GeoHash算法原理及实现

我不是女神ヾ 2023-02-09 13:28 49阅读 0赞

### GeoHash原理与Java实现 ###

*   *  GeoHash算法原理
     *   *  样例数据（基于15次区域分割）
         *  GeoHash算法思想
         *  GeoHash算法原理
         *  后续问题
     *  GeoHash算法代码实现
     *  写在最后

## GeoHash算法原理 ##

> GeoHash是目前比较主流实现位置服务的技术，Geohash算法将经纬度二维数据编码为一个字符串，本质是一个降维的过程

### 样例数据（基于15次区域分割） ###

<table> 
 <thead> 
  <tr> 
   <th>位置</th> 
   <th>经纬度</th> 
   <th>Geohash</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>北京站</td> 
   <td>116.433589,39.910508</td> 
   <td>wx4g19</td> 
  </tr> 
  <tr> 
   <td>天安门</td> 
   <td>116.403874,39.913884</td> 
   <td>wx4g0f</td> 
  </tr> 
  <tr> 
   <td>首都机场</td> 
   <td>116.606819,40.086109</td> 
   <td>wx4uj3</td> 
  </tr> 
 </tbody> 
</table>

### GeoHash算法思想 ###

我们知道，经度范围是东经180到西经180，纬度范围是南纬90到北纬90，我们设定西经为负，南纬为负，所以地球上的经度范围就是\[-180， 180\]，纬度范围就是\[-90，90\]。如果以本初子午线、赤道为界，地球可以分成4个部分。

GeoHash的思想就是将地球划分的四部分映射到二维坐标上。

> \[-90˚,0˚)代表0，(0˚,90˚\]代表1，\[-180˚,0)代表0，(0˚,180˚\]代表1  
> 映射到二维空间划分为四部分则如下图

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70]  
但是这么粗略的划分没有什么意义，想要更精确的使用GeoHash就需要再进一步`二分切割`  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70 1]  
通过上图可看出，进一步`二分切割`将原本大略的划分变为细致的区域划分，这样就会更加精确。GeoHash算法就是基于这种思想，递归划分的次数越多，所计算出的数据越精确。

### GeoHash算法原理 ###

> GeoHash算法大体上分为三步：1. 计算经纬度的二进制、2. 合并经纬度的二进制、3. 通过Base32对合并后的二进制进行编码。

1.  计算经纬度的二进制

//根据经纬度和范围，获取对应的二进制
    	private BitSet getBits(double l, double floor, double ceiling) { 
    		BitSet buffer = new BitSet(numbits);
    		for (int i = 0; i < numbits; i++) { 
    			double mid = (floor + ceiling) / 2;
    			if (l >= mid) { 
    				buffer.set(i);
    				floor = mid;
    			} else { 
    				ceiling = mid;
    			}
    		}
    		return buffer;
    	}

上述代码numbits为：`private static int numbits = 3 * 5; //经纬度单独编码长度`也就是说将地球进行15次`二分切割`

> 注： 这里需要对BitSet类进行一下剖析，没了解过该类的话指定懵。

了解BitSet只需了去了解它的set()、get()方法就足够了

*  BitSet的set方法

/** * Sets the bit at the specified index to {@code true}. * * @param bitIndex a bit index * @throws IndexOutOfBoundsException if the specified index is negative * @since JDK1.0 */
        public void set(int bitIndex) { 
            if (bitIndex < 0)
                throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
    
            int wordIndex = wordIndex(bitIndex);
            expandTo(wordIndex);
    
            words[wordIndex] |= (1L << bitIndex); // Restores invariants
    
            checkInvariants();
        }

set方法内`wordIndex(bitIndex)`底层将bitIndex右移6位然后返回，`ADDRESS_BITS_PER_WORD`为常量6

/** * Given a bit index, return word index containing it. */
       private static int wordIndex(int bitIndex) { 
           return bitIndex >> ADDRESS_BITS_PER_WORD;
       }

set方法内的`expandTo(wordIndex)`只是一个判断数组是否需要扩容的方法

/** * Ensures that the BitSet can accommodate a given wordIndex, * temporarily violating the invariants. The caller must * restore the invariants before returning to the user, * possibly using recalculateWordsInUse(). * @param wordIndex the index to be accommodated. */
        private void expandTo(int wordIndex) { 
            int wordsRequired = wordIndex+1;
            if (wordsInUse < wordsRequired) { 
                ensureCapacity(wordsRequired);
                wordsInUse = wordsRequired;
            }
        }

set内重要的一行代码`words[wordIndex] |= (1L << bitIndex)`,这里只解释一下`|=`

> `a|=b`就是a=a|b,就是说将a、b转为二进制按位与，同0为0，否则为1

*  BitSet的get方法

/** * Returns the value of the bit with the specified index. The value * is {@code true} if the bit with the index {@code bitIndex} * is currently set in this {@code BitSet}; otherwise, the result * is {@code false}. * * @param bitIndex the bit index * @return the value of the bit with the specified index * @throws IndexOutOfBoundsException if the specified index is negative */
       public boolean get(int bitIndex) { 
           if (bitIndex < 0)
               throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
    
           checkInvariants();
    
           int wordIndex = wordIndex(bitIndex);
           return (wordIndex < wordsInUse)
               && ((words[wordIndex] & (1L << bitIndex)) != 0);
       }

