python - OpenCV中轮廓之间的仿射变换

Question

我有一个从胶片扫描的海 basemap 像的历史时间序列，需要配准。

from pylab import *
import cv2
import urllib

urllib.urlretrieve('http://geoport.whoi.edu/images/frame014.png','frame014.png');
urllib.urlretrieve('http://geoport.whoi.edu/images/frame015.png','frame015.png');

gray1=cv2.imread('frame014.png',0)
gray2=cv2.imread('frame015.png',0)
figure(figsize=(14,6))
subplot(121);imshow(gray1,cmap=cm.gray);
subplot(122);imshow(gray2,cmap=cm.gray);

我想使用每张图像左侧的黑色区域进行配准，因为该区域在相机内部，应该及时修复。所以我只需要计算黑色区域之间的仿射变换。

我通过阈值化和找到最大轮廓来确定这些区域:

def find_biggest_contour(gray,threshold=40):
    # threshold a grayscale image 
    ret,thresh = cv2.threshold(gray,threshold,255,1)
    # find the contours
    contours,h = cv2.findContours(thresh,mode=cv2.RETR_LIST,method=cv2.CHAIN_APPROX_NONE)
    # measure the perimeter
    perim = [cv2.arcLength(cnt,True) for cnt in contours]
    # find contour with largest perimeter
    i=perim.index(max(perim))
    return contours[i]

c1=find_biggest_contour(gray1)
c2=find_biggest_contour(gray2)

x1=c1[:,0,0];y1=c1[:,0,1]
x2=c2[:,0,0];y2=c2[:,0,1]

figure(figsize=(8,8))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1,y1,'b-')
imshow(gray2,cmap=cm.gray, alpha=0.5);plot(x2,y2,'g-')
axis([0,1500,1000,0]);

蓝色是第一帧最长的轮廓，绿色是第二帧最长的轮廓。

确定蓝色和绿色轮廓之间的旋转和偏移的最佳方法是什么？

我只想在台阶周围的某个区域使用轮廓的右侧，比如箭头之间的区域。

当然，如果有更好的方法来注册这些图像，我很乐意听到。我已经在原始图像上尝试了标准的特征匹配方法，但效果不够好。

Answer 1

按照 Shambool 的建议方法，这就是我想出的方法。我使用了 Ramer-Douglas-Peucker 算法来简化感兴趣区域的轮廓，并确定了两个转折点。我打算使用两个转折点来获得我的三个未知数(xoffset、yoffset 和旋转角度)，但是第二个转折点向右有点太远了，因为 RDP 简化了该区域中更平滑的曲线。因此，我使用了通往第一个转折点的线段的角度。在 image1 和 image2 之间区分这个角度给了我旋转角度。我仍然对这个解决方案并不完全满意。它对这两个图像运行良好，但我不确定它是否能在整个图像序列上运行良好。走着瞧。

将轮廓拟合到黑色边框的已知形状会更好。

# select region of interest from largest contour 
ind1=where((x1>190.) & (y1>200.) & (y1<900.))[0]
ind2=where((x2>190.) & (y2>200.) & (y2<900.))[0]
figure(figsize=(10,10))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1[ind1],y1[ind1],'b-')
imshow(gray2,cmap=cm.gray, alpha=0.5);plot(x2[ind2],y2[ind2],'g-')
axis([0,1500,1000,0])

def angle(x1,y1):
    #  Returns angle of each segment along an (x,y) track
    return array([math.atan2(y,x) for (y,x) in zip(diff(y1),diff(x1))])

def simplify(x,y, tolerance=40, min_angle = 60.*pi/180.): 
    """
    Use the Ramer-Douglas-Peucker algorithm to simplify the path
    http://en.wikipedia.org/wiki/Ramer-Douglas-Peucker_algorithm
    Python implementation: https://github.com/sebleier/RDP/
    """
    from RDP import rdp   
    points=vstack((x,y)).T
    simplified = array(rdp(points.tolist(), tolerance))
    sx, sy = simplified.T

    theta=abs(diff(angle(sx,sy)))
    # Select the index of the points with the greatest theta
    # Large theta is associated with greatest change in direction.
    idx = where(theta>min_angle)[0]+1
    return sx,sy,idx

sx1,sy1,i1 = simplify(x1[ind1],y1[ind1])
sx2,sy2,i2 = simplify(x2[ind2],y2[ind2])
fig = plt.figure(figsize=(10,6))
ax =fig.add_subplot(111)

ax.plot(x1, y1, 'b-', x2, y2, 'g-',label='original path')
ax.plot(sx1, sy1, 'ko-', sx2, sy2, 'ko-',lw=2, label='simplified path')
ax.plot(sx1[i1], sy1[i1], 'ro', sx2[i2], sy2[i2], 'ro', 
    markersize = 10, label='turning points')
ax.invert_yaxis()
plt.legend(loc='best')

# determine x,y offset between 1st turning points, and 
# angle from difference in slopes of line segments approaching 1st turning point
xoff = sx2[i2[0]] - sx1[i1[0]]
yoff = sy2[i2[0]] - sy1[i1[0]]
iseg1 = [i1[0]-1, i1[0]]
iseg2 = [i2[0]-1, i2[0]]
ang1 = angle(sx1[iseg1], sy1[iseg1])
ang2 = angle(sx2[iseg2], sy2[iseg2])
ang = -(ang2[0] - ang1[0])
print xoff, yoff, ang*180.*pi

-28 14 5.07775871644

# 2x3 affine matrix M
M=array([cos(ang),sin(ang),xoff,-sin(ang),cos(ang),yoff]).reshape(2,3)
print M

[[  9.99959685e-01   8.97932821e-03  -2.80000000e+01]
 [ -8.97932821e-03   9.99959685e-01   1.40000000e+01]]

# warp 2nd image into coordinate frame of 1st
Minv = cv2.invertAffineTransform(M)
gray2b = cv2.warpAffine(gray2,Minv,shape(gray2.T))

figure(figsize=(10,10))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1[ind1],y1[ind1],'b-')
imshow(gray2b,cmap=cm.gray, alpha=0.5);
axis([0,1500,1000,0]);
title('image1 and transformed image2 overlain with 50% transparency');