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：
geohash.class.php
View Code
View Code
1 <?php
2 /**
3 * Geohash generation class
4 * http://blog.dixo.net/downloads/
5 *
6 * This file copyright (C) 2008 Paul Dixon (paul@elphin.com)
7 *
8 * This program is free software; you can redistribute it and/or
9 * modify it under the terms of the GNU General Public License
10 * as published by the Free Software Foundation; either version 3
11 * of the License, or (at your option) any later version.
12 *
13 * This program is distributed in the hope that it will be useful,
14 * but WITHOUT ANY WARRANTY; without even the implied warranty of
15 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
16 * GNU General Public License for more details.
17 *
18 * You should have received a copy of the GNU General Public License
19 * along with this program; if not, write to the Free Software
20 * Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
21 */
22
23
24
25 /**
26 * Encode and decode geohashes
27 *
28 */
29 class Geohash
30 {
31 private $coding="0123456789bcdefghjkmnpqrstuvwxyz";
32 private $codingMap=array();
33
34 public function Geohash()
35 {
36 //build map from encoding char to 0 padded bitfield
37 for($i=0; $i<32; $i++)
38 {
39 $this->codingMap[substr($this->coding,$i,1)]=str_pad(decbin($i), 5, "0", STR_PAD_LEFT);
40 }
41
42 }
43
44 /**
45 * Decode a geohash and return an array with decimal lat,long in it
46 */
47 public function decode($hash)
48 {
49 //decode hash into binary string
50 $binary="";
51 $hl=strlen($hash);
52 for($i=0; $i<$hl; $i++)
53 {
54 $binary.=$this->codingMap[substr($hash,$i,1)];
55 }
56
57 //split the binary into lat and log binary strings
58 $bl=strlen($binary);
59 $blat="";
60 $blong="";
61 for ($i=0; $i<$bl; $i++)
62 {
63 if ($i%2)
64 $blat=$blat.substr($binary,$i,1);
65 else
66 $blong=$blong.substr($binary,$i,1);
67
68 }
69
70 //now concert to decimal
71 $lat=$this->binDecode($blat,-90,90);
72 $long=$this->binDecode($blong,-180,180);
73
74 //figure out how precise the bit count makes this calculation
75 $latErr=$this->calcError(strlen($blat),-90,90);
76 $longErr=$this->calcError(strlen($blong),-180,180);
77
78 //how many decimal places should we use? There's a little art to
79 //this to ensure I get the same roundings as geohash.org
80 $latPlaces=max(1, -round(log10($latErr))) - 1;
81 $longPlaces=max(1, -round(log10($longErr))) - 1;
82
83 //round it
84 $lat=round($lat, $latPlaces);
85 $long=round($long, $longPlaces);
86
87 return array($lat,$long);
88 }
89
90
91 /**
92 * Encode a hash from given lat and long
93 */
94 public function encode($lat,$long)
95 {
96 //how many bits does latitude need?
97 $plat=$this->precision($lat);
98 $latbits=1;
99 $err=45;
100 while($err>$plat)
101 {
102 $latbits++;
103 $err/=2;
104 }
105
106 //how many bits does longitude need?
107 $plong=$this->precision($long);
108 $longbits=1;
109 $err=90;
110 while($err>$plong)
111 {
112 $longbits++;
113 $err/=2;
114 }
115
116 //bit counts need to be equal
117 $bits=max($latbits,$longbits);
118
119 //as the hash create bits in groups of 5, lets not
120 //waste any bits - lets bulk it up to a multiple of 5
121 //and favour the longitude for any odd bits
122 $longbits=$bits;
123 $latbits=$bits;
124 $addlong=1;
125 while (($longbits+$latbits)%5 != 0)
126 {
127 $longbits+=$addlong;
128 $latbits+=!$addlong;
129 $addlong=!$addlong;
130 }
131
132
133 //encode each as binary string
134 $blat=$this->binEncode($lat,-90,90, $latbits);
135 $blong=$this->binEncode($long,-180,180,$longbits);
136
137 //merge lat and long together
138 $binary="";
139 $uselong=1;
140 while (strlen($blat)+strlen($blong))
141 {
142 if ($uselong)
143 {
144 $binary=$binary.substr($blong,0,1);
145 $blong=substr($blong,1);
146 }
147 else
148 {
149 $binary=$binary.substr($blat,0,1);
150 $blat=substr($blat,1);
151 }
152 $uselong=!$uselong;
153 }
154
155 //convert binary string to hash
156 $hash="";
157 for ($i=0; $i<strlen($binary); $i+=5)
158 {
