GeoHash实现周边推荐
旅游,外卖等需要定位的项目中一般会有周边推荐的需求,如推荐出周边五公里的景点。目前实现的算法也有很多,这里简单的说下GeoHash的实现原理以及Java的实现代码。
GeoHash基础
- GeoHash使用一个字符串来表示经度和纬度。这样在做周边搜索的时候可以在一列上加索引。
- GeoHash表示的不是一个点,而是一个矩形区域。
- 精度范围为(-180,180),纬度范围为(-90,90)
GeoHash转换过程
转换
GeoHash就是将经纬度信息,转换为可以排序、比较的字符串编码。 首先将纬度范围(-90,90)分为两个区间(-90,0),(0,90),如果目标纬度在前一个区间,编码为0,否则编码为1。以此类推,直到精度符合要求为止。
给出一个经纬度(39.92324,116.3906),转换过程如下
纬度范围 | 划分区间0 | 划分区间1 | 39.92324所属区间 |
(-90, 90) | (-90, 0.0) | (0.0, 90) | 1 |
(0.0, 90) | (0.0, 45.0) | (45.0, 90) | 0 |
(0.0, 45.0) | (0.0, 22.5) | (22.5, 45.0) | 1 |
(22.5, 45.0) | (22.5, 33.75) | (33.75, 45.0) | 1 |
(33.75, 45.0) | (33.75, 39.375) | (39.375, 45.0) | 1 |
(39.375, 45.0) | (39.375, 42.1875) | (42.1875, 45.0) | 0 |
(39.375, 42.1875) | (39.375, 40.7812) | (40.7812, 42.1875) | 0 |
(39.375, 40.7812) | (39.375, 40.0781) | (40.0781, 40.7812) | 0 |
(39.375, 40.0781) | (39.375, 39.7265) | (39.7265, 40.0781) | 1 |
(39.7265, 40.0781) | (39.7265, 39.9023) | (39.9023, 40.0781) | 1 |
(39.9023, 40.0781) | (39.9023, 39.9902) | (39.9902, 40.0781) | 0 |
(39.9023, 39.9902) | (39.9023, 39.9462) | (39.9462, 39.9902) | 0 |
(39.9023, 39.9462) | (39.9023, 39.9243) | (39.9243, 39.9462) | 0 |
(39.9023, 39.9243) | (39.9023, 39.9133) | (39.9133, 39.9243) | 1 |
(39.9133, 39.9243) | (39.9133, 39.9188) | (39.9188, 39.9243) | 1 |
(39.9188, 39.9243) | (39.9188, 39.9215) | (39.9215, 39.9243) | 1 |
(39.9215, 39.9243) | (39.9215, 39.9229) | (39.9229, 39.9243) | 1 |
(39.9229, 39.9243) | (39.9229, 39.9236) | (39.9236, 39.9243) | 0 |
(39.9229, 39.9236) | (39.9229, 39.92325) | (39.92325, 39.9236) | 0 |
(39.9229, 39.92325) | (39.9229, 39.923075) | (39.923075, 39.92325) | 1 |
因此可以得到纬度的编码为 1011 1000 1100 0111 1001 经度也用同样的算法,对(-180, 180)依次细分,得到116.3906的编码为1101 0010 1100 0100 0100。 接下来将经度和纬度的编码合并,奇数位是纬度,偶数位是经度,得到编码 11100 11101 00100 01111 00000 01101 01011 00001。
合并
将 11100 11101 00100 01111 00000 01101 01011 00001转化为对应的十进制为 28 29 4 15 0 13 11 1
用0-9、b-z(去掉a, i, l, o)这32个字母进行base32编码
十进制 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
base32 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | b | c | d | e | f | g |
十进制 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 |
base32 | h | j | k | m | n | p | q | r | s | t | u | v | w | x | y | z |
可以得到(39.92324, 116.3906)的编码为wx4g0ec1
误差
下表列出了不同的编码长度对应的精度:
geohash length | lat bits | lng bits | lat error | lng error | km error |
1 | 2 | 3 | ±23 | ±23 | ±2500 |
2 | 5 | 5 | ±2.8 | ±5.6 | ±630 |
3 | 7 | 8 | ±0.70 | ±0.7 | ±78 |
4 | 10 | 10 | ±0.087 | ±0.18 | ±20 |
5 | 12 | 13 | ±0.022 | ±0.022 | ±2.4 |
6 | 15 | 15 | ±0.0027 | ±0.0055 | ±0.61 |
7 | 17 | 18 | ±0.00068 | ±0.0068 | ±0.076 |
8 | 20 | 20 | ±0.000085 | ±0.00017 | ±0.019 |
