前言:
当前咱们对“图像翻转算法”可能比较着重,看官们都想要分析一些“图像翻转算法”的相关文章。那么小编也在网摘上搜集了一些有关“图像翻转算法””的相关资讯,希望同学们能喜欢,姐妹们快快来学习一下吧!图像旋转是非常常见的图像变换,通常应用于图像矫正,在OpenCV可以使用密集仿射变换函数cv::warpAffine()实现图像旋转。为了理解图像旋转的原理,本文实现了一个图像旋转算法。
图像旋转是指将图像绕某个中心点旋转一定角度后,得到一幅新的图像。图像旋转的示意图如图1所示。其中,四边形ABCD表示需要旋转的图像区域,它经过旋转角度后得到的图像区域为四边形 A'B'C'D'。点p(x,y)为图像内任意一点,它经过旋转角度后对应的点为p'(x',y')。
图像是如何进行旋转的?通常这个过程有三个步骤。
第一步,把图像内的坐标点绕旋转中心点旋转到对应的坐标上。由于图像是通过二维数组进行保存的,所以图像的坐标点一定要落在坐标系的第一象限内,并且要保证它们是整数坐标点。通常情况下,进行旋转后得到的坐标点不是整数点也不一定在第一象限内,因此需要对旋转后得到的点进行平移和取整,使得它们都是落在第一象限内的整数点。
图像内任意一点p(x,y)绕某个旋转点(X,Y)逆时针旋转角度后得到点p'(x',y')的计算公式如下:
点旋转的C++实现代码如下。(OpenCV3.45+VS2019)
//将点point2绕点point1逆时针旋转angle度后得到新的点newPointvoid rotatePoint(cv::Point& point1, cv::Point& point2, cv::Point& newPoint, double angle){ int dx, dy; double dx1, dy1; dy1 = -((double)point2.x - point1.x) * sin(angle) + ((double)point2.y - point1.y) * cos(angle); dx1 = ((double)point2.x - point1.x) * cos(angle) + ((double)point2.y - point1.y) * sin(angle); if (dx1 - (int)dx1 > 0.5) //做一个四舍五入取整 dx = (int)dx1 + 1; else { if (dx1 - (int)dx1 < -0.5) dx = (int)dx1 - 1; else dx = (int)(dx1); } if (dy1 - (int)dy1 > 0.5) //做一个四舍五入取整 dy = (int)dy1 + 1; else { if (dy1 - (int)dy1 < -0.5) dy = (int)dy1 - 1; else dy = (int)(dy1); } newPoint.x = point1.x + dx; newPoint.y = point1.y + dy;}
用来平移坐标点的代码如下。
void translationPoint(cv::Point& point, int x, int y) //平移运算{ point.x = point.x + x; point.y = point.y + y;}
注:平移量x与y的大小,可以根据旋转后图像的四个顶点A'、B'、C'、D'获得。
第二步,把图像对应的坐标像素大小赋给旋转后的坐标。即,图像内任意点p(x,y)对应的像素值为I(x,y),那它旋转后得到的点p'(x',y')的像素值I(x',y')=I(x,y)。
如下图2所示。当我们需要旋转的图像区域在图片内时(这区域也可以是整张图片),如何确定旋转区域ABCD是很重要的,只有这样才能判断整张图片内的哪些点是四边形ABCD区域内的。
我们以图片的左上顶点为原点建立如图2所示的坐标系,其中四边形ABCD的四个顶点是已知的,分别为A(x0,y0)、B(x1,y1)、C(x2,y2)、D(x3,y3)。这时根据两点式可得到四条边的直线方程如下:
根据线性规划的知识,可以通过直线方程来表示四边形ABCD的区域。
注:因为四边形ABCD内的任意点p在直线AB上方,所以直线AB方程大于等于0;点p在直线BC左侧,所以直线BC方程小于等于0。同理可得,直线CD方程小于等于0、直线AD方程大于等于0。
实现代码块如下:
std::vector<cv::Point> newPoints; cv::Point newP; for (int i = 0; i < 4; ++i) { if (points[i] != point) //判断输入的4个顶点是否与旋转点point相同 { rotatePoint(point, points[i], newP, angle); //顶点points[i]与旋转点point不同,则进行旋转计算 newPoints.push_back(newP); } else { newPoints.push_back(points[i]); } } //获取经旋转后,新图像的大小,其中w表示图像宽长,h表示图像高长。 