蟒蛇与谷歌地球坐标距离计算

Question

嗨我有一个名为地标的kml文件和4个节点（地标），位于谷歌地球上的一个区域。每个地标节点都有一个经度和纬度。通过上面的代码中，我能提取数据

(u'node0:', 21.78400936610002, 38.2874355527483) 
(u'node1:', 21.78453228393861, 38.28690995466475) 
(u'node2:', 21.7848823502596, 38.2869152766261) 
(u'node3:', 21.78459887820567, 38.28740826552452) 

我想是计算NODE0和node2,3,4之间的距离......（保持不变NODE0在距离函数），然后打印结果。

我想使用的功能是：

import math 

R = 6371 # km 
dLat = (lat2-lat1) # Make sure it's in radians, not degrees 
dLon = (lon2-lon1) # Idem 
a = math.sin(dLat/2) * math.sin(dLat/2) + 
    math.cos(lat1) * math.cos(lat2) * 
    math.sin(dLon/2) * math.sin(dLon/2) 
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a)) 
d = R * c; 

from xml.dom import minidom 

xmldoc = minidom.parse("placemarks.kml") 
kml = xmldoc.getElementsByTagName("kml")[0] 
document = kml.getElementsByTagName("Document")[0] 
placemarks = document.getElementsByTagName("Placemark") 

for placemark in placemarks: 
    nodename = placemark.getElementsByTagName("name")[0].firstChild.data 
    lst = nodename.split(":") 
    coords = placemark.getElementsByTagName("coordinates")[0].firstChild.data 
    lst1 = coords.split(",") 
    longitude = float(lst1[0]) 
    latitude = float(lst1[1]) 

def calc_distance(longitude, latitude) 
    print(nodename + ":",longitude, latitude) 

Answer 1

使用：

import math 

def distance(origin, destination): 
    lat1, lon1 = origin 
    lat2, lon2 = destination 
    radius = 6371 # km 

    dlat = math.radians(lat2-lat1) 
    dlon = math.radians(lon2-lon1) 
    a = math.sin(dlat/2) * math.sin(dlat/2) + math.cos(math.radians(lat1)) \ 
     * math.cos(math.radians(lat2)) * math.sin(dlon/2) * math.sin(dlon/2) 
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a)) 
    d = radius * c 

    return d 

pos = [ 
     (u'node0:', 21.78400936610002, 38.2874355527483), 
     (u'node1:', 21.78453228393861, 38.28690995466475), 
     (u'node2:', 21.7848823502596, 38.2869152766261), 
     (u'node3:', 21.78459887820567, 38.28740826552452)] 

node0 = [p for p in pos if p[0] == 'node0:'][0] 
dist=(distance(node0[1:3], p[1:3]) for p in pos if p[0] != 'node0:') 
for d in dist: print(d) 

Answer 2

这是你用很少的修改做什么（我觉得）你想代码：

# adapted from haversine.py <https://gist.github.com/rochacbruno/2883505> 
# see also <http://en.wikipedia.org/wiki/Haversine_formula> 
from math import atan2, cos, sin, sqrt, radians 

def calc_distance(origin, destination): 
    """great-circle distance between two points on a sphere 
     from their longitudes and latitudes""" 
    lat1, lon1 = origin 
    lat2, lon2 = destination 
    radius = 6371 # km. earth 

    dlat = radians(lat2-lat1) 
    dlon = radians(lon2-lon1) 
    a = (sin(dlat/2) * sin(dlat/2) + cos(radians(lat1)) * cos(radians(lat2)) * 
     sin(dlon/2) * sin(dlon/2)) 
    c = 2 * atan2(sqrt(a), sqrt(1-a)) 
    d = radius * c 

    return d 

from xml.dom import minidom 

xmldoc = minidom.parse("placemarks.kml") 
kml = xmldoc.getElementsByTagName("kml")[0] 
document = kml.getElementsByTagName("Document")[0] 
placemarks = document.getElementsByTagName("Placemark") 

nodes = {} 
for placemark in placemarks: 
    nodename = placemark.getElementsByTagName("name")[0].firstChild.data[:-1] 
    coords = placemark.getElementsByTagName("coordinates")[0].firstChild.data 
    lst1 = coords.split(",") 
    longitude = float(lst1[0]) 
    latitude = float(lst1[1]) 
    nodes[nodename] = (latitude, longitude) 

## used for testing due to lack of kml file... 
#nodes = { 
# u'node0': (21.78400936610002, 38.2874355527483), 
# u'node1': (21.78453228393861, 38.28690995466475), 
# u'node2': (21.7848823502596, 38.2869152766261), 
# u'node3': (21.78459887820567, 38.28740826552452)} 

node0coords = nodes[u'node0'] 
for nodename in sorted(name for name in nodes if name != u'node0'): 
    longitude, latitude = nodes[nodename] 
    print('{}: ({:3.14f}, {:3.14f}), distance from node0: {:.3f} km'.format(
      nodename, longitude, latitude, 
      calc_distance(node0coords, nodes[nodename]))) 

输出：

node1: (21.78453228393861, 38.28690995466475), distance from node0: 0.080 km 
node2: (21.78488235025960, 38.28691527662610), distance from node0: 0.111 km 
node3: (21.78459887820567, 38.28740826552452), distance from node0: 0.066 km