159 $n=bindec(substr($binary,$i,5));
160 $hash=$hash.$this->coding[$n];
161 }
162
163
164 return $hash;
165 }
166
167 /**
168 * What's the maximum error for $bits bits covering a range $min to $max
169 */
170 private function calcError($bits,$min,$max)
171 {
172 $err=($max-$min)/2;
173 while ($bits--)
174 $err/=2;
175 return $err;
176 }
177
178 /*
179 * returns precision of number
180 * precision of 42 is 0.5
181 * precision of 42.4 is 0.05
182 * precision of 42.41 is 0.005 etc
183 */
184 private function precision($number)
185 {
186 $precision=0;
187 $pt=strpos($number,'.');
188 if ($pt!==false)
189 {
190 $precision=-(strlen($number)-$pt-1);
191 }
192
193 return pow(10,$precision)/2;
194 }
195
196
197 /**
198 * create binary encoding of number as detailed in http://en.wikipedia.org/wiki/Geohash#Example
199 * removing the tail recursion is left an exercise for the reader
200 */
201 private function binEncode($number, $min, $max, $bitcount)
202 {
203 if ($bitcount==0)
204 return "";
205
206 #echo "$bitcount: $min $max<br>";
207
208 //this is our mid point - we will produce a bit to say
209 //whether $number is above or below this mid point
210 $mid=($min+$max)/2;
211 if ($number>$mid)
212 return "1".$this->binEncode($number, $mid, $max,$bitcount-1);
213 else
214 return "0".$this->binEncode($number, $min, $mid,$bitcount-1);
215 }
216
217
218 /**
219 * decodes binary encoding of number as detailed in http://en.wikipedia.org/wiki/Geohash#Example
220 * removing the tail recursion is left an exercise for the reader
221 */
222 private function binDecode($binary, $min, $max)
223 {
224 $mid=($min+$max)/2;
225
226 if (strlen($binary)==0)
227 return $mid;
228
229 $bit=substr($binary,0,1);
230 $binary=substr($binary,1);
231
232 if ($bit==1)
233 return $this->binDecode($binary, $mid, $max);
234 else
235 return $this->binDecode($binary, $min, $mid);
236 }
237 }
238
239
240
241
242
243
244 ?>
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
View Code
1 import java.io.File;
2 import java.io.FileInputStream;
3 import java.util.BitSet;
4 import java.util.HashMap;
5
6
7 public class Geohash {
8
9 private static int numbits = 6 * 5;
10 final static char[] digits = { '0', '1', '2', '3', '4', '5', '6', '7', '8',
11 '9', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j', 'k', 'm', 'n', 'p',
12 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };
13
14 final static HashMap<Character, Integer> lookup = new HashMap<Character, Integer>();
15 static {
16 int i = 0;
17 for (char c : digits)
18 lookup.put(c, i++);
19 }
20
21 public static void main(String[] args) throws Exception{
22
23 System.out.println(new Geohash().encode(45, 125));
24
25 }
26
27 public double[] decode(String geohash) {
28 StringBuilder buffer = new StringBuilder();
29 for (char c : geohash.toCharArray()) {
30
31 int i = lookup.get(c) + 32;
32 buffer.append( Integer.toString(i, 2).substring(1) );
33 }
34
35 BitSet lonset = new BitSet();
36 BitSet latset = new BitSet();
37
38 //even bits
39 int j =0;
40 for (int i=0; i< numbits*2;i+=2) {
41 boolean isSet = false;
42 if ( i < buffer.length() )
43 isSet = buffer.charAt(i) == '1';
44 lonset.set(j++, isSet);
45 }
46
47 //odd bits
48 j=0;
49 for (int i=1; i< numbits*2;i+=2) {
50 boolean isSet = false;
51 if ( i < buffer.length() )
52 isSet = buffer.charAt(i) == '1';
53 latset.set(j++, isSet);
54 }
55
56 double lon = decode(lonset, -180, 180);
57 double lat = decode(latset, -90, 90);
58
59 return new double[] {lat, lon};
60 }
61
62 private double decode(BitSet bs, double floor, double ceiling) {
63 double mid = 0;
64 for (int i=0; i<bs.length(); i++) {
65 mid = (floor + ceiling) / 2;
66 if (bs.get(i))
67 floor = mid;
68 else
69 ceiling = mid;
70 }
71 return mid;
72 }
73
74
75 public String encode(double lat, double lon) {
76 BitSet latbits = getBits(lat, -90, 90);
77 BitSet lonbits = getBits(lon, -180, 180);
78 StringBuilder buffer = new StringBuilder();
79 for (int i = 0; i < numbits; i++) {
80 buffer.append( (lonbits.get(i))?'1':'0');
