geohash算法原理及实现方式
3、geohash的php 、python、java、C#实现代码
4、观点讨论
首先，geohash用一个字符串表示经度和纬度两个坐标。某些情况下无法在两列上同时应用索引 （例如MySQL 4之前的版本，Google App Engine的数据层等），利用geohash，只需在一列上应用索引即可。
其次，geohash表示的并不是一个点，而是一个矩形区域。比如编码wx4g0ec19，它表示的是一个矩形区域。 使用者可以发布地址编码，既能表明自己位于北海公园附近，又不至于暴露自己的精确坐标，有助于隐私保护。
第三，编码的前缀可以表示更大的区域。例如wx4g0ec1，它的前缀wx4g0e表示包含编码wx4g0ec1在内的更大范围。 这个特性可以用于附近地点搜索。首先根据用户当前坐标计算geohash（例如wx4g0ec1）然后取其前缀进行查询 （SELECT * FROM place WHERE geohash LIKE 'wx4g0e%'），即可查询附近的所有地点。
Geohash比直接用经纬度的高效很多。
Geohash的最简单的解释就是：将一个经纬度信息，转换成一个可以排序，可以比较的字符串编码
首先将纬度范围(-90, 90)平分成两个区间(-90,0)、(0, 90)，如果目标纬度位于前一个区间，则编码为0，否则编码为1。
由于39.92324属于(0, 90)，所以取编码为1。然后再将(0, 90)分成 (0, 45), (45, 90)两个区间，而39.92324位于(0, 45)，所以编码为0。
以此类推，直到精度符合要求为止，得到纬度编码为1011 1000 1100 0111 1001。
| 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 |
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。
| 经度范围 | 划分区间0 | 划分区间1 | 116.3906所属区间 |
| (-180, 180) | (-180, 0.0) | (0.0, 180) | 1 |
| (0.0, 180) | (0.0, 90.0) | (90.0, 180) | 1 |
| (90.0, 180) | (90.0, 135.0) | (135.0, 180) | 0 |
| (90.0, 135.0) | (90.0, 112.5) | (112.5, 135.0) | 1 |
| (112.5, 135.0) | (112.5, 123.75) | (123.75, 135.0) | 0 |
| (112.5, 123.75) | (112.5, 118.125) | (118.125, 123.75) | 0 |
| (112.5, 118.125) | (112.5, 115.312) | (115.312, 118.125) | 1 |
| (115.312, 118.125) | (115.312, 116.718) | (116.718, 118.125) | 0 |
| (115.312, 116.718) | (115.312, 116.015) | (116.015, 116.718) | 1 |
| (116.015, 116.718) | (116.015, 116.367) | (116.367, 116.718) | 1 |
| (116.367, 116.718) | (116.367, 116.542) | (116.542, 116.718) | 0 |
| (116.367, 116.542) | (116.367, 116.455) | (116.455, 116.542) | 0 |
| (116.367, 116.455) | (116.367, 116.411) | (116.411, 116.455) | 0 |
| (116.367, 116.411) | (116.367, 116.389) | (116.389, 116.411) | 1 |
| (116.389, 116.411) | (116.389, 116.400) | (116.400, 116.411) | 0 |
| (116.389, 116.400) | (116.389, 116.394) | (116.394, 116.400) | 0 |
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。
最后，用0-9、b-z（去掉a, i, l, o）这32个字母进行base32编码，得到(39.92324, 116.3906)的编码为wx4g0ec1。
| 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 |
解码算法与编码算法相反，先进行base32解码，然后分离出经纬度，最后根据二进制编码对经纬度范围进行细分即可，这里不再赘述。
php版本的实现方式：http://blog.dixo.net/downloads/geohash-php-class/ 我下载了一个上传的
php：
<?php
/**
* Geohash generation class
* http://blog.dixo.net/downloads/
*
* This file copyright (C) 2008 Paul Dixon (paul@elphin.com)
*
* This program is free software; you can redistribute it and/or
* modify it under the terms of the GNU General Public License
* as published by the Free Software Foundation; either version 3
* of the License, or (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
*/
/**
* Encode and decode geohashes
*
*/
class Geohash
{
private $coding="0123456789bcdefghjkmnpqrstuvwxyz";
private $codingMap=array();
public function Geohash()
{
//build map from encoding char to 0 padded bitfield
for($i=0; $i<32; $i++)
{
$this->codingMap[substr($this->coding,$i,1)]=str_pad(decbin($i), 5, "0", STR_PAD_LEFT);
}
}
/**
* Decode a geohash and return an array with decimal lat,long in it
*/
public function decode($hash)
{
//decode hash into binary string
$binary="";
$hl=strlen($hash);
for($i=0; $i<$hl; $i++)
{
$binary.=$this->codingMap[substr($hash,$i,1)];
}
//split the binary into lat and log binary strings
$bl=strlen($binary);
$blat="";
$blong="";
for ($i=0; $i<$bl; $i++)
{
if ($i%2)
$blat=$blat.substr($binary,$i,1);
else
$blong=$blong.substr($binary,$i,1);
}
//now concert to decimal
$lat=$this->binDecode($blat,-90,90);
$long=$this->binDecode($blong,-180,180);
//figure out how precise the bit count makes this calculation
$latErr=$this->calcError(strlen($blat),-90,90);
$longErr=$this->calcError(strlen($blong),-180,180);
//how many decimal places should we use? There's a little art to
//this to ensure I get the same roundings as geohash.org
$latPlaces=max(1, -round(log10($latErr))) - 1;
$longPlaces=max(1, -round(log10($longErr))) - 1;
//round it
$lat=round($lat, $latPlaces);
$long=round($long, $longPlaces);
return array($lat,$long);
}
/**
* Encode a hash from given lat and long
*/
public function encode($lat,$long)
{
//how many bits does latitude need?
