将负Sobel梯度值转换为0到6.28的范围

Question

我尝试创建一个具有十二个方向特征提取功能的图像处理程序（来自此url源https://www.ijcaonline.org/journal/number3/pxc387173.pdf）

我用这段代码得到了梯度结果

def sobel(citra_thresh):
h, w = citra_thresh.shape
# define filters
horizontal = np.array([[-1, 0, 1], [-2, 0, 2], [-1, 0, 1]]) 
vertical = np.array([[-1, -2, -1], [0, 0, 0], [1, 2, 1]]) 
newhorizontalImage = np.zeros((h, w))
newverticalImage = np.zeros((h, w))
newgradientImage = np.zeros((h, w))
newdirectionImage = np.zeros((h,w))

for i in range(1, h - 1):
    for j in range(1, w - 1):
        horizontalGrad = (horizontal[0, 0] * citra_thresh[i - 1, j - 1]) + \
                         (horizontal[0, 1] * citra_thresh[i - 1, j]) + \
                         (horizontal[0, 2] * citra_thresh[i - 1, j + 1]) + \
                         (horizontal[1, 0] * citra_thresh[i, j - 1]) + \
                         (horizontal[1, 1] * citra_thresh[i, j]) + \
                         (horizontal[1, 2] * citra_thresh[i, j + 1]) + \
                         (horizontal[2, 0] * citra_thresh[i + 1, j - 1]) + \
                         (horizontal[2, 1] * citra_thresh[i + 1, j]) + \
                         (horizontal[2, 2] * citra_thresh[i + 1, j + 1])

        #if (horizontalGrad == 0):
         #   horizontalGrad = 0.0001
        newhorizontalImage[i - 1, j - 1] = abs(horizontalGrad)
        verticalGrad = (vertical[0, 0] * citra_thresh[i - 1, j - 1]) + \
                       (vertical[0, 1] * citra_thresh[i - 1, j]) + \
                       (vertical[0, 2] * citra_thresh[i - 1, j + 1]) + \
                       (vertical[1, 0] * citra_thresh[i, j - 1]) + \
                       (vertical[1, 1] * citra_thresh[i, j]) + \
                       (vertical[1, 2] * citra_thresh[i, j + 1]) + \
                       (vertical[2, 0] * citra_thresh[i + 1, j - 1]) + \
                       (vertical[2, 1] * citra_thresh[i + 1, j]) + \
                       (vertical[2, 2] * citra_thresh[i + 1, j + 1])
        newverticalImage[i - 1, j - 1] = abs(verticalGrad)
        mag = np.sqrt(pow(horizontalGrad, 2.0) + pow(verticalGrad, 2.0))
        newgradientImage[i - 1, j - 1] = mag
        newgradientImage = newgradientImage.astype(np.int)

        direct = np.arctan2((verticalGrad),(horizontalGrad))
        #tes=(horizontalGrad/verticalGrad)
        #direct = (np.arctan(tes))
        newdirectionImage[i - 1, j - 1] = direct
        #newdirectionImage = newdirectionImage.astype(np.float)
        newdirectionImage = np.around(newdirectionImage,decimals=2)
return newdirectionImage

我试图获取这张图片的渐变

我得到了这个结果

期刊希望渐变值应介于0到6.28之间。

但是我得到了负值梯度。如何将它们转换为0到6.28范围？