81 buffer.append( (latbits.get(i))?'1':'0');
82 }
83 return base32(Long.parseLong(buffer.toString(), 2));
84 }
85
86 private BitSet getBits(double lat, double floor, double ceiling) {
87 BitSet buffer = new BitSet(numbits);
88 for (int i = 0; i < numbits; i++) {
89 double mid = (floor + ceiling) / 2;
90 if (lat >= mid) {
91 buffer.set(i);
92 floor = mid;
93 } else {
94 ceiling = mid;
95 }
96 }
97 return buffer;
98 }
99
100 public static String base32(long i) {
101 char[] buf = new char[65];
102 int charPos = 64;
103 boolean negative = (i < 0);
104 if (!negative)
105 i = -i;
106 while (i <= -32) {
107 buf[charPos--] = digits[(int) (-(i % 32))];
108 i /= 32;
109 }
110 buf[charPos] = digits[(int) (-i)];
111
112 if (negative)
113 buf[--charPos] = '-';
114 return new String(buf, charPos, (65 - charPos));
115 }
116
117 }
C#：
1 using System; 2 3 namespace sharonjl.utils 4 { 5 public static class Geohash 6 { 7 #region Direction enum 8 9 public enum Direction 10 { 11 Top = 0, 12 Right = 1, 13 Bottom = 2, 14 Left = 3 15 } 16 17 #endregion 18 19 private const string Base32 = "0123456789bcdefghjkmnpqrstuvwxyz"; 20 private static readonly int[] Bits = new[] {16, 8, 4, 2, 1}; 21 22 private static readonly string[][] Neighbors = { 23 new[] 24 { 25 "p0r21436x8zb9dcf5h7kjnmqesgutwvy", // Top 26 "bc01fg45238967deuvhjyznpkmstqrwx", // Right 27 "14365h7k9dcfesgujnmqp0r2twvyx8zb", // Bottom 28 "238967debc01fg45kmstqrwxuvhjyznp", // Left 29 }, new[] 30 { 31 "bc01fg45238967deuvhjyznpkmstqrwx", // Top 32 "p0r21436x8zb9dcf5h7kjnmqesgutwvy", // Right 33 "238967debc01fg45kmstqrwxuvhjyznp", // Bottom 34 "14365h7k9dcfesgujnmqp0r2twvyx8zb", // Left 35 } 36 }; 37 38 private static readonly string[][] Borders = { 39 new[] {"prxz", "bcfguvyz", "028b", "0145hjnp"}, 40 new[] {"bcfguvyz", "prxz", "0145hjnp", "028b"} 41 }; 42 43 public static String CalculateAdjacent(String hash, Direction direction) 44 { 45 hash = hash.ToLower(); 46 47 char lastChr = hash[hash.Length - 1]; 48 int type = hash.Length%2; 49 var dir = (int) direction; 50 string nHash = hash.Substring(0, hash.Length - 1); 51 52 if (Borders[type][dir].IndexOf(lastChr) != -1) 53 { 54 nHash = CalculateAdjacent(nHash, (Direction) dir); 55 } 56 return nHash + Base32[Neighbors[type][dir].IndexOf(lastChr)]; 57 } 58 59 public static void RefineInterval(ref double[] interval, int cd, int mask) 60 { 61 if ((cd & mask) != 0) 62 { 63 interval[0] = (interval[0] + interval[1])/2; 64 } 65 else 66 { 67 interval[1] = (interval[0] + interval[1])/2; 68 } 69 } 70 71 public static double[] Decode(String geohash) 72 { 73 bool even = true; 74 double[] lat = {-90.0, 90.0}; 75 double[] lon = {-180.0, 180.0}; 76 77 foreach (char c in geohash) 78 { 79 int cd = Base32.IndexOf(c); 80 for (int j = 0; j < 5; j++) 81 { 82 int mask = Bits[j]; 83 if (even) 84 { 85 RefineInterval(ref lon, cd, mask); 86 } 87 else 88 { 89 RefineInterval(ref lat, cd, mask); 90 } 91 even = !even; 92 } 93 } 94 95 return new[] {(lat[0] + lat[1])/2, (lon[0] + lon[1])/2}; 96 } 97 98 public static String Encode(double latitude, double longitude, int precision = 12) 99 { 100 bool even = true; 101 int bit = 0; 102 int ch = 0; 103 string geohash = ""; 104 105 double[] lat = {-90.0, 90.0}; 106 double[] lon = {-180.0, 180.0}; 107 108 if (precision < 1 || precision > 20) precision = 12; 109 110 while (geohash.Length < precision) 111 { 112 double mid; 113 114 if (even) 115 { 116 mid = (lon[0] + lon[1])/2; 117 if (longitude > mid) 118 { 119 ch |= Bits[bit]; 120 lon[0] = mid; 121 } 122 else 123 lon[1] = mid; 124 } 125 else 126 { 127 mid = (lat[0] + lat[1])/2; 128 if (latitude > mid) 129 { 130 ch |= Bits[bit]; 131 lat[0] = mid; 132 } 133 else 134 lat[1] = mid; 135 } 136 137 even = !even; 138 if (bit < 4) 139 bit++; 140 else 141 { 142 geohash += Base32[ch]; 143 bit = 0; 144 ch = 0; 145 } 146 } 147 return geohash; 148 } 149 } 150 }
C#代码来自：https://github.com/sharonjl/geohash-net
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