$plat=$this->precision($lat);
$latbits=1;
$err=45;
while($err>$plat)
{
$latbits++;
$err/=2;
}
//how many bits does longitude need?
$plong=$this->precision($long);
$longbits=1;
$err=90;
while($err>$plong)
{
$longbits++;
$err/=2;
}
//bit counts need to be equal
$bits=max($latbits,$longbits);
//as the hash create bits in groups of 5, lets not
//waste any bits - lets bulk it up to a multiple of 5
//and favour the longitude for any odd bits
$longbits=$bits;
$latbits=$bits;
$addlong=1;
while (($longbits+$latbits)%5 != 0)
{
$longbits+=$addlong;
$latbits+=!$addlong;
$addlong=!$addlong;
}
//encode each as binary string
$blat=$this->binEncode($lat,-90,90, $latbits);
$blong=$this->binEncode($long,-180,180,$longbits);
//merge lat and long together
$binary="";
$uselong=1;
while (strlen($blat)+strlen($blong))
{
if ($uselong)
{
$binary=$binary.substr($blong,0,1);
$blong=substr($blong,1);
}
else
{
$binary=$binary.substr($blat,0,1);
$blat=substr($blat,1);
}
$uselong=!$uselong;
}
//convert binary string to hash
$hash="";
for ($i=0; $i<strlen($binary); $i+=5)
{
$n=bindec(substr($binary,$i,5));
$hash=$hash.$this->coding[$n];
}
return $hash;
}
/**
* What's the maximum error for $bits bits covering a range $min to $max
*/
private function calcError($bits,$min,$max)
{
$err=($max-$min)/2;
while ($bits--)
$err/=2;
return $err;
}
/*
* returns precision of number
* precision of 42 is 0.5
* precision of 42.4 is 0.05
* precision of 42.41 is 0.005 etc
*/
private function precision($number)
{
$precision=0;
$pt=strpos($number,'.');
if ($pt!==false)
{
$precision=-(strlen($number)-$pt-1);
}
return pow(10,$precision)/2;
}
/**
* create binary encoding of number as detailed in http://en.wikipedia.org/wiki/Geohash#Example
* removing the tail recursion is left an exercise for the reader
*/
private function binEncode($number, $min, $max, $bitcount)
{
if ($bitcount==0)
return "";
#echo "$bitcount: $min $max<br>";
//this is our mid point - we will produce a bit to say
//whether $number is above or below this mid point
$mid=($min+$max)/2;
if ($number>$mid)
return "1".$this->binEncode($number, $mid, $max,$bitcount-1);
else
return "0".$this->binEncode($number, $min, $mid,$bitcount-1);
}
/**
* decodes binary encoding of number as detailed in http://en.wikipedia.org/wiki/Geohash#Example
* removing the tail recursion is left an exercise for the reader
*/
private function binDecode($binary, $min, $max)
{
$mid=($min+$max)/2;
if (strlen($binary)==0)
return $mid;
$bit=substr($binary,0,1);
$binary=substr($binary,1);
if ($bit==1)
return $this->binDecode($binary, $mid, $max);
else
return $this->binDecode($binary, $min, $mid);
}
}
?>python：
python版本的geohash：python-geohash
java：
java版本的geohash,实现：http://code.google.com/p/geospatialweb/source/browse/#svn/trunk/geohash/src
import java.io.File;
import java.io.FileInputStream;
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 static void main(String[] args) throws Exception{
System.out.println(new Geohash().encode(45, 125));
}
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));
}
}C#：
using System;
namespace sharonjl.utils
{
public static class Geohash
{
#region Direction enum
public enum Direction
{
Top = 0,
Right = 1,
Bottom = 2,
Left = 3
}
#endregion
private const string Base32 = "0123456789bcdefghjkmnpqrstuvwxyz";
private static readonly int[] Bits = new[] {16, 8, 4, 2, 1};