int w = 0, h = 0; int suw[4] = { newPoints[1].x - newPoints[0].x,newPoints[1].x - newPoints[3].x, newPoints[2].x - newPoints[0].x,newPoints[2].x - newPoints[3].x }; int suh[4] = { newPoints[2].y - newPoints[0].y ,newPoints[2].y - newPoints[1].y, newPoints[3].y - newPoints[0].y,newPoints[3].y - newPoints[1].y }; w = absMax4(suw); h = absMax4(suh); //获取需要旋转的四边形区域的外接矩形表示区域范围(x_min,y_min)、(x_max,y_max) int y_max, y_min, x_max, x_min; int points_x[4] = { points[0].x,points[1].x,points[2].x,points[3].x }; int points_y[4] = { points[0].y,points[1].y,points[2].y,points[3].y }; y_max = Max4(points_y); y_min = Min4(points_y); x_max = Max4(points_x); x_min = Min4(points_x); //计算向x轴的平移量dx,向y轴的平移量dy int dx, dy; int a[4] = { newPoints[0].x,newPoints[1].x,newPoints[2].x,newPoints[3].x }; int b[4] = { newPoints[0].y,newPoints[1].y,newPoints[2].y,newPoints[3].y }; dx = Min4(a); dy = Min4(b); //初始化输出矩阵 if(inputMat.type() == CV_8UC1) cv::Mat(h, w, CV_8UC1, cv::Scalar::all(255)).copyTo(outputMat); if(inputMat.type() == CV_8UC3) cv::Mat(h, w, CV_8UC3, cv::Scalar(255, 255, 255)).copyTo(outputMat); //实现I(x',y')=I(x,y) double z1, z2, z3, z4; for (int i = y_min; i < y_max; ++i) { for (int j = x_min; j < x_max; ++j) { //四边形顶点A为points[0],顶点B为points[1],顶点C为points[2],顶点D为points[3]. z1 = i - (double)points[0].y - (j - (double)points[0].x) * ((double)points[0].y - points[1].y) / ((double)points[0].x - points[1].x); z2 = j - (double)points[1].x - (i - (double)points[1].y) * ((double)points[1].x - points[2].x) / ((double)points[1].y - points[2].y); z3 = i - (double)points[2].y - (j - (double)points[2].x) * ((double)points[2].y - points[3].y) / ((double)points[2].x - points[3].x); z4 = j - (double)points[0].x - (i - (double)points[0].y) * ((double)points[0].x - points[3].x) / ((double)points[0].y - points[3].y); if (z1 >= 0 && z2 <= 0 && z3 <= 0 && z4 >= 0) { cv::Point point0(j, i); rotatePoint(point, point0, point0, angle); //将点point0绕点point旋转angle度得到新的点point0 translationPoint(point0, -dx, -dy); //平移 if (point0.x >= 0 && point0.x < w && point0.y >= 0 && point0.y < h) { if (inputMat.type() == CV_8UC1) { uchar* str = inputMat.ptr<uchar>(i); outputMat.at<uchar>(point0.y, point0.x) = str[j]; } if (inputMat.type() == CV_8UC3) { cv::Vec3b* str = inputMat.ptr<cv::Vec3b>(i); outputMat.at<cv::Vec3b>(point0.y, point0.x) = str[j]; } } } } }
第三步,对旋转后的图像进行插值。由于在第一步中对旋转后的点进行了取整,这难免会使得新图像存在间隙,所以需要对这些间隙进行填充。在OpenCV中常用的插值方法有以下5种:
在本文中采用的插值方法与最近邻插值类似,即把最近四个方向(上下左右)的平均值作为插值。
