Java经纬度生成 判断一个点是否在一个面上
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2022-04-02 18:49:31
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经纬度生成
让前端圈一个面给你，然后文件读取出来，生成范围内的点。
/**
* 〈一句话功能简述〉<br>
* 〈〉
*
* @author maruko
* @date 2021/4/23 19:02
* @since 1.0.0
*/
public class Test {
public static void main(String[] args) {
Double[][] point=new Double[10000][2];
int num=0;
Double[][] jwds = jwdFile("D:\\桌面文件\\bj/bj.txt");
double minLon = 75d;
double maxLon = 130d;
double minLat = 19d;
double maxLat = 55d;
for (int i = 0; i < 10000; i++) {
Double[] jwd = randomLonLat(minLon, maxLon, minLat, maxLat);
boolean ptInPoly = isPtInPoly(jwd, jwds);
if(ptInPoly){
// point[num]=jwd;
num++;
System.err.println(jwd[0]+","+jwd[1]);
}
}
}
/* public static void main(String[] args) {
Double[][] ps = new Double[][] { new Double[]{114.309914d,30.599556d},
new Double[]{114.295688,30.592879},
new Double[]{114.292812,30.587726},
new Double[]{114.292812,30.587726},
new Double[]{114.30058,30.580318},
new Double[]{114.303606,30.586959},
new Double[]{114.304534,30.594751},
new Double[]{114.30838,30.590131},
new Double[]{114.308651,30.584182},
new Double[]{114.304495,30.584015},
new Double[]{114.301301,30.578759},
new Double[]{114.309437,30.578528},
new Double[]{114.323282,30.592786}};
Double[] n1 = new Double[]{114.303217,30.583553};
Double[] n2 = new Double[]{114.307336,30.597592};
Double[] n3 = new Double[]{114.286565,30.590056};
Double[] y1 = new Double[]{114.227342,30.587987};
Double[] y2 = new Double[]{120.1866 , 30.2672};
Double[] y4 = new Double[]{120.1869 , 30.2718};
System.out.println( "n1:" + isPtInPoly(n1[0] , n1[1] , ps));
System.out.println( "n2:" + isPtInPoly(n2[0] , n2[1] , ps));
System.out.println( "n3:" + isPtInPoly(n3[0] , n3[1] , ps));
System.out.println( "y1:" + isPtInPoly(y1[0] , y1[1] , ps));
System.out.println( "y2:" + isPtInPoly(y2[0] , y2[1] , ps));
System.out.println( "y4:" + isPtInPoly(y4[0] , y4[1] , ps));
}
*/
/**
* 随机生成经度或者维度
*
* @param MinLon 最小经度
* @param MaxLon 最大经度
* @param MinLat 最小纬度
* @param MaxLat 最大维度
* @return
*/
public static Double[] randomLonLat(double MinLon, double MaxLon, double MinLat, double MaxLat) {
Double[] jwd = new Double[2];
BigDecimal db = new BigDecimal(Math.random() * (MaxLon - MinLon) + MinLon);
jwd[0] = Double.parseDouble(db.setScale(6, BigDecimal.ROUND_HALF_UP).toString());
db = new BigDecimal(Math.random() * (MaxLat - MinLat) + MinLat);
jwd[1] = Double.parseDouble(db.setScale(6, BigDecimal.ROUND_HALF_UP).toString());
return jwd;
}
/**
* 读取经纬度范围文件
*
* @param filePath
* @return
*/
public static Double[][] jwdFile(String filePath) {
//D:\桌面文件\bj/bj.txt
File file = new File(filePath);
BufferedReader reader = null;
StringBuffer sbf = new StringBuffer();
try {
reader = new BufferedReader(new FileReader(file));
String tempStr;
while ((tempStr = reader.readLine()) != null) {
sbf.append(tempStr);
}
reader.close();
// System.out.println(sbf.toString());
String[] split = sbf.toString().split(";");
Double[][] str = new Double[split.length][2];
for (int i = 0; i < split.length; i++) {
String[] split1 = split[i].split(",");
for (int j = 0; j < split1.length; j++) {
str[i][j] = Double.parseDouble(split1[j]);
}
}
return str;
// for (int i = 0; i < split.length; i++) {
// str[i][0] = doubles[i];
// }
// for (String s : split) {
// System.out.println(s);
// }
// List<Object> list = JSON.parseArray(sbf.toString());
// for (Object o : list) {
// System.out.println(o.toString());
// }
// List<Object> bj = new ArrayList<>();
// List<Object> list1 = list.subList(0, 3049);
// List<Object> list = Arrays.asList(split);
// Map map = GeoJsonUtil.cerateFqh(list);
// ObjectMapper mapper = new ObjectMapper();
// try {
// mapper.writeValue(new File("D:\\桌面文件\\bj/2020.geojson"), str);
// } catch (
// Exception e) {
// e.printStackTrace();
// }
// System.out.println(list.size());
} catch (IOException e) {
e.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException e1) {
e1.printStackTrace();
}
}
}
return null;
}
/**
* 判断一个点是否在一个面上面
*
* @param jwd
* @param jwds
* @return
*/
public static boolean isPtInPoly(Double[] jwd, Double[][] jwds) {
Double longitude = jwd[0];
Double latitude = jwd[1];
int iSum, iCount, iIndex;
double dLon1 = 0, dLon2 = 0, dLat1 = 0, dLat2 = 0, dLon;
if (jwds.length < 3) {
return false;
}
iSum = 0;
iCount = jwds.length;
for (iIndex = 0; iIndex < iCount; iIndex++) {
if (iIndex == iCount - 1) {
dLon1 = jwds[iIndex][0];
dLat1 = jwds[iIndex][1];
dLon2 = jwds[0][0];
dLat2 = jwds[0][1];
} else {
dLon1 = jwds[iIndex][0];
dLat1 = jwds[iIndex][1];
dLon2 = jwds[iIndex + 1][0];
dLat2 = jwds[iIndex + 1][1];
}
// 以下语句判断A点是否在边的两端点的水平平行线之间，在则可能有交点，开始判断交点是否在左射线上
if (((latitude >= dLat1) && (latitude < dLat2)) || ((latitude >= dLat2) && (latitude < dLat1))) {
if (Math.abs(dLat1 - dLat2) > 0) {
//得到 A点向左射线与边的交点的x坐标：
dLon = dLon1 - ((dLon1 - dLon2) * (dLat1 - latitude)) / (dLat1 - dLat2);
// 如果交点在A点左侧（说明是做射线与 边的交点），则射线与边的全部交点数加一：
if (dLon < longitude) {
iSum++;
}
}
}
}
if ((iSum % 2) != 0) {
return true;
}
return false;
}
}