电⼦围栏
1、了个电⼦围栏算法，也就是多边形点，在图形内还是图形外
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import json
import math
lnglatlist = []
data = '[{"name":"武汉市三环","points":[{"lng":114.193437,"lat":30.513069},{"lng":114.183376,"lat":30.509211},{"lng":114.188191,"lat":30.505291},{"lng":114.187975,"lat":30.504731},{"lng":114.201773,"lat":30.492782},{"lng":114.213559,"lat":30.4885 data = json.loads(data)
if'points'in data[0]:
for point in data[0]['points']:
activitiestodo中文什么意思#print(str(point['lng'])+" "+str(point['lat']))
lnglat = []
lnglat.append(float(str(point['lng'])))
lnglat.append(float(str(point['lat'])))
lnglatlist.append(lnglat)
def windingNumber(point, poly):
poly.append(poly[0])
px = point[0]
py = point[1]
sum = 0
length = len(poly)-1
for index in range(0,length):
sx = poly[index][0]
sy = poly[index][1]
tx = poly[index+1][0]
ty = poly[index+1][1]
#点与多边形顶点重合或在多边形的边上
if((sx - px) * (px - tx) >= 0 and (sy - py) * (py - ty) >= 0 and (px - sx) * (ty - sy) == (py - sy) * (tx - sx)):
return"on"
#点与相邻顶点连线的夹⾓
angle = math.atan2(sy - py, sx - px) - math.atan2(ty - py, tx - px)
#确保夹⾓不超出取值范围（-π到π）
if(angle >= math.pi):
angle = angle - math.pi * 2
elif(angle <= -math.pi):
angle = angle + math.pi * 2
sum += angle
#计算回转数并判断点和多边形的⼏何关系
result = 'out'if int(sum / math.pi) == 0 else'in'
return result
point = [114.193437,30.513068]
print(windingNumber(point,lnglatlist))
转⾃
2、我的项⽬json⽂件⽐较复杂，需要⽤个⼆维数据来存储坐标数据
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import json
import math
def windingNumber(point, poly):
poly.append(poly[0])cathy这个英文名老气吗
px = point[0]
py = point[1]
sum = 0
length = len(poly)-1
for index in range(0,length):
sx = poly[index][0]
sy = poly[index][1]
tx = poly[index+1][0]
ty = poly[index+1][1]
#点与多边形顶点重合或在多边形的边上
if((sx - px) * (px - tx) >= 0 and (sy - py) * (py - ty) >= 0 and (px - sx) * (ty - sy) == (py - sy) * (tx - sx)):
return"on"
#点与相邻顶点连线的夹⾓
angle = math.atan2(sy - py, sx - px) - math.atan2(ty - py, tx - px)
#确保夹⾓不超出取值范围（-π到π）
if(angle >= math.pi):
angle = angle - math.pi * 2
elif(angle <= -math.pi):
angle = angle + math.pi * 2
sum += angle
#计算回转数并判断点和多边形的⼏何关系
result = 'out'if int(sum / math.pi) == 0 else'in'
return result
if__name__ == "__main__":
with open("railwayFence.json", 'r') as f:
data = json.ad())
data = data['railwayFence']
python解析json文件
#print(len(data))
#print(data[65]['areaName'])
lnglatlist = [[]]*len(data)
point_test = [115.259161584,38.813623816]
point_test1 = [115.243922249,38.836012517]
for i in range(0,len(data)):
for point in data[i]['coordinates']:
#print(str(point['L'])+" "+str(point['B']))
lnglat = []
lnglat.append(float(str(point['L'])))
lnglat.append(float(str(point['B'])))
lnglatlist[i].append(lnglat)
ret = windingNumber(point_test1,lnglatlist[i])
if ret == 'in'or ret == 'on':
break
print ret
3、解析出json中的坐标点，可以描出电⼦围栏，打点测试就很⽅便了
<
截取⼏个点，格式如下
38.836013 115.243822
38.836012 115.243818
38.836013 115.243813
38.836013 115.243809
38.836015 115.243805
38.836017 115.243801
38.836019 115.243898
38.836022 115.243895
38.836023 115.243895
38.836092 115.243850
38.836160 115.243806
38.836189 115.243788
38.836218 115.243769
38.836416 115.243642
38.836613 115.243515
38.837036 115.243243
plot.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import matplotlib.pyplot as plt
import numpy as np
import json
with open("railwayFence.json", 'r') as f:
data = json.ad())
data = data['railwayFence']
fo = open("tmp", "w")
for i in range(0,len(data)):
for point in data[i]['coordinates']:
d = '%(x).9f %(y).9f\n'%{'x':float(str(point['B'])),'y':float(str(point['L']))}
fo.write( d )
fo.close()
data = np.loadtxt('tmp')