> get方法用一句话概括就是：如果传入的下标有值，返回true；反之为false

以天安门坐标为例：`39.913884, 116.403874`

BitSet latbits = getBits(lat, -90, 90);
       	BitSet lonbits = getBits(lon, -180, 180);
       	// 纬度
       	for (int i = 0; i < numbits; i++) { 
       		System.out.print(latbits.get(i) + " ");
       	}
       	// 经度
        for (int i = 0; i < numbits; i++) { 
       		System.out.print(lonbits.get(i) + " ");
       	}

纬度经过转换为：

true false true true true false false false true true false false false true false

转为二进制：

1 0 1 1 1 0 0 0 1 1 0 0 0 1 0

经度经过转换为：

true true false true false false true false true true false false false true true

转为二进制：

1 1 0 1 0 0 1 0 1 1 0 0 0 1 1

1.  合并经纬度二进制  
    合并原则：经度占偶数位，纬度占奇数位。也就是说经纬度交替合并，首位0位置为经度的0位置

合并后二进制编码为：

11100 11101 00100 01111 00000 01110

1.  使用Base32对合并后的经纬度二进制进行编码

*  代码实现

// Base32进行编码
    	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');
    		}
    		String code = base32(Long.parseLong(buffer.toString(), 2));
    		return code;
    	}

本文案例经纬度编码后

wx4g0f

### 后续问题 ###

如果要使用此功能实现附近的人。假如红点为使用者，经过Geohash算法分割后只会推荐同区域`0011`中的绿点，但是如下图所示，蓝色点相对于绿色点更接近用户，所以区域划分的弊端就展现在这里。  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70 2]  
针对上述问题，我们可以人为获取红色用户所在的`0011`区域周边八个区域中的用户，即获取`0011`的同时还要获取`0100`,`0110`,`1100`,`0001`,`1001`,`0000`,`0010`,`1000`

*  代码实现

public ArrayList<String> getArroundGeoHash(double lat, double lon) { 
    		ArrayList<String> list = new ArrayList<>();
    		double uplat = lat + minLat;
    		double downLat = lat - minLat;
    
    		double leftlng = lon - minLng;
    		double rightLng = lon + minLng;
    
    		String leftUp = encode(uplat, leftlng);
    		list.add(leftUp);
    
    		String leftMid = encode(lat, leftlng);
    		list.add(leftMid);
    
    		String leftDown = encode(downLat, leftlng);
    		list.add(leftDown);
    
    		String midUp = encode(uplat, lon);
    		list.add(midUp);
    
    		String midMid = encode(lat, lon);
    		list.add(midMid);
    
    		String midDown = encode(downLat, lon);
    		list.add(midDown);
    
    		String rightUp = encode(uplat, rightLng);
    		list.add(rightUp);
    
    		String rightMid = encode(lat, rightLng);
    		list.add(rightMid);
    
    		String rightDown = encode(downLat, rightLng);
    		list.add(rightDown);
    
    		return list;
    	}

然后根据球体两点间的距离计算红色用户与周边区域用户距离，从而进行附近的人功能实现

*  通过两经纬度计算距离java代码实现

static double getDistance(double lat1, double lon1, double lat2, double lon2) { 
    		// 经纬度（角度）转弧度。弧度用作参数，以调用Math.cos和Math.sin
    		double radiansAX = Math.toRadians(lon1); // A经弧度
    		double radiansAY = Math.toRadians(lat1); // A纬弧度
    		double radiansBX = Math.toRadians(lon2); // B经弧度
    		double radiansBY = Math.toRadians(lat2); // B纬弧度
    
    		// 公式中“cosβ1cosβ2cos（α1-α2）+sinβ1sinβ2”的部分，得到∠AOB的cos值
    		double cos = Math.cos(radiansAY) * Math.cos(radiansBY) * Math.cos(radiansAX - radiansBX)
    				+ Math.sin(radiansAY) * Math.sin(radiansBY);
    		double acos = Math.acos(cos); // 反余弦值
    		return EARTH_RADIUS * acos; // 最终结果
    	}

