网友提问(透视变化)

您好:
    我在网上看到您的opencv透视变换的博客,https://www.cnblogs.com/jsxyhelu/p/4219564.html
    我是opencv小菜鸟一个,现在想要得到一个图片变形之后保存,整个图片信息不丢失,即四个角的信息不丢失应该怎么做?原图中某一点在新图中坐标应该怎么计算?万望不吝赐教,不胜感激,万分感谢。

你好:
  我按照您的代码和网上找到的python代码(https://blog.csdn.net/fengxueniu/article/details/77964375)修改并实现的基本的透视变换,但基本原理是将原图test0.png加边界,边界值与透视之后的背景一致,是图片看起来信息为丢失。但图像旋转之后的预期效果如test1.png所示,而实际生成的图片如test2.png所示,现在想要实现从test0.png到test1.png的变换并求test0.png图中某一点坐标值在test1.png中位置,应该怎么做?对于透视变换不是特别理解。万望不吝赐教,万分感谢。
  图片与代码附注如下。

   =》 
现有代码
 #-*- coding:utf-8 -*-
#-*- coding:utf-8 -*-
import cv2
import numpy as np
import random
def rad(x):
return x*np.pi/180
img = cv2.imread("test0.png")
cv2.imshow("original", img)
#扩展图像,保证内容不超出可视范围
img = cv2.copyMakeBorder(img,100,100,100,100,cv2.BORDER_CONSTANT,(255, 255, 255))
w,h=img.shape[0:2]
cv2.imshow("original1", img)
#anglex=45
#angley = 45
#anglez = 0
#fov = 42
anglex=random.uniform(30,60)
angley =random.uniform(30,60)
anglez =random.uniform(30,60)
fov = random.uniform(30,60)

print(anglex,angley,anglez,fov)
#镜头与图像间的距离,21为半可视角,算z的距离是为了保证在此可视角度下恰好显示整幅图像
z=np.sqrt(w**2 + h**2)/2/np.tan(rad(fov/2))
#齐次变换矩阵
rx = np.array([[1, 0, 0, 0],
[0, np.cos(anglex*np.pi/180), -np.sin(anglex*np.pi/180), 0],
[0, -np.sin(anglex*np.pi/180), np.cos(anglex*np.pi/180), 0,],
[0, 0, 0, 1]], np.float32)
ry = np.array([[np.cos(angley*np.pi/180), 0, np.sin(angley*np.pi/180), 0],
[0, 1, 0, 0],
[-np.sin(angley*np.pi/180),0, np.cos(angley*np.pi/180), 0,],
[0, 0, 0, 1]], np.float32)
rz = np.array([[np.cos(anglez*np.pi/180), np.sin(anglez*np.pi/180), 0, 0],
[-np.sin(anglez*np.pi/180), np.cos(anglez*np.pi/180), 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]], np.float32)
r = rx.dot(ry).dot(rz)
#四对点的生成
pcenter = np.array([h/2, w/2, 0, 0], np.float32)
p1 = np.array([0,0, 0,0], np.float32) - pcenter
p2 = np.array([w,0, 0,0], np.float32) - pcenter
p3 = np.array([0,h, 0,0], np.float32) - pcenter
p4 = np.array([w,h, 0,0], np.float32) - pcenter
dst1 = r.dot(p1)
dst2 = r.dot(p2)
dst3 = r.dot(p3)
dst4 = r.dot(p4)
list_dst = [dst1, dst2, dst3, dst4]
org = np.array([[0,0],
[w,0],
[0,h],
[w,h]], np.float32)
dst = np.zeros((4,2), np.float32)
#投影至成像平面
for i in range(4):
dst[i,0] = list_dst[i][0]*z/(z-list_dst[i][2]) + pcenter[0]
dst[i,1] = list_dst[i][1]*z/(z-list_dst[i][2]) + pcenter[1]
warpR = cv2.getPerspectiveTransform(org, dst)
print(org)
print(dst)
result = cv2.warpPerspective(img, warpR, (h,w),(0,255,0))
print(result.shape)
cv2.imshow("result", result)
cv2.imencode('.png',result)[1].tofile('test2.png')
c=cv2.waitKey(0)
cv2.destroyAllWindows()
这个问题的出现,在于对透视变化原理的理解;可能还有一些调试的技巧。
int _tmain(int argc_TCHARargv[])
{
    Mat src  = imread("E:/sandbox/test0.png");
    copyMakeBorder(src,src,10,10,10,10,BORDER_CONSTANT);
    if (!src.data)
        return 0;
    vector<Pointnot_a_rect_shape;
    not_a_rect_shape.push_back(Point(10,10));
    not_a_rect_shape.push_back(Point(74,10));
    not_a_rect_shape.push_back(Point(74,77));
    not_a_rect_shape.push_back(Point(10,77));
 
 
    cv::Point2f src_vertices[4];
    src_vertices[0] = not_a_rect_shape[0];
    src_vertices[1] = not_a_rect_shape[1];
    src_vertices[2] = not_a_rect_shape[2];
    src_vertices[3] = not_a_rect_shape[3];
    Point2f dst_vertices[4];
    dst_vertices[0] = Point(0, 10);
    dst_vertices[1] = Point(64,0);
    dst_vertices[2] = Point(84,77);
    dst_vertices[3] = Point(10,87);
    Mat warpMatrix = getPerspectiveTransform(src_verticesdst_vertices);
    cv::Mat rotated;
    warpPerspective(srcrotatedwarpMatrixrotated.size(), INTER_LINEARBORDER_CONSTANT);
    // Display the image
    cv::namedWindow"Original Image");
    cv::imshow"Original Image",src);
    cv::namedWindow"warp perspective");
    cv::imshow"warp perspective",rotated);
    
    cv::waitKey();
    return 0;
}
结果:





posted @ 2018-05-24 19:46 jsxyhelu 阅读(...) 评论(...) 编辑 收藏