plt.plot(data[:,0],data[:,1])
#plt.annotate('test point', xy=(39.82775139,116.250658818),arrowprops=dict(facecolor='red', shrink=
0)) #plt.annotate('test point', xy=(39.823400546,116.25345992),arrowprops=dict(facecolor='red', shrink=0)) plt.annotate('test point', xy=(39.813623816,116.259161584),arrowprops=dict(facecolor='red', shrink=0)) plt.show()
python plot.py
4、计算单个围栏
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import json
import math
lnglatlist = []
with open("65.json", 'r') as f:
data = json.ad())
#print(data)
print(data['areaName'])
#data = json.loads(data)
if'coordinates'in data:
for point in data['coordinates']:
#print(str(point['L'])+" "+str(point['B']))
lnglat = []
lnglat.append(float(str(point['L'])))
lnglat.append(float(str(point['B'])))
lnglatlist.append(lnglat)
def windingNumber(point, poly):
poly.append(poly[0])
px = point[0]
vbs正则表达式获取匹配结果py = point[1]
sum = 0
length = len(poly)-1
for index in range(0,length):
sx = poly[index][0]
sy = poly[index][1]
tx = poly[index+1][0]
ty = poly[index+1][1]
#点与多边形顶点重合或在多边形的边上
if((sx - px) * (px - tx) >= 0 and (sy - py) * (py - ty) >= 0 and (px - sx) * (ty - sy) == (py - sy) * (tx - sx)): return"on"
#点与相邻顶点连线的夹⾓
angle = math.atan2(sy - py, sx - px) - math.atan2(ty - py, tx - px)
#确保夹⾓不超出取值范围（-π到π）
if(angle >= math.pi):
angle = angle - math.pi * 2
elif(angle <= -math.pi):
angle = angle + math.pi * 2
sum += angle
#计算回转数并判断点和多边形的⼏何关系
result = 'out'if int(sum / math.pi) == 0 else'in'
return result
point = [115.259161584,38.813623816]
print(windingNumber(point,lnglatlist))
5、展⽰单个围栏
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import matplotlib.pyplot as plt
import numpy as np
import json
'''
with open("railwayFence.json", 'r') as f:
data = json.ad())
data = data['railwayFence']
fo = open("tmp", "w")
for i in range(0,len(data)):
for point in data[i]['coordinates']:
d = '%(x).9f %(y).9f\n'%{'x':float(str(point['B'])),'y':float(str(point['L']))}
fo.write( d )
fo.close()
'''
data = np.loadtxt('65.txt')
plt.plot(data[:,0],data[:,1])
#plt.annotate('test point', xy=(38.82775139,115.250658818),arrowprops=dict(facecolor='red', shrink=0))
#plt.annotate('test point', xy=(38.823400546,115.25345992),arrowprops=dict(facecolor='red', shrink=0)) plt.annotate('test point', xy=(38.813623816,115.259161584),arrowprops=dict(facecolor='red', shrink=0)) plt.show()
6、⽤c语⾔解析json也⾏
main.c
/*
reference documentation
blog.csdn/stsahana/article/details/79638992
blog.csdn/fengxinlinux/article/details/53121287
*/
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include "cJSON.h"
char *json_file_path = "./railwayFence.json";
char *json_loader(char *path)
{
FILE *f;
long len;
char *content;
f=fopen(path,"rb");
fseek(f,0,SEEK_END);
len=ftell(f);
fseek(f,0,SEEK_SET);
content=(char*)malloc(len+1);
fread(content,1,len,f);
fclose(f);
javaweb课程
return content;
}
int main(void)
{
FILE *fp1;
fp1=fopen("","w+");
char *json_str = json_loader(json_file_path);
cJSON *root=cJSON_Parse(json_str);
if (!root) {
printf("Error before: [%s]\n",cJSON_GetErrorPtr());
}
//railwayFence array
cJSON *railwayFenceArray = cJSON_GetObjectItem(root, "railwayFence");
if(!railwayFenceArray){
printf("Error before: [%s]\n",cJSON_GetErrorPtr());
}
int railwayFence_array_size = cJSON_GetArraySize(railwayFenceArray);
printf("railwayFence array size is %d\n",railwayFence_array_size);
int i = 0;
char *p  = NULL;
cJSON *it;
for(i=0; i< railwayFence_array_size; i++) {
FILE *fp;
char  buffer[32];
sprintf( buffer, "./data/%d.txt", i );
fp=fopen(buffer,"w+");
cJSON *railwayFenceItem = cJSON_GetArrayItem(railwayFenceArray,i);