## GeoHash算法代码实现 ##

public class GeoHash { 
    	public static final double MINLAT = -90;
    	public static final double MAXLAT = 90;
    	public static final double MINLNG = -180;
    	public static final double MAXLNG = 180;
    
    	private static int numbits = 3 * 5; //经纬度单独编码长度
    
    	private static double minLat;
    	private static double minLng;
    
    	private 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'};
    
    	//定义编码映射关系
    	final static HashMap<Character, Integer> lookup = new HashMap<Character, Integer>();
    
    	//初始化编码映射内容
    	static { 
    		int i = 0;
    		for (char c : digits)
    			lookup.put(c, i++);
    	}
    
    	public GeoHash() { 
    		setMinLatLng();
    	}
    
    	// Base32进行编码
    	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');
    		}
    		String code = base32(Long.parseLong(buffer.toString(), 2));
    		return code;
    	}
    
    	public ArrayList<String> getArroundGeoHash(double lat, double lon) { 
    		ArrayList<String> list = new ArrayList<>();
    		double uplat = lat + minLat;
    		double downLat = lat - minLat;
    
    		double leftlng = lon - minLng;
    		double rightLng = lon + minLng;
    
    		String leftUp = encode(uplat, leftlng);
    		list.add(leftUp);
    
    		String leftMid = encode(lat, leftlng);
    		list.add(leftMid);
    
    		String leftDown = encode(downLat, leftlng);
    		list.add(leftDown);
    
    		String midUp = encode(uplat, lon);
    		list.add(midUp);
    
    		String midMid = encode(lat, lon);
    		list.add(midMid);
    
    		String midDown = encode(downLat, lon);
    		list.add(midDown);
    
    		String rightUp = encode(uplat, rightLng);
    		list.add(rightUp);
    
    		String rightMid = encode(lat, rightLng);
    		list.add(rightMid);
    
    		String rightDown = encode(downLat, rightLng);
    		list.add(rightDown);
    
    		return list;
    	}
    
    	//根据经纬度和范围，获取对应的二进制
    	private BitSet getBits(double l, double floor, double ceiling) { 
    		BitSet buffer = new BitSet(numbits);
    		for (int i = 0; i < numbits; i++) { 
    			double mid = (floor + ceiling) / 2;
    			if (l >= mid) { 
    				buffer.set(i);
    				floor = mid;
    			} else { 
    				ceiling = mid;
    			}
    		}
    		return buffer;
    	}
    
    	//将经纬度合并后的二进制进行指定的32位编码
    	private 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));
    	}
    
    	private void setMinLatLng() { 
    		minLat = MAXLAT - MINLAT;
    		for (int i = 0; i < numbits; i++) { 
    			minLat /= 2.0;
    		}
    		minLng = MAXLNG - MINLNG;
    		for (int i = 0; i < numbits; i++) { 
    			minLng /= 2.0;
    		}
    	}
    
    	//根据二进制和范围解码
    	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;
    	}
    
    	//对编码后的字符串解码
    	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();
    
    		//偶数位，经度
    		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);
    		}
    
    		//奇数位，纬度
    		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};
    	}
    
    	public static void main(String[] args) { 
    		GeoHash geoHash = new GeoHash();
    		// 北京站
    		String encode = geoHash.encode(39.910508, 116.433589);
    		System.out.println(encode);
    
    		// 天安门
    		System.out.println(geoHash.encode(39.913884, 116.403874));
    
    		// 首都机场
    		System.out.println(geoHash.encode(40.086109, 116.606819));
    
    		BitSet latbits = geoHash.getBits(39.913884, -90, 90);
    		BitSet lonbits = geoHash.getBits(116.403874, -180, 180);
    
    // for (int i=0; i< latbits.length(); i++) { 
    // System.out.println(latbits.get(i));
    // }
    
    		for (int i = 0; i < numbits; i++) { 
    // System.out.print(latbits.get(i));
    			System.out.print(latbits.get(i) ? '1' : '0');
    			System.out.print(" ");
    		}
    
    		System.out.println();
    		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');
    		}
    
    		System.out.println(buffer.toString());
    
    		System.out.println(geoHash.encode(39.913884, 116.403874));
    	}
    
    }

## 写在最后 ##

如果嫌GeoHash算法麻烦，但是还想用它，没关系。  
Redis知道你懒[Redis官网GeoHash用法][Redis_GeoHash]

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20200519092805300.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg=,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20200519095313362.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg=,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg_size_16_color_FFFFFF_t_70 2]: https://img-blog.csdnimg.cn/20200519160012889.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L00yODM1OTIzMzg=,size_16,color_FFFFFF,t_70
[Redis_GeoHash]: https://redis.io/commands/GEOHASH