opencv python 我为你提供二个点start_point = (0, 180) end_point = (1920, 350)，沿着这二点的线，裁剪图片frame

好的，您可以使用 OpenCV 和 Python 实现这个功能。下面是一个示例代码，您可以根据需要进行调整：

import cv2

# 读取图片
frame = cv2.imread('your_image.jpg')

# 定义裁剪的起始点和结束点
start_point = (0, 180)
end_point = (1920, 350)

# 绘制矩形框
cv2.rectangle(frame, start_point, end_point, (0, 255, 0), 3)

# 裁剪图片
cropped = frame[start_point[1]:end_point[1], start_point[0]:end_point[0]]

# 显示原图和裁剪后的图片
cv2.imshow("Original Image", frame)
cv2.imshow("Cropped Image", cropped)
cv2.waitKey(0)

在这个示例中，我们首先读取了一张图片。然后，我们定义了起始点和结束点，这里是 (0, 180) 和 (1920, 350)。接下来，我们使用 `cv2.rectangle()` 函数绘制了一个矩形框，用于表示裁剪的区域。最后，我们使用 Numpy 数组切片的方式，对原图进行裁剪，并将裁剪后的图片显示出来。

注意，这个示例代码中的裁剪区域是矩形，如果您需要沿着任意的线裁剪图片，则需要使用 OpenCV 中的其他函数。

相关问题

import cv2 import math def cal_ang(start, center, end): point_1 = start point_2 = center point_3 = end a = math.sqrt( (point_2[0] - point_3[0]) * (point_2[0] - point_3[0]) + (point_2[1] - point_3[1]) * (point_2[1] - point_3[1])) b = math.sqrt( (point_1[0] - point_3[0]) * (point_1[0] - point_3[0]) + (point_1[1] - point_3[1]) * (point_1[1] - point_3[1])) c = math.sqrt( (point_1[0] - point_2[0]) * (point_1[0] - point_2[0]) + (point_1[1] - point_2[1]) * (point_1[1] - point_2[1])) A = math.degrees(math.acos((a * a - b * b - c * c) / (-2 * b * c))) B = math.degrees(math.acos((b * b - a * a - c * c) / (-2 * a * c))) C = math.degrees(math.acos((c * c - a * a - b * b) / (-2 * a * b))) return B img = cv2.imread('46.png') gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY) ret,thresh = cv2.threshold(gray, 70, 255, cv2.THRESH_BINARY) contours,hierarchy=cv2.findContours(thresh,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_NONE) hull = cv2.convexHull(contours[0],returnPoints=False) defects = cv2.convexityDefects(contours[0],hull) start = end = (0,0) for i in range(0,defects.shape[0]): s,e,f,d = defects[i,0] start = tuple(contours[0][s][0]) end = tuple(contours[0][e][0]) far = tuple(contours[0][f][0]) if d > 5000: cv2.line(img,start,end,[0,255,0],2) cv2.circle(img,end,5,[0,0,255],-1) cv2.circle(img,start,5,[0,0,255],-1) break cv2.imshow('find', img) center,radius = cv2.minEnclosingCircle(contours[0]) cv2.circle(img,(int(center[0]),int(center[1])),8,(255,0,255),-1) cv2.circle(img,end,8,[255,0,0],-1) cv2.circle(img,start,8,[255,0,0],-1) cv2.line(img,start,(int(center[0]),int(center[1])),[0,0,255],2) cv2.line(img,end,(int(center[0]),int(center[1])),[0,0,255],2) angle = cal_ang(start,center,end) print('angle = %0.2f' % angle) length = (1 - angle / 360.0) * math.pi * radius * 2 print((angle / 360.0)) print('radius = %0.2f' % radius) strL = 'length=%0.2f' % length cv2.putText(img,strL,(int(center[0]-40),int(center[1]+40)),0,0.8,(0,255,0),2) cv2.imshow('result', img) angle_1 = cal_ang(start, center, ((center[0]+100),(center[1]))) angle_2 = cal_ang(end, center, ((center[0]+100),(center[1]))) cv2.ellipse(img,(int(center[0]),int(center[1])),(int(radius),int(radius)),0,-angle_1,0,(255,0,255),2, cv2.LINE_AA) cv2.ellipse(img,(int(center[0]),int(center[1])),(int(radius),int(radius)),0,0,angle_2,(255,0,255),2,cv2.LINE_AA) cv2.imshow('result', img) cv2.imwrite('result.png',img) cv2.waitKey(0) cv2.destroyAllWindows(),将这段代码转换为c++

以下是将Python代码转换为C++的结果。注意，由于数据类型和库函数的区别，代码可能需要进行微调才能正常运行。

#include <iostream>
#include <opencv2/opencv.hpp>
#include <math.h>

using namespace std;
using namespace cv;

double cal_ang(Point start, Point center, Point end) {
    Point2f point_1 = start;
    Point2f point_2 = center;
    Point2f point_3 = end;
    double a = sqrt(pow(point_2.x - point_3.x, 2) + pow(point_2.y - point_3.y, 2));
    double b = sqrt(pow(point_1.x - point_3.x, 2) + pow(point_1.y - point_3.y, 2));
    double c = sqrt(pow(point_1.x - point_2.x, 2) + pow(point_1.y - point_2.y, 2));
    double A = acos((a * a - b * b - c * c) / (-2 * b * c)) * 180 / CV_PI;
    double B = acos((b * b - a * a - c * c) / (-2 * a * c)) * 180 / CV_PI;
    double C = acos((c * c - a * a - b * b) / (-2 * a * b)) * 180 / CV_PI;
    return B;
}

int main() {
    Mat img = imread("46.png");
    Mat gray;
    cvtColor(img, gray, COLOR_BGR2GRAY);
    threshold(gray, gray, 70, 255, THRESH_BINARY);
    vector<vector<Point>> contours;
    vector<Vec4i> hierarchy;
    findContours(gray, contours, hierarchy, RETR_EXTERNAL, CHAIN_APPROX_NONE);
    vector<vector<Point>> hull(contours.size());
    vector<vector<int>> hullsI(contours.size());
    vector<vector<Vec4i>> defects(contours.size());
    convexHull(contours[0], hull[0], false);
    convexHull(contours[0], hullsI[0], false);
    if (hullsI[0].size() > 0) {
        Point2f* pts = new Point2f[hullsI[0].size()];
        for (size_t i = 0; i < hullsI[0].size(); i++) {
            pts[i] = contours[0][hullsI[0][i]];
        }
        int n = hullsI[0].size();
        convexityDefects(pts, n, hullsI[0], defects[0]);
        delete[] pts;
    }
    Point start, end;
    for (int i = 0; i < defects[0].size(); i++) {
        Vec4i& v = defects[0][i];
        int startidx = v[0];
        Point ptStart(contours[0][startidx]);
        int endidx = v[1];
        Point ptEnd(contours[0][endidx]);
        int faridx = v[2];
        Point ptFar(contours[0][faridx]);
        if (v[3] > 5000) {
            line(img, ptStart, ptEnd, Scalar(0, 255, 0), 2);
            circle(img, ptEnd, 5, Scalar(0, 0, 255), -1);
            circle(img, ptStart, 5, Scalar(0, 0, 255), -1);
            start = ptStart;
            end = ptEnd;
            break;
        }
    }
    Point2f center;
    float radius = 0;
    minEnclosingCircle(contours[0], center, radius);
    circle(img, center, 8, Scalar(255, 0, 255), -1);
    circle(img, end, 8, Scalar(255, 0, 0), -1);
    circle(img, start, 8, Scalar(255, 0, 0), -1);
    line(img, start, Point(center), Scalar(0, 0, 255), 2);
    line(img, end, Point(center), Scalar(0, 0, 255), 2);
    double angle = cal_ang(start, center, end);
    cout << "angle = " << angle << endl;
    double length = (1 - angle / 360.0) * CV_PI * radius * 2;
    cout << angle / 360.0 << endl;
    cout << "radius = " << radius << endl;
    stringstream ss;
    ss << "length=" << fixed << setprecision(2) << length;
    String strL = ss.str();
    putText(img, strL, Point(center.x - 40, center.y + 40), 0, 0.8, Scalar(0, 255, 0), 2);
    double angle_1 = cal_ang(start, center, Point(center.x + 100, center.y));
    double angle_2 = cal_ang(end, center, Point(center.x + 100, center.y));
    ellipse(img, center, Size(radius, radius), 0, -angle_1, 0, Scalar(255, 0, 255), 2, LINE_AA);
    ellipse(img, center, Size(radius, radius), 0, 0, angle_2, Scalar(255, 0, 255), 2, LINE_AA);
    imshow("result", img);
    imwrite("result.png", img);
    waitKey(0);
    destroyAllWindows();
    return 0;
}

opencv python 人意给出斜率和两点 如何经过这两点画出一条贯穿图像的直线

要在OpenCV中使用斜率和两点来画出一条贯穿图像的直线，可以按照以下步骤进行操作：

1. 计算斜率：使用两点的坐标计算直线的斜率，公式为： slope = (y2 - y1) / (x2 - x1)

2. 计算截距：使用斜率和其中一个点的坐标计算直线的截距，公式为： intercept = y1 - slope * x1

3. 画出直线：使用cv2.line()函数画出直线，需要提供起点和终点的坐标，即可将直线画在图像上。

下面是一个示例代码，其中(x1,y1)和(x2,y2)是两个点的坐标：

import cv2

# 读取图像
img = cv2.imread('image.jpg')

# 计算斜率和截距
x1, y1 = 100, 100
x2, y2 = 200, 200
slope = (y2 - y1) / (x2 - x1)
intercept = y1 - slope * x1

# 画出直线
start_point = (0, int(intercept))
end_point = (img.shape[1], int(slope * img.shape[1] + intercept))
color = (0, 0, 255)
thickness = 2
cv2.line(img, start_point, end_point, color, thickness)

# 显示图像
cv2.imshow('image', img)
cv2.waitKey(0)
cv2.destroyAllWindows()

这样就可以在图像上画出一条贯穿两点的直线了。