p = cJSON_PrintUnformatted(railwayFenceItem);
it = cJSON_Parse(p);
if(!it)
continue ;
cJSON *areaName,*dangerType;
areaName = cJSON_GetObjectItem(it, "areaName");
printf("areaName is %s\n",areaName->valuestring);
dangerType = cJSON_GetObjectItem(it, "dangerType");
printf("dangerType is %s\n",dangerType->valuestring);
//Coordinate array
cJSON *CoordinateArray = cJSON_GetObjectItem(railwayFenceItem, "coordinates");
if(!CoordinateArray){
printf("Error before: [%s]\n",cJSON_GetErrorPtr());
}
int Coordinate_array_size = cJSON_GetArraySize(CoordinateArray);
printf("Coordinate array size is %d\n",Coordinate_array_size);
int j = 0;
char *q  = NULL;
cJSON *jt;
for(j=0; j< Coordinate_array_size; j++) {
cJSON *CoordinateItem = cJSON_GetArrayItem(CoordinateArray,j);
q = cJSON_PrintUnformatted(CoordinateItem);
jt = cJSON_Parse(q);
if(!jt)
continue ;
cJSON *B,*L;
B = cJSON_GetObjectItem(jt, "B");
printf("B is %f\n",B->valuedouble);
L = cJSON_GetObjectItem(jt, "L");
printf("L is %f\n",L->valuedouble);
fprintf(fp1,"%f %f\n",B->valuedouble,L->valuedouble);
fprintf(fp,"%f %f\n",B->valuedouble,L->valuedouble);
free(q); cJSON_Delete(jt);
}
free(p); cJSON_Delete(it);
}
if(root) {
cJSON_Delete(root);
//return 0;
}
return0;
}
Makefile
OBJ= main
all: ${OBJ}
main:
gcc -g -o main main.c cJSON.c -lm
clean:
rm -f ${OBJ}
.PHONY: ${OBJ}
7、nodejs解析json⽂件更简单
main.js
var fs = require("fs");
var contents = fs.readFileSync("railwayFence.json");
var obj = JSON.parse(contents);
//console.log("<<<<<<<<<<<<<<<<<<<<"+JSON.stringify(obj));
for(var i in obj.railwayFence){
console.log('>>>>>>>>>>>>>>>>>>>>>>>>put areaName: ' + obj.railwayFence[i].areaName)
console.log('>>>>>>>>>>>>>>>>>>>>>>>>put dangerTypeName: ' + obj.railwayFence[i].danggerTypeName) for(var j in obj.railwayFence[i].coordinates){
console.log('>>>>>>>>>>>>>>>>>>>>>>>>put B: ' + obj.railwayFence[i].coordinates[j].B)
console.log('>>>>>>>>>>>>>>>>>>>>>>>>put L: ' + obj.railwayFence[i].coordinates[j].L)
}
}
8、同事的c算法，记录⼀下
/**
* 功能：判断点是否在多边形内
* ⽅法：求解通过该点的⽔平线（射线）与多边形各边的交点
* 结论：单边交点为奇数，成⽴!
* 参数：p 指定的某个点
ptPolygon 多边形的各个顶点坐标（⾸末点可以不⼀致）
nCount 多边形定点的个数
* 说明：
*/
// 注意：在有些情况下x值会计算错误，可把double类型改为long类型即可解决。
int PtInPolygon(Point_t* p, Point_t* ptPolygon, int nCount)
{
int nCross = 0, i;
double x;
Point_t p1, p2;entware常用插件
for (i = 0; i < nCount; i++)
{
p1 = ptPolygon[i];
p2 = ptPolygon[(i + 1) % nCount];
// 求解 y=p->y 与 p1p2 的交点
if ( p1.y == p2.y ) // p1p2 与 y=p->y平⾏
continue;
if ( p->y < min(p1.y, p2.y) ) // 交点在p1p2延长线上
continue;
if ( p->y >= max(p1.y, p2.y) ) // 交点在p1p2延长线上
continue;
// 求交点的 X 坐标 --------------------------------------------------------------
x = (double)(p->y - p1.y) * (double)(p2.x - p1.x) / (double)(p2.y - p1.y) + p1.x;
if ( x > p->x )
{
nCross++; // 只统计单边交点
}
}
// 单边交点为偶数，点在多边形之外 ---
return (nCross % 2 == 1);
}
/*******************************************************************************
* 名称: pointInPolygon
* 功能: 当前平⾯坐标点是否在区域⾯内
* 形参: point：需判断点
*      points：区域⾯点集合
*      count：区域⾯点个数
* 返回: 判断结果 1在区域内 0在区域外
* 说明: ⽆
******************************************************************************/
int pointInPolygon(Point_t *point, Point_t *points,int count)
{
int  i,j=count-1 ;
int  oddNodes = 0 ;
double  x,y;
if(point == NULL || points == NULL)
return -1;
x = point->x, y = point->y;
for (i=0;i<count; i++)
{
if((points[i].y< y && points[j].y>=y
|| points[j].y<y && points[i].y>=y)
&& (points[i].x<=x || points[j].x<=x))
{
oddNodes^=(points[i].x+(y-points[i].y)/(points[j].y-points[i].y)*(points[j].x-points[i].x)<x);  }
j=i;
}
return oddNodes;
}
怎么处理都⾏，怎么⽅便怎么⽤。
end

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