//灰度图(CV_8UC1)的插值代码for (int i = 1; i < outputMat.rows - 1; ++i) { for (int j = 1; j < outputMat.cols - 1; ++j) { if (outputMat.at<uchar>(i, j) == 255) { int sum = 0; uchar* str1 = outputMat.ptr<uchar>(i - 1); sum = str1[j - 1] + str1[j] + str1[j + 1]; uchar* str2 = outputMat.ptr<uchar>(i); sum = sum + str2[j - 1] + str2[j + 1]; uchar* str3 = outputMat.ptr<uchar>(i + 1); sum = sum + str3[j - 1] + str3[j] + str3[j + 1]; sum = sum / 8; outputMat.at<uchar>(i, j) = (uchar)sum; } } }///彩色图(CV_8UC3)的插值代码for (int i = 1; i < outputMat.rows - 1; ++i) { for (int j = 1; j < outputMat.cols - 1; ++j) { if (outputMat.at<cv::Vec3b>(i, j) == cv::Vec3b(255, 255, 255)) { int sum[3] = { 0,0,0 }; uchar r, g, b; for (int k = 0; k < 3; k++) { cv::Vec3b* str1 = outputMat.ptr<cv::Vec3b>(i - 1); sum[k] = str1[j - 1][k] + str1[j][k] + str1[j + 1][k]; cv::Vec3b* str2 = outputMat.ptr<cv::Vec3b>(i); sum[k] = sum[k] + str2[j - 1][k] + str2[j + 1][k]; cv::Vec3b* str3 = outputMat.ptr<cv::Vec3b>(i + 1); sum[k] = sum[k] + str3[j - 1][k] + str3[j][k] + str3[j + 1][k]; sum[k] = sum[k] / 8; } r = (uchar)sum[0]; g = (uchar)sum[1]; b = (uchar)sum[2]; outputMat.at<cv::Vec3b>(i, j) = cv::Vec3b(r, g, b); } } }
整个算法的完整代码如下:
#include<iostream>#include<opencv2/opencv.hpp> //计算点point2绕点point1逆时针旋转angle度后得到新的点newPointvoid rotatePoint(cv::Point& point1, cv::Point& point2, cv::Point& newPoint, double angle){ int dx, dy; double dx1, dy1; dy1 = -((double)point2.x - point1.x) * sin(angle) + ((double)point2.y - point1.y) * cos(angle); dx1 = ((double)point2.x - point1.x) * cos(angle) + ((double)point2.y - point1.y) * sin(angle); if (dx1 - (int)dx1 > 0.5) //做一个四舍五入 dx = (int)dx1 + 1; else { if (dx1 - (int)dx1 < -0.5) dx = (int)dx1 - 1; else dx = (int)(dx1); } if (dy1 - (int)dy1 > 0.5) //做一个四舍五入 dy = (int)dy1 + 1; else { if (dy1 - (int)dy1 < -0.5) dy = (int)dy1 - 1; else dy = (int)(dy1); } newPoint.x = point1.x + dx; newPoint.y = point1.y + dy;}void translationPoint(cv::Point& point, int x, int y) //平移运算{ point.x = point.x + x; point.y = point.y + y;}int Max4(int a[4]) //获取四个数中的最大值{ int max = a[0]; for (int i = 1; i < 4; i++) { if (max < a[i]) max = a[i]; } return max;}int Min4(int a[4]) //获取四个数中的最小值{ int min = a[0]; for (int i = 1; i < 4; i++) { if (min > a[i]) min = a[i]; } return min;}int absMax4(int a[4]){ int max = 0, m; for (int i = 0; i < 4; i++) { if (a[i] < 0) m = -a[i]; else m = a[i]; if (max < m) max = m; } return max;}void rotateImage(cv::Mat inputMat, cv::Mat& outputMat, std::vector<cv::Point> points, cv::Point point, double angle){ std::vector<cv::Point> newPoints; cv::Point newP; for (int i = 0; i < 4; ++i) { if (points[i] != point) //判断输入的4个顶点是否与旋转点point相同 { rotatePoint(point, points[i], newP, angle); //顶点points[i]与旋转点point不同,则进行旋转计算 newPoints.push_back(newP); } else { newPoints.push_back(points[i]); } } //获取经旋转后,新图像的大小,其中w表示图像宽长,h表示图像高长。 