最近项目上有这样的需求，需要依据设备的经纬度坐标计算距离及角度。经验证后效果较好，并分享。

1 经纬度转换距离代码

#!/usr/bin/env python

# -*- coding: utf-8 -*-

__author__ = 'Seven'

import math

# 计算距离

def getDistance(latA, lonA, latB, lonB):

ra = 6378140 # 赤道半径

rb = 6356755 # 极半径

flatten = (ra - rb) / ra # Partial rate of the earth

# change angle to radians

radLatA = math.radians(latA)

radLonA = math.radians(lonA)

radLatB = math.radians(latB)

radLonB = math.radians(lonB)

pA = math.atan(rb / ra * math.tan(radLatA))

pB = math.atan(rb / ra * math.tan(radLatB))

x = math.acos(math.sin(pA) * math.sin(pB) + math.cos(pA) * math.cos(pB) * math.cos(radLonA - radLonB))

c1 = (math.sin(x) - x) * (math.sin(pA) + math.sin(pB)) ** 2 / math.cos(x / 2) ** 2

c2 = (math.sin(x) + x) * (math.sin(pA) - math.sin(pB)) ** 2 / math.sin(x / 2) ** 2

dr = flatten / 8 * (c1 - c2)

distance = ra * (x + dr)

distance = round(distance / 1000, 4)

return f'{distance}km'

2 经纬度转化角度代码

#!/usr/bin/env python

# -*- coding: utf-8 -*-

__author__ = 'Seven'

import math

# 计算角度

def getDegree(latA, lonA, latB, lonB):

radLatA = math.radians(latA)

radLonA = math.radians(lonA)

radLatB = math.radians(latB)

radLonB = math.radians(lonB)

dLon = radLonB - radLonA

y = math.sin(dLon) * math.cos(radLatB)

x = math.cos(radLatA) * math.sin(radLatB) - math.sin(radLatA) * math.cos(radLatB) * math.cos(dLon)

brng = math.degrees(math.atan2(y, x))

brng = round((brng + 360) % 360, 4)

brng = int(brng)

if (brng == 0.0) or ((brng == 360.0)):

return '正北方向'

elif brng == 90.0:

return '正东方向'

elif brng == 180.0:

return '正南方向'

elif brng == 270.0:

return '正西方向'

elif 0 < brng < 90:

return f'北偏东{brng}'

elif 90 < brng < 180:

return f'东偏南{brng - 90}'

elif 180 < brng < 270:

return f'西偏南{270 - brng}'

elif 270 < brng < 360:

return f'北偏西{brng - 270}'

else:

pass

3 验证

选取深圳野生动物园(22.599578, 113.973129)为起点，深圳坪山站(22.6986848, 114.3311032)为终点，结合百度地图、谷歌地图等进行效果验证。

程序运行结果如下：

百度测距为38.3km

Google地图手动测距为39.31km

距离与角度均无问题。