使用python在Land中查找最近的位置坐标

Question

地理编码 api 不返回海洋/海洋坐标的位置信息。 对于这些记录，我想找到具有有效位置信息的最近可能坐标（即最近的陆地坐标）下面是通过传递坐标获取位置信息的代码

import requests    
request_url = "https://api.mapbox.com/geocoding/v5/mapbox.places/{0}%2C{1}.json?access_token={2}&types=country&limit=1".format(lng,lat,key)
response = requests.get(request_url)
output = response.json()

我不知道找到最近的位置。 我也是 Python 新手

示例输出：

{'type': 'FeatureCollection',
 'query': [32.12, 54.21],
 'features': [{'id': 'country.10008046970720960',
   'type': 'Feature',
   'place_type': ['country'],
   'relevance': 1,
   'properties': {'short_code': 'ru', 'wikidata': 'Q159'},
   'text': 'Russia',
   'place_name': 'Russia',
   'bbox': [19.608673, 41.185353, 179.9, 81.961618],
   'center': [37.61667, 55.75],
   'geometry': {'type': 'Point', 'coordinates': [37.61667, 55.75]}}],
 'attribution': 'NOTICE: © 2020 Mapbox and its suppliers. All rights reserved. Use of this data is subject to the Mapbox Terms of Service (https://www.mapbox.com/about/maps/). This response and the information it contains may not be retained. POI(s) provided by Foursquare.'}

坐标为海洋时的输出：

{'type': 'FeatureCollection',
 'query': [0, 0],
 'features': [],
 'attribution': 'NOTICE: © 2020 Mapbox and its suppliers. All rights reserved. Use of this data is subject to the Mapbox Terms of Service (https://www.mapbox.com/about/maps/). This response and the information it contains may not be retained. POI(s) provided by Foursquare.'}

Answer 1

使用Haversine公式根据纬度和经度找到最近的点（例如城市）。

def dist_between_two_lat_lon(*args):
    from math import asin, cos, radians, sin, sqrt
    lat1, lat2, long1, long2 = map(radians, args)

    dist_lats = abs(lat2 - lat1) 
    dist_longs = abs(long2 - long1) 
    a = sin(dist_lats/2)**2 + cos(lat1) * cos(lat2) * sin(dist_longs/2)**2
    c = asin(sqrt(a)) * 2
    radius_earth = 6378 # the "Earth radius" R varies from 6356.752 km at the poles to 6378.137 km at the equator.
    return c * radius_earth

def find_closest_lat_lon(data, v):
    try:
        return min(data, key=lambda p: dist_between_two_lat_lon(v['lat'],p['lat'],v['lon'],p['lon']))
    except TypeError:
        print('Not a list or not a number.')
    
# city = {'lat_key': value, 'lon_key': value}  # type:dict()
new_york = {'lat': 40.712776, 'lon': -74.005974}
washington = {'lat': 47.751076,  'lon': -120.740135}
san_francisco = {'lat': 37.774929, 'lon': -122.419418}

city_list = [new_york, washington, san_francisco]

city_to_find = {'lat': 29.760427, 'lon': -95.369804}  # Houston
print(find_closest_lat_lon(city_list, city_to_find))

哪个产量：

{'lat': 47.751076, 'lon': -120.740135}  # Corresponds to Washington

假设您从 mapbox 获得了四个 json 答案，并将它们保存在一个列表中：

json_answers = list()  # = []

json_answers.append({'type': 'FeatureCollection', 
'query': [32.12, 54.21],
'features': [{'id': 'country.10008046970720960',
'type': 'Feature',
'place_type': ['country'],
'relevance': 1,
'properties': {'short_code': 'ru', 'wikidata': 'Q159'},
'text': 'Russia',
'place_name': 'Russia',
'bbox': [19.608673, 41.185353, 179.9, 81.961618],
'center': [37.61667, 55.75],
'geometry': {'type': 'Point', 'coordinates': [37.61667, 55.75]}}],
'attribution': 'NOTICE: ...'})

# I changed only the 'coordinates' value for this example
json_answers.append({'type': 'FeatureCollection', 
'query': [32.12, 54.21],
'features': [{'id': 'country.10008046970720960',
'type': 'Feature',
'place_type': ['country'],
'relevance': 1,
'properties': {'short_code': 'ru', 'wikidata': 'Q159'},
'text': 'Russia',
'place_name': 'Russia',
'bbox': [19.608673, 41.185353, 179.9, 81.961618],
'center': [37.61667, 55.75],
'geometry': {'type': 'Point', 'coordinates': [38.21667, 56.15]}}],
'attribution': 'NOTICE: ...'})

# I changed only the 'coordinates' value for this example
json_answers.append({'type': 'FeatureCollection', 
'query': [32.12, 54.21],
'features': [{'id': 'country.10008046970720960',
'type': 'Feature',
'place_type': ['country'],
'relevance': 1,
'properties': {'short_code': 'ru', 'wikidata': 'Q159'},
'text': 'Russia',
'place_name': 'Russia',
'bbox': [19.608673, 41.185353, 179.9, 81.961618],
'center': [37.61667, 55.75],
'geometry': {'type': 'Point', 'coordinates': [33.21667, 51.15]}}],
'attribution': 'NOTICE: ...'})

# The last answer is "null"
json_answers.append({'type': 'FeatureCollection',
'query': [0, 0],
'features': [],
'attribution': 'NOTICE: ...'})

coord_list = []
for answer in json_answers:
    if answer['features']:  # check if ['features'] is not empty
        # I'm not sure if it's [lat, lon] or [lon, lat] (you can verify it on mapbox)
        print(f"Coordinates in [lat, lon]: {answer['features'][0]['geometry']['coordinates']}")
        lat = answer['features'][0]['geometry']['coordinates'][0]
        lon = answer['features'][0]['geometry']['coordinates'][1]

        temp_dict = {'lat': lat, 'lon': lon}
        coord_list.append(temp_dict)

print(f"coord_list = {coord_list}")

point_to_find = {'lat': 37.41667, 'lon': 55.05}  # Houston
print(f"point_to_find = {point_to_find}")
print(f"find_closest_lat_lon = {find_closest_lat_lon(coord_list, point_to_find)}")

其中产生：

{'lat': 47.751076, 'lon': -120.740135}
Coordinates in [lat, lon]: [37.61667, 55.75]
Coordinates in [lat, lon]: [38.21667, 56.15]
Coordinates in [lat, lon]: [33.21667, 51.15]

coord_list = [{'lat': 37.61667, 'lon': 55.75}, {'lat': 38.21667, 'lon': 56.15}, {'lat': 33.21667, 'lon': 51.15}]

point_to_find = {'lat': 37.41667, 'lon': 55.05}

find_closest_lat_lon = {'lat': 38.21667, 'lon': 56.15}

Answer 2

在 python 中使用reverse_geocode库来获取离国家最近的城市。

例子：

import reverse_geocode

coordinates = (-37.81, 144.96), (31.76, 35.21)

reverse_geocode.search(coordinates)

结果：

[{'city': 'Melbourne', 'code': 'AU', 'country': 'Australia'}, {'city': 'Jerusalem', 'code': 'IL', 'country': '以色列'}]

Answer 3

这是一个未优化的解决方案。

函数背后发生了什么：

1. 在一个点上运行GeoPy反向查找。
2. 如果找到该点，则返回其国家/地区名称。
3. 如果未找到该点，则在world_geometry变量中搜索最近的陆world_geometry 。
4. 对该最近点执行反向查找。
5. 返回该点的国家名称（如果存在）或地点名称（如果没有国家名称）。

from geopy.geocoders import Nominatim
from shapely.ops import nearest_points

def country_lookup(query, geocoder, land_geometry):
    
    try:
        loc = geocoder.reverse((query.y, query.x))
        return loc.raw['address']['country']
    except (KeyError, AttributeError):
        _, p2 = nearest_points(query, land_geometry)
        loc = geocoder.reverse((p2.y, p2.x)).raw['address']
        if 'country' in loc.keys():
            return loc['country']
        else:
            return loc['locality']
        
# get world (or any land) geometry, instantiate geolocator service
world = gp.read_file(gp.datasets.get_path('naturalearth_lowres'))
world_geometry = world.geometry.unary_union
geolocator = Nominatim(user_agent="GIW")

# Create a column of country names from points in a GDF's geometry.
gdf['country'] = gdf.geometry.apply(country_lookup, args=(geolocator, world_geometry))

结果的准确性取决于您提供的土地几何图形的准确性。 比如geopandas的世界几何就很不错。 除了巴哈马群岛中一些最小的岛屿之外，我能够找到所有国家/地区的名称。 那些它找不到的被函数标记为“百慕大三角”，这对我来说已经足够了。

Answer 4

不同的试用包是 reverse_geocoder，它将返回最近的城市、州和国家。 似乎比 reverse_geocode 包更好。

import reverse_geocoder as rg
coordinates = (29,-84.1),(37,-125) #Both located in the ocean
rg.search(coordinates) 

输出：

[OrderedDict([('lat', '29.67106'),
              ('lon', '-83.38764'),
              ('name', 'Steinhatchee'),
              ('admin1', 'Florida'),
              ('admin2', 'Taylor County'),
              ('cc', 'US')]),
 OrderedDict([('lat', '38.71519'),
              ('lon', '-123.45445'),
              ('name', 'Sea Ranch'),
              ('admin1', 'California'),
              ('admin2', 'Sonoma County'),
              ('cc', 'US')])]