private static readonly string[][] Neighbors = {
new[]
{
"p0r21436x8zb9dcf5h7kjnmqesgutwvy", // Top
"bc01fg45238967deuvhjyznpkmstqrwx", // Right
"14365h7k9dcfesgujnmqp0r2twvyx8zb", // Bottom
"238967debc01fg45kmstqrwxuvhjyznp", // Left
}, new[]
{
"bc01fg45238967deuvhjyznpkmstqrwx", // Top
"p0r21436x8zb9dcf5h7kjnmqesgutwvy", // Right
"238967debc01fg45kmstqrwxuvhjyznp", // Bottom
"14365h7k9dcfesgujnmqp0r2twvyx8zb", // Left
}
};
private static readonly string[][] Borders = {
new[] {"prxz", "bcfguvyz", "028b", "0145hjnp"},
new[] {"bcfguvyz", "prxz", "0145hjnp", "028b"}
};
public static String CalculateAdjacent(String hash, Direction direction)
{
hash = hash.ToLower();
char lastChr = hash[hash.Length - 1];
int type = hash.Length%2;
var dir = (int) direction;
string nHash = hash.Substring(0, hash.Length - 1);
if (Borders[type][dir].IndexOf(lastChr) != -1)
{
nHash = CalculateAdjacent(nHash, (Direction) dir);
}
return nHash + Base32[Neighbors[type][dir].IndexOf(lastChr)];
}
public static void RefineInterval(ref double[] interval, int cd, int mask)
{
if ((cd & mask) != 0)
{
interval[0] = (interval[0] + interval[1])/2;
}
else
{
interval[1] = (interval[0] + interval[1])/2;
}
}
public static double[] Decode(String geohash)
{
bool even = true;
double[] lat = {-90.0, 90.0};
double[] lon = {-180.0, 180.0};
foreach (char c in geohash)
{
int cd = Base32.IndexOf(c);
for (int j = 0; j < 5; j++)
{
int mask = Bits[j];
if (even)
{
RefineInterval(ref lon, cd, mask);
}
else
{
RefineInterval(ref lat, cd, mask);
}
even = !even;
}
}
return new[] {(lat[0] + lat[1])/2, (lon[0] + lon[1])/2};
}
public static String Encode(double latitude, double longitude, int precision = 12)
{
bool even = true;
int bit = 0;
int ch = 0;
string geohash = "";
double[] lat = {-90.0, 90.0};
double[] lon = {-180.0, 180.0};
if (precision < 1 || precision > 20) precision = 12;
while (geohash.Length < precision)
{
double mid;
if (even)
{
mid = (lon[0] + lon[1])/2;
if (longitude > mid)
{
ch |= Bits[bit];
lon[0] = mid;
}
else
lon[1] = mid;
}
else
{
mid = (lat[0] + lat[1])/2;
if (latitude > mid)
{
ch |= Bits[bit];
lat[0] = mid;
}
else
lat[1] = mid;
}
even = !even;
if (bit < 4)
bit++;
else
{
geohash += Base32[ch];
bit = 0;
ch = 0;
}
}
return geohash;
}
}
}geohash演示：http://openlocation.org/geohash/geohash-js/
各种版本下载：打包下载
引用阿里云以为技术专家的博客上的讨论：
这一点是有些用户对geohash的误解，虽然geo确实尽可能的将位置相近的点hash到了一起，可是这并不是严格意义上的(实际上也并不可能，因为毕竟多一维坐标)，
例如在方格4的左下部分的点和大方格1的右下部分的点离的很近，可是它们的geohash值一定是相差的相当远，因为头一次的分块就相差太大了，很多时候我们对geohash的值进行简单的排序比较，结果貌似真的能够找出相近的点，并且似乎还是按照距离的远近排列的，可是实际上会有一些点被漏掉了。
上述这个问题，可以通过搜索一个格子，周围八个格子的数据，统一获取后再进行过滤。这样就在编码层次解决了这个问题。
2.既然不能做到将相近的点hash值也相近，那么geohash的意义何在呢？
我觉得geohash还是相当有用的一个算法，毕竟这个算法通过无穷的细分，能确保将每一个小块的geohash值确保在一定的范围之内，这样就为灵活的周边查找和范围查找提供了可能。
常见的一些应用场景
A、如果想查询附近的点？如何操作
查出改点的gehash值，然后到数据库里面进行前缀匹配就可以了。
B、如果想查询附近点，特定范围内，例如一个点周围500米的点，如何搞？
可以查询结果，在结果中进行赛选，将geohash进行解码为经纬度，然后进行比较
*在纬度相等的情况下：
*经度每隔0.00001度，距离相差约1米；
*每隔0.0001度，距离相差约10米；
*每隔0.001度，距离相差约100米；
*每隔0.01度，距离相差约1000米；
*每隔0.1度，距离相差约10000米。
*在经度相等的情况下：
*纬度每隔0.00001度，距离相差约1.1米；
*每隔0.0001度，距离相差约11米；
*每隔0.001度，距离相差约111米；
*每隔0.01度，距离相差约1113米；
*每隔0.1度，距离相差约11132米。
Geohash，如果geohash的位数是6位数的时候，大概为附近1千米…
参考资料：
http://iamzhongyong.iteye.com/blog/1399333
http://tech.idv2.com/2011/06/17/location-search/
http://blog.sina.com.cn/s/blog_62ba0fdd0100tul4.html
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