geohash length | width | height |
1 | 5009.4km | 4992.6km |
2 | 1252.3km | 624.1km |
3 | 156.5km | 156km |
4 | 39.1km | 19.5km |
5 | 4.9km | 4.9km |
6 | 1.2km | 609.4m |
7 | 152.9m | 152.4m |
8 | 38.2m | 19m |
9 | 4.8m | 4.8m |
10 | 1.2m | 59.5m |
11 | 14.9cm | 14.9cm |
12 | 3.7cm | 1.9cm |
Java实现代码
import java.util.BitSet;
import java.util.HashMap;
public class GeoHash {
private static int numbits = 6 * 5;
final static char[] digits = { '0', '1', '2', '3', '4', '5', '6', '7', '8',
'9', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j', 'k', 'm', 'n', 'p',
'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };
final static HashMap<Character, Integer> lookup = new HashMap<Character, Integer>();
static {
int i = 0;
for (char c : digits)
lookup.put(c, i++);
}
public double[] decode(String geohash) {
StringBuilder buffer = new StringBuilder();
for (char c : geohash.toCharArray()) {
int i = lookup.get(c) + 32;
buffer.append( Integer.toString(i, 2).substring(1) );
}
BitSet lonset = new BitSet();
BitSet latset = new BitSet();
//even bits
int j =0;
for (int i=0; i< numbits*2;i+=2) {
boolean isSet = false;
if ( i < buffer.length() )
isSet = buffer.charAt(i) == '1';
lonset.set(j++, isSet);
}
//odd bits
j=0;
for (int i=1; i< numbits*2;i+=2) {
boolean isSet = false;
if ( i < buffer.length() )
isSet = buffer.charAt(i) == '1';
latset.set(j++, isSet);
}
double lon = decode(lonset, -180, 180);
double lat = decode(latset, -90, 90);
return new double[] {lat, lon};
}
private double decode(BitSet bs, double floor, double ceiling) {
double mid = 0;
for (int i=0; i<bs.length(); i++) {
mid = (floor + ceiling) / 2;
if (bs.get(i))
floor = mid;
else
ceiling = mid;
}
return mid;
}
public String encode(double lat, double lon) {
BitSet latbits = getBits(lat, -90, 90);
BitSet lonbits = getBits(lon, -180, 180);
StringBuilder buffer = new StringBuilder();
for (int i = 0; i < numbits; i++) {
buffer.append( (lonbits.get(i))?'1':'0');
buffer.append( (latbits.get(i))?'1':'0');
}
return base32(Long.parseLong(buffer.toString(), 2));
}
private BitSet getBits(double lat, double floor, double ceiling) {
BitSet buffer = new BitSet(numbits);
for (int i = 0; i < numbits; i++) {
double mid = (floor + ceiling) / 2;
if (lat >= mid) {
buffer.set(i);
floor = mid;
} else {
ceiling = mid;
}
}
return buffer;
}
public static String base32(long i) {
char[] buf = new char[65];
int charPos = 64;
boolean negative = (i < 0);
if (!negative)
i = -i;
while (i <= -32) {
buf[charPos--] = digits[(int) (-(i % 32))];
i /= 32;
}
buf[charPos] = digits[(int) (-i)];
if (negative)
buf[--charPos] = '-';
return new String(buf, charPos, (65 - charPos));
}
}