求任意多边形面积 python实现

Posted 2020-08-30

数学解决方法:

　　多边形外选取一点，连接各点构成三角形，计算求和......  详细链接  http://blog.csdn.net/hemmingway/article/details/7814494

 

　　已知三角形三边长，求三角形面积------>海伦公式  链接:http://www.zybang.com/question/29d209e9732d1f1f6a6de35b94edd3ba.html

 

由上实现下面python代码   源代码出处  链接: http://blog.csdn.net/zfq43/article/details/8595288

 

# -*- coding: cp936 -*-
import math


class Point():
    def __init__(self,x,y):
        self.x = x
        self.y = y




def GetAreaOfPolyGon(points):

    area = 0
    if(len(points)<3):
        
         raise Exception("error")

    p1 = points[0]
    for i in range(1,len(points)-1):
        p2 = points[1]
        p3 = points[2]

        #计算向量
        vecp1p2 = Point(p2.x-p1.x,p2.y-p1.y)
        vecp2p3 = Point(p3.x-p2.x,p3.y-p2.y)


        
        #判断顺时针还是逆时针，顺时针面积为正，逆时针面积为负
        vecMult = vecp1p2.x*vecp2p3.y - vecp1p2.y*vecp2p3.x   #判断正负方向比较有意思
        sign = 0
        if(vecMult>0):
            sign = 1
        elif(vecMult<0):
            sign = -1

        triArea = GetAreaOfTriangle(p1,p2,p3)*sign
        area += triArea
    return abs(area)


def GetAreaOfTriangle(p1,p2,p3):
    ‘‘‘计算三角形面积   海伦公式‘‘‘
    area = 0
    p1p2 = GetLineLength(p1,p2)
    p2p3 = GetLineLength(p2,p3)
    p3p1 = GetLineLength(p3,p1)
    s = (p1p2 + p2p3 + p3p1)/2
    area = s*(s-p1p2)*(s-p2p3)*(s-p3p1)   #海伦公式
    area = math.sqrt(area)
    return area

def GetLineLength(p1,p2):
    ‘‘‘计算边长‘‘‘
    length = math.pow((p1.x-p2.x),2) + math.pow((p1.y-p2.y),2)  #pow  次方
    length = math.sqrt(length)   
    return length    

def main():
    p1 = Point(1,1)
    p2 = Point(2,1)
    p3 = Point(2,2)
    p4 = Point(1,2)
    points = [p1,p2,p3,p4]
    area = GetAreaOfPolyGon(points)
    print(math.ceil(area))
    assert math.ceil(area)==1


if __name__ == ‘__main__‘:
    main()