知源图四点坐标和目标图四点坐标,求透视变换矩阵
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最近在搞图像处理,碰到了透视变换的问题。
同事给我一些代码,里边有误差,挺严重,让我帮他想想哪里出错了。捣鼓了很久,我猜测肯定是透视变换矩阵求错了,然后我的透视变换之旅就开始了。
后来问题解决了,发现他的矩阵和我求得矩阵一摸一样,他的代码并没有错误,是他采用的图片在做广角变换的时候有误差,导致程序结果误差。
首先感谢xiaowei_cqu,是她的一篇博客(http://blog.csdn.net/xiaowei_cqu/article/details/26471527)教会了我变换原理。
后来去用opencv来验证,发现函数调来调去太麻烦了,代码量也不小。
可是我只想要变换矩阵,自己写个不就好了!然后根据opencv源码里的原理,配上高斯消元自己写了个qwq。
talk is cheap,show me the code!!!
//Calculates coefficients of perspective transformation /* Calculates coefficients of perspective transformation * which maps (xi,yi) to (ui,vi), (i=1,2,3,4): * * c00*xi + c01*yi + c02 * ui = --------------------- * c20*xi + c21*yi + c22 * * c10*xi + c11*yi + c12 * vi = --------------------- * c20*xi + c21*yi + c22 * * Coefficients are calculated by solving linear system: * / x0 y0 1 0 0 0 -x0*u0 -y0*u0 \ /c00\ /u0* | x1 y1 1 0 0 0 -x1*u1 -y1*u1 | |c01| |u1| * | x2 y2 1 0 0 0 -x2*u2 -y2*u2 | |c02| |u2| * | x3 y3 1 0 0 0 -x3*u3 -y3*u3 |.|c10|=|u3|, * | 0 0 0 x0 y0 1 -x0*v0 -y0*v0 | |c11| |v0| * | 0 0 0 x1 y1 1 -x1*v1 -y1*v1 | |c12| |v1| * | 0 0 0 x2 y2 1 -x2*v2 -y2*v2 | |c20| |v2| * \ 0 0 0 x3 y3 1 -x3*v3 -y3*v3 / \c21/ \v3/ * * where: * cij - matrix coefficients, c22 = 1 */ void Gauss(double A[][9], int equ, int var, double* ans) { //epu:A‘s row var:A‘s col-1 int row, col; for (row = 0, col = 0; col<var&&row<equ; col++, row++) { int max_r = row; for (int i = row + 1; i<equ; i++) { if ((1e-12)<fabs(A[i][col]) - fabs(A[max_r][col])) { max_r = i; } } if (max_r != row) for (int j = 0; j<var + 1; j++) swap(A[row][j], A[max_r][j]); for (int i = row + 1; i<equ; i++) { if (fabs(A[i][col])<(1e-12)) continue; double tmp = -A[i][col] / A[row][col]; for (int j = col; j<var + 1; j++) { A[i][j] += tmp*A[row][j]; } } } for (int i = var - 1; i >= 0; i--) { //计算唯一解。 double tmp = 0; for (int j = i + 1; j<var; j++) { tmp += A[i][j] * (*(ans + j)); } ans[i] = (A[i][var] - tmp) / A[i][i]; } } void byx_getPerspectiveTransform(PointD * src, PointD * dst, double* ret){ double x0 = src[0].x, x1 = src[1].x, x2 = src[3].x, x3 = src[2].x; double y0 = src[0].y, y1 = src[1].y, y2 = src[3].y, y3 = src[2].y; double u0 = dst[0].x, u1 = dst[1].x, u2 = dst[3].x, u3 = dst[2].x; double v0 = dst[0].y, v1 = dst[1].y, v2 = dst[3].y, v3 = dst[2].y; double A[8][9] = { { x0, y0, 1, 0, 0, 0, -x0*u0, -y0*u0, u0 }, { x1, y1, 1, 0, 0, 0, -x1*u1, -y1*u1, u1 }, { x2, y2, 1, 0, 0, 0, -x2*u2, -y2*u2, u2 }, { x3, y3, 1, 0, 0, 0, -x3*u3, -y3*u3, u3 }, { 0, 0, 0, x0, y0, 1, -x0*v0, -y0*v0, v0 }, { 0, 0, 0, x1, y1, 1, -x1*v1, -y1*v1, v1 }, { 0, 0, 0, x2, y2, 1, -x2*v2, -y2*v2, v2 }, { 0, 0, 0, x3, y3, 1, -x3*v3, -y3*v3, v3 }, }; double ans[8] = { 0 }; Gauss(A, 8, 8, ans); *(ret + 0) = ans[0]; *(ret + 1) = ans[1]; *(ret + 2) = 0; *(ret + 3) = ans[2]; *(ret + 4) = ans[3]; *(ret + 5) = ans[4]; *(ret + 6) = 0; *(ret + 7) = ans[5]; *(ret + 8) = 0; *(ret + 9) = 0; *(ret + 10) = 1; *(ret + 11) = 0; *(ret + 12) = ans[6]; *(ret + 13) = ans[7]; *(ret + 14) = 0; *(ret + 15) = 1; } PointD byx_Transform(PointD p,double * mat){ PointD ret; double D = p.x*mat[12]+p.y*mat[13]+mat[15]; ret.x=(p.x*mat[0]+p.y*mat[1]+mat[3])/D; ret.y=(p.x*mat[4]+p.y*mat[5]+mat[7])/D; return ret; }
ps:变换矩阵是3*3的,我同事需要4*4的(第三行和第三列自行忽略)
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