int w = 0, h = 0; int suw[4] = { newPoints[1].x - newPoints[0].x,newPoints[1].x - newPoints[3].x, newPoints[2].x - newPoints[0].x,newPoints[2].x - newPoints[3].x }; int suh[4] = { newPoints[2].y - newPoints[0].y ,newPoints[2].y - newPoints[1].y, newPoints[3].y - newPoints[0].y,newPoints[3].y - newPoints[1].y }; w = absMax4(suw); h = absMax4(suh); //获取需要旋转的四边形区域的外接矩形表示区域范围(x_min,y_min)、(x_max,y_max) int y_max, y_min, x_max, x_min; int points_x[4] = { points[0].x,points[1].x,points[2].x,points[3].x }; int points_y[4] = { points[0].y,points[1].y,points[2].y,points[3].y }; y_max = Max4(points_y); y_min = Min4(points_y); x_max = Max4(points_x); x_min = Min4(points_x); //计算向x轴的平移量dx,向y轴的平移量dy int dx, dy; int a[4] = { newPoints[0].x,newPoints[1].x,newPoints[2].x,newPoints[3].x }; int b[4] = { newPoints[0].y,newPoints[1].y,newPoints[2].y,newPoints[3].y }; dx = Min4(a); dy = Min4(b); //初始化输出矩阵 if(inputMat.type() == CV_8UC1) cv::Mat(h, w, CV_8UC1, cv::Scalar::all(255)).copyTo(outputMat); if(inputMat.type() == CV_8UC3) cv::Mat(h, w, CV_8UC3, cv::Scalar(255, 255, 255)).copyTo(outputMat); //实现I(x',y')=I(x,y) double z1, z2, z3, z4; for (int i = y_min; i < y_max; ++i) { for (int j = x_min; j < x_max; ++j) { //四边形顶点A为points[0],顶点B为points[1],顶点C为points[2],顶点D为points[3]. //直线AB z1 = i - (double)points[0].y - (j - (double)points[0].x) * ((double)points[0].y - points[1].y) / ((double)points[0].x - points[1].x); //直线BC z2 = j - (double)points[1].x - (i - (double)points[1].y) * ((double)points[1].x - points[2].x) / ((double)points[1].y - points[2].y); //直线CD z3 = i - (double)points[2].y - (j - (double)points[2].x) * ((double)points[2].y - points[3].y) / ((double)points[2].x - points[3].x); //直线AD z4 = j - (double)points[0].x - (i - (double)points[0].y) * ((double)points[0].x - points[3].x) / ((double)points[0].y - points[3].y); if (z1 >= 0 && z2 <= 0 && z3 <= 0 && z4 >= 0) { cv::Point point0(j, i); rotatePoint(point, point0, point0, angle); //将点point0绕点point旋转angle度得到新的点point0 translationPoint(point0, -dx, -dy); if (point0.x >= 0 && point0.x < w && point0.y >= 0 && point0.y < h) { if (inputMat.type() == CV_8UC1) { uchar* str = inputMat.ptr<uchar>(i); outputMat.at<uchar>(point0.y, point0.x) = str[j]; } if (inputMat.type() == CV_8UC3) { cv::Vec3b* str = inputMat.ptr<cv::Vec3b>(i); outputMat.at<cv::Vec3b>(point0.y, point0.x) = str[j]; } } } } } if (inputMat.type() == CV_8UC1) { //插值 for (int i = 1; i < outputMat.rows - 1; ++i) { for (int j = 1; j < outputMat.cols - 1; ++j) { if (outputMat.at<uchar>(i, j) == 255) { int sum = 0; uchar* str1 = outputMat.ptr<uchar>(i - 1); sum = str1[j - 1] + str1[j] + str1[j + 1]; uchar* str2 = outputMat.ptr<uchar>(i); sum = sum + str2[j - 1] + str2[j + 1]; uchar* str3 = outputMat.ptr<uchar>(i + 1); sum = sum + str3[j - 1] + str3[j] + str3[j + 1]; sum = sum / 8; outputMat.at<uchar>(i, j) = (uchar)sum; } } } } if (inputMat.type() == CV_8UC3) { //插值 for (int i = 1; i < outputMat.rows - 1; ++i) { for (int j = 1; j < outputMat.cols - 1; ++j) { if (outputMat.at<cv::Vec3b>(i, j) == cv::Vec3b(255, 255, 255)) { int sum[3] = { 0,0,0 }; uchar r, g, b; for (int k = 0; k < 3; k++) { cv::Vec3b* str1 = outputMat.ptr<cv::Vec3b>(i - 1); sum[k] = str1[j - 1][k] + str1[j][k] + str1[j + 1][k]; cv::Vec3b* str2 = outputMat.ptr<cv::Vec3b>(i); sum[k] = sum[k] + str2[j - 1][k] + str2[j + 1][k]; cv::Vec3b* str3 = outputMat.ptr<cv::Vec3b>(i + 1); sum[k] = sum[k] + str3[j - 1][k] + str3[j][k] + str3[j + 1][k]; sum[k] = sum[k] / 8; } r = (uchar)sum[0]; g = (uchar)sum[1]; b = (uchar)sum[2]; outputMat.at<cv::Vec3b>(i, j) = cv::Vec3b(r, g, b); } } } }}int main(){ cv::Mat srcImage = cv::imread("F:/图像处理/图片一/thee2.jpg"); //cv::Mat srcImage = cv::imread("F:/图像处理/图片一/thee2.jpg", 0); if (srcImage.empty()) { printf("图片读取失败!\n"); return -1; } //需要旋转的图像区域四个顶点 std::vector<cv::Point> points; points.push_back(cv::Point(0, 0)); points.push_back(cv::Point(srcImage.cols, 0)); points.push_back(cv::Point(srcImage.cols, srcImage.rows)); points.push_back(cv::Point(0, srcImage.rows)); cv::Mat outputImage; rotateImage(srcImage, outputImage, points, cv::Point(40, 70), 0.4); cv::imshow("原图", srcImage); cv::imshow("旋转得到的图像", outputImage); cv::waitKey(0); return 0;}
算法测试:
输入图像(大小460X613):
//旋转区域1,其中srcImage为输入图像std::vector<cv::Point> points; points.push_back(cv::Point(0, 0)); //顶点A points.push_back(cv::Point(srcImage.cols, 0)); //顶点B points.push_back(cv::Point(srcImage.cols, srcImage.rows)); //顶点C points.push_back(cv::Point(0, srcImage.rows));//旋转点为Point(40, 70),旋转角度0.4
旋转得到的结果如下:
//旋转区域2,其中srcImage为输入图像std::vector<cv::Point> points; points.push_back(cv::Point(70, 0)); //顶点A points.push_back(cv::Point(srcImage.cols-50, 0)); points.push_back(cv::Point(srcImage.cols-100, srcImage.rows-100)); points.push_back(cv::Point(0, srcImage.rows-50));//旋转点为Point(30, 0),旋转角度0.4
旋转得到的结果如下:
标签: #图像翻转算法 #图像旋转c程序 #图像旋转c程序是什么