Question

我正在使用python中的Lucas-Kanade进行分配，但我无法理解如何将输出（u和v）转换为像素位置，如第一个示例（移动汽车）中{{{ 3}}。在示例中，cv2.calcOpticalFlowPyrLK的输出会自动转换为像素位置，但我的代码会给出u和v值，这些值通常非常小（小于1）。我无法发布我写的代码，但它与tutorial page非常相似：

import numpy as np
import scipy.signal as si
from PIL import Image

def gauss_kern():
   h1 = 15
   h2 = 15
   x, y = np.mgrid[0:h2, 0:h1]
   x = x-h2/2
   y = y-h1/2
   sigma = 1.5
   g = np.exp( -( x**2 + y**2 ) / (2*sigma**2) );
   return g / g.sum()

def deriv(im1, im2):
   g = gauss_kern()
   Img_smooth = si.convolve(im1,g,mode='same')
   fx,fy=np.gradient(Img_smooth)  
   ft = si.convolve2d(im1, 0.25 * np.ones((2,2))) + \
       si.convolve2d(im2, -0.25 * np.ones((2,2)))

   fx = fx[0:fx.shape[0]-1, 0:fx.shape[1]-1]  
   fy = fy[0:fy.shape[0]-1, 0:fy.shape[1]-1];
   ft = ft[0:ft.shape[0]-1, 0:ft.shape[1]-1];
   return fx, fy, ft

import matplotlib.pyplot as plt
import numpy as np
import scipy.signal as si
from PIL import Image
import deriv
import numpy.linalg as lin

def lk(im1, im2, i, j, window_size) :
   fx, fy, ft = deriv.deriv(im1, im2)
   halfWindow = np.floor(window_size/2)
   curFx = fx[i-halfWindow-1:i+halfWindow,
              j-halfWindow-1:j+halfWindow]
   curFy = fy[i-halfWindow-1:i+halfWindow,
              j-halfWindow-1:j+halfWindow]
   curFt = ft[i-halfWindow-1:i+halfWindow,
              j-halfWindow-1:j+halfWindow]
   curFx = curFx.T
   curFy = curFy.T
   curFt = curFt.T

   curFx = curFx.flatten(order='F')
   curFy = curFy.flatten(order='F')
   curFt = -curFt.flatten(order='F')

   A = np.vstack((curFx, curFy)).T
   U = np.dot(np.dot(lin.pinv(np.dot(A.T,A)),A.T),curFt)
   return U[0], U[1]

Answer 1

这是预期的。我正在研究同一个问题。如果检查calcOpticalFlowPyrLK，您会发现它也有一些小的变化量。只要确保在每次迭代中舍入这些浮点数就不会丢失它们。

Lucas-Kanade光流：将U和V转换为像素位置

1 个答案: