经纬度距离计算-优快云博客

  # 根据经纬度计算两点间距离 
 
  # 经度 long 纬度 lat 
 
  def 
  GetDistance( lng1, lat1, lng2, lat2):  
 
  u
  '''''计算两点间球面距离 单位为m''' 
 
  EARTH_RADIUS 
  = 
  6378.137 
  # 地球周长/2*pi 此处地球周长取40075.02km pi=3.1415929134165665  
 
  from 
  math 
  import 
  asin,sin,cos,acos,radians, degrees,
  pow
  ,sqrt, hypot,pi  
 
  # 方法0  
 
  # 最简单的求平面两点间距离 误差比较大  
 
  d
  =
  hypot(lng2
  -
  lng1,lat2
  -
  lat1)
  *
  40075.02
  /
  360
  *
  1000 
 
  print 
  'd0='
  ,d  
 
  # 方法1  
 
  # ''' d＝111.12cos{1/[sinΦAsinΦB+cosΦAcosΦBcos(λB—λA)]} '''  
 
  # ''' 其中A点经度、纬度分别为λA和ΦA，B点的经度、纬度分别为λB和ΦB，d为距离。'''  
 
  m
  =
  cos(radians(lat1))
  *
  cos(radians(lat2))
  *
  cos(radians(lng2
  -
  lng1))  
 
  x
  =
  1
  /
  (sin(radians(lat1))
  *
  sin(radians(lat2)) 
  +
  m)  
 
  d
  =
  111.12
  *
  cos(radians(
  1
  /
  x))  
 
  print 
  'd1='
  ,d  
 
  # 方法2  
 
  # 据说来源于 google maps 的脚本  
 
  # 见 http://en.wikipedia.org/wiki/Great-circle_distance 中的 haversine  
 
  radLat1 
  = 
  radians(lat1) 
  # a点纬度(单位是弧度)  
 
  radLat2 
  = 
  radians(lat2) 
  # b点纬度(单位是弧度  
 
  a 
  = 
  radLat1 
  - 
  radLat2 
  # 两点间的纬度弧度差  
 
  b 
  = 
  radians(lng1) 
  - 
  radians(lng2) 
  # 两点间的经度弧度差  
 
  s 
  = 
  2 
  * 
  asin(sqrt(
  pow
  (sin(a
  /
  2
  ),
  2
  ) 
  + 
  cos(radLat1)
  *
  cos(radLat2)
  *
  pow
  (sin(b
  /
  2
  ),
  2
  ))) 
  # 两点间的弧度  
 
  s 
  = 
  s 
  * 
  EARTH_RADIUS  
 
  # s = round(s * 10000) / 10000 # 四舍五入保留小数点后4位 
 
  d
  =
  s
  *
  1000 
 
  print 
  'd2='
  ,d  
 
  # 方法3  
 
  # ''' 经纬坐标为P(x1,y1) Q(x2,y2) '''  
 
  # D=arccos[cosy1*cosy2*cos(x1-x2)+siny1*siny2]*2*PI*R/360  
 
  d
  =
  acos(cos(radians(lat1))
  *
  cos(radians(lat2))
  *
  cos(radians(lng1
  -
  lng2))
  +
  sin(radians(lat1))
  *
  sin(radians(lat2)))
  *
  EARTH_RADIUS
  *
  1000 
 
  print 
  'd3='
  ,d  
 
  return 
  d  
 
  GetDistance(
  116.95400
  ,
  39.95400
  ,
  116.95300
  ,
  39.95300
  )