良好的方向覆盖，但在我的实施角度中，从1＆gt; n don＆＃ 39;覆盖除了半个方向。甚至垂直方向也未正确覆盖。这可以在该图中示出，其包含比例3的过滤器组和从1到8的取向：
来自上方过滤器的覆盖率很明显，过滤器覆盖了空间的两侧，而不是纸张所说的。通过使用范围为-4->的9个取向可以使这更加明确。 4.下图包含一张图像中的所有滤镜，以显示它如何覆盖光谱的两侧（此图像是通过从所有滤镜的每个位置获取最大值来创建的）：
中间列（方向$ \ pi / 2 $）：在方向上的第一个数字中 - >＆gt; 8可以看出过滤器在方向$ \ pi / 2 $处消失。这是正常的吗？当我在一个图像中组合所有滤镜（所有5个刻度和9个方向）时，也可以看到这一点：

更新在空间域中添加滤波器的脉冲响应，正如您所看到的那样，-4＆amp;中存在明显的失真。 4个方向：

Answer 1

经过大量的代码分析后，我发现我的实现是正确的，但是getPolar函数搞砸了，所以上面的代码应该可以正常工作。如果有人在寻找它，那么这是一个没有getPolar函数的新代码：

number_scales = 5          # scale resolution
number_orientations = 8    # orientation resolution
N = 128                    # image dimensions
def getFilter(f_0, theta_0):
    # filter configuration
    scale_bandwidth =  0.996 * math.sqrt(2/3)
    angle_bandwidth =  0.996 * (1/math.sqrt(2)) * (np.pi/number_orientations)

    # x,y grid
    extent = np.arange(-N/2, N/2 + N%2)
    x, y = np.meshgrid(extent,extent)

    mid = int(N/2)
    ## orientation component ##
    theta = np.arctan2(y,x)
    center_angle = ((np.pi/number_orientations) * theta_0) if (f_0 % 2) \
                else ((np.pi/number_orientations) * (theta_0+0.5))

    # calculate (theta-center_theta), we calculate cos(theta-center_theta) 
    # and sin(theta-center_theta) then use atan to get the required value,
    # this way we can eliminate the angular distance wrap around problem
    costheta = np.cos(theta)
    sintheta = np.sin(theta)
    ds = sintheta * math.cos(center_angle) - costheta * math.sin(center_angle)    
    dc = costheta * math.cos(center_angle) + sintheta * math.sin(center_angle)  
    dtheta = np.arctan2(ds,dc)

    orientation_component =  np.exp(-0.5 * (dtheta/angle_bandwidth)**2)

    ## frequency componenet ##
    # go to polar space
    raw = np.sqrt(x**2+y**2)
    # set origin to 1 as in the log space zero is not defined
    raw[mid,mid] = 1
    # go to log space
    raw = np.log2(raw)

    center_scale = math.log2(N) - f_0
    draw = raw-center_scale
    frequency_component = np.exp(-0.5 * (draw/ scale_bandwidth)**2)

    # reset origin to zero (not needed as it is already 0?)
    frequency_component[mid,mid] = 0

    return frequency_component * orientation_component

实施日志Gabor过滤器库

1 个答案: