思路
1.输入三个点
2.确定三个点的位置
- 缺失对角的为 x1 点,另外两个分别为 x2, x3
- 三个确定的点组成一个三角形,计算两点之间的距离
- 距离转角度
- 角度最大的为 x1
3.x4 = x2+ x3 – x1
y4 = y2+ y3 – y1
import cv2
import numpy as np
import os
import math
def cal_ang(point_1 , point_2, point_3):
point = { }
""" 根据三点坐标计算夹角 :param point_1: 点1坐标 :param point_2: 点2坐标 :param point_3: 点3坐标 :return: 返回任意角的夹角值,这里只是返回点2的夹角 """
#计算两点之间的距离
a=math.sqrt((point_2[0]-point_3[0])**2+(point_2[1]-point_3[1])**2)
b=math.sqrt((point_1[0]-point_3[0])**2+(point_1[1]-point_3[1])**2)
c=math.sqrt((point_1[0]-point_2[0])**2+(point_1[1]-point_2[1])**2)
#计算线段b, c之间的夹角, 点2对应的夹角
A=math.degrees(math.acos((a*a-b*b-c*c)/(-2*b*c)))# 2,3 2
#计算线段a, c之间的夹角, 点3对应的夹角
B=math.degrees(math.acos((b*b-a*a-c*c)/(-2*a*c)))# 1,3 3
#计算线段a, b之间的夹角, 点1对应的夹角
C=math.degrees(math.acos((c*c-a*a-b*b)/(-2*a*b)))# 1,2 1
point["bndbox[0]"] = A
point["bndbox[1]"] = B
point["bndbox[2]"] = C
#对夹角进行排序
point = dict(sorted(point.items(), key=lambda e: e[1]))
keys = list(point.keys())
if keys[2] == 'bndbox[0]':
return [point_1, point_2, point_3]
elif keys[2] == 'bndbox[1]':
return [point_2, point_1, point_3]
elif keys[2] == 'bndbox[2]':
return [point_3, point_2, point_1]
point = [[40, 0], [22, 6], [0, 71]]
points = cal_ang(point[0] , point[1], point[2])
x4 = points[1][0] + points[2][0] - points[0][0]
y4 = points[1][1] + points[2][1] - points[0][1]
#注意:可能x4, y4会超过边界;