Geohash算法的实现

原创
2013/04/25 09:14
阅读数 1.2K


geohash的算法

下面以(39.92324, 116.3906)为例,介绍一下geohash的编码算法。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。 不过由于geohash表示的是区间,编码越长越精确,但不可能解码出完全一致的地址。

#define BASE32 "0123456789bcdefghjkmnpqrstuvwxyz"
      static void encode_geohash(double latitude, double longitude, int precision, char *geohash) {
      int is_even=1, i=0;
      double lat[2], lon[2], mid;
      char bits[] = {16,8,4,2,1};
      int bit=0, ch=0;
      lat[0] = -90.0; lat[1] = 90.0;
      lon[0] = -180.0; lon[1] = 180.0;
      while (i < precision) {
      if (is_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;
      }
      is_even = !is_even;
      if (bit < 4)
      bit++;
      else {
      geohash[i++] = BASE32[ch];
      bit = 0;
      ch = 0;
      }
      }
      geohash[i] = 0;
      }
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