Question

我从国际空间站（ISS）获得了一些坐标，我想知道当记录坐标时ISS是在陆地还是在海洋上，因此我应该离线进行此操作，但是我不确定该采用哪种方法使用。来自python标准库的一部分，我只能使用以下库：

numpy
scipy
tensorflow
pandas
opencv-python
opencv-contrib-python
evdev
matplotlib
logzero
pyephem
scikit-image
scikit-learn
reverse-geocoder

如果您知道如何使用其他一些库来执行此操作，那还是很好。

使用此代码，我得到坐标并将其写入文件：

import logging
import logzero
from logzero import logger
import os
import ephem
import time

dir_path = os.path.dirname(os.path.realpath(__file__))
logzero.logfile(dir_path+"/coordinates.csv")

# Set a custom formatter
formatter = logging.Formatter('%(name)s - %(asctime)-15s - %(levelname)s: %(message)s');
logzero.formatter(formatter)

name = "ISS (ZARYA)"
line1 = "1 25544U 98067A   18282.18499736  .00001222  00000-0  25998-4 0  9992"
line2 = "2 25544  51.6418 170.6260 0003545 261.4423 234.4561 15.53790940136242"
iss = ephem.readtle(name, line1, line2)

iss.compute()

latitude = iss.sublat
longitude = iss.sublong

# Save the data to the file
logger.info("%s,%s", latitude, longitude )

你们有什么主意吗？预先感谢。

Answer 1

mpl_toolkits.basemap也许可以提供帮助。

from mpl_toolkits.basemap import Basemap
bm = Basemap()   # default: projection='cyl'
print(bm.is_land(99.0, 13.0))  #True
print(bm.is_land(0.0, 0.0)) # False

文档：here及其相关方法如下：

is_land（xpt，ypt） 如果给定的x，y点（在投影坐标中）在陆地上，则返回True，否则返回False。土地的定义基于与类实例关联的GSHHS海岸线多边形。陆地区域内的湖泊上的点数不算作陆地点。

注意：您可能需要注意与底图对象一起使用的投影。

Answer 2

最后，我只能使用这些库解决问题。我使用这个网站geoplaner来获得海洋形状的粗略轮廓（这确实很粗糙，因为我是手工完成的，但是对于我的目的来说效果很好，我认为在线应该有一些更准确的多边形但是我对如何使用它们迷失了。）

我对每个海洋都这样做了（请注意，我使用的坐标不能完全覆盖海洋，例如，我避开了南洋）：

    atlanticOcean = [(-24.6,68.5), (25.3,69.8), (5.7,61.4), (4.6,52.2), (-6.3,48.4),
            (-9.45,43.5), (-9.63,37.6), (-6.3,35.5), (-10.5,31.1), (-10.5,28.4),
            (-16.1,24.5), (-17.2,14.7), (-8.2,4.1), (6.3,3.6), (9.9,3.4),
            (9,-1.7), (13.8,-12.6), (11.7,-16.5), (14.5,-22.3), (16.1,-28.67),
            (18.9,-34.5), (18.9,-55.7), (-66,-55.7), (-68.5,-50.4), (-58.6,-39.3), (-48.1,-28.2),
            (-48.1,-25.7), (-41.6,-22.7), (-38.7,-17.4), (-39.5,-13.7), (-36.9,-12.5),
            (-34.9,-10.4), (-35.0,-5.5), (-50,-0.1), (-53,5.5), (-57.2,6.1),
            (-62.8,10.9), (-67.8,10.9), (-74.2,10.8), (-76.9,8.5), (-81.6,9.4),
            (-82.7,14), (-87.4,16.1), (-86.3,21.6), (-90.2,21.7), (-91.2,19.2),
            (-95.7,18.8), (-97.1,25.5), (-91.0,28.9), (-84,29.7), (-82.9,27.3),
            (-80.9,24.9), (-79.3,26.7), (-81.1,31.3), (-75.4,35.2), (-73.8,40.3),
            (-69.6,41.4), (-65.1,43.5), (-60,45.8), (-52.2,47.1), (-54.9,52.9),
            (-44.5,60.1), (-38.8,65.1)]

indianOcean =  [(21.40,-34.15), (27.37,-33.71), (40.03,-15.61), (39.68,-3.50), (51.80,10.16), 
                (58.84,22.26), (65.69,25.18), (71.32,19.83), (77.47,6.86), (80.24,12.53),
                (80.90,15.85), (89.05,22.12), (91.38,22.08), (94.54,17.74), (94.02,16.02),
                (97.00,16.82), (98.19,8.33), (100.78,3.18), (94.98,6.29), (105.0,-6.52),
                (118.16,-9.26), (123.52,-11.25), (129.93,-11.08), (128.62,-14.51), (125.89,-3.57),
                 (118.51,-20.37), (113.06,-22.18), (115.26,-34.44), (123.52,-34.88), (130.99,-32.09),
                (137.23,-36.59), (137.50,-66.47), (102.26,-65.79), (85.65,-66.22), (75.01,-69.50),
                (69.04,-67.67), (54.18,-65.76), (37.48,-68.65)]

现在，太平洋变得更加复杂，因为它在地图的两侧延伸，并且您可以有两个连续的经度分别为-179和179的点，这导致该多边形在xy平面中的表现不佳。我所做的就是将它一分为二，所以我得到了：

pacificEast = [(149.9,-37.8),(153.9,-28.5),(143.2,-11.5),(152.1,-0.9),(127.9,5.7),
                (122.9,23.8),(123.4,31),(128.9,33.7),(129.8,29.4),(141.6,35),
                (142.8,41),(148,43.3),(144.6,45.5),(146.2,49.3),(144.9,54.2),
                (136.8,55.2),(143.1,59.1),(153.7,59.2),(159.4,61.6),(160.3,60.5),
                (161.4,60.3),(155.4,57),(156.6,50.3),(160.8,52.8),(164.1,55.8),
                (163.8,58.1),(167.3,60.1),(170.7,59.8), (179.9,-77.1),
                (166.4,-77.1), (173.8,-71.8), (142.9,-66.8), (146.9,-44.8)]

pacificWest = [(-179.9,62.2),(-179.7,64.7),
                (-177.3,65.3),(-173.6,63.4),(-166,62.2),(-165.8,60.9),(-168.4,60.4),
                (-166.6,58.9),(-158.5,57.8),(-153.1,57),(-144.8,59.9),(-136.1,56.9),
                (-131.7,51.9),(-125.2,48.4),(-124.5,44.6),(-124.4,40.7),(-117.6,32.7),
                (-110.7,23.2),(-105.8,19.7),(-96.1,15.3),(-87.9,12.4),(-83.7,7.3),
                (-78.7,6.1),(-80.2,0.9),(-82.2,-0.6),(-81.2,-6.3),(-76.7,-14.4),
                (-70.4,-18.9),(-73.7,-36.7),(-76,-46.2),(-75.1,-53),(-73.4,-55.1),
                (-66.6,-56.3),(-64.6,-55),(-59.6,-63.4),(-68.4,-65.7),(-75.8,-72.2),
                (-98.6,-71.8),(-126.8,-73.2),(-146.8,-75.7),(-162.6,-78.4),(-179.9,-77.1)]

据我了解，使用matplotlib可以使用path从顶点（坐标列表）创建多边形，然后可以使用contains_point()函数检查该点是否在任一多边形（因此在“海洋”中）（或不在“陆地”中）：

    p1 = path.Path(atlanticOcean)
    p2 = path.Path(indianOcean)
    p3 = path.Path(pacificEast)
    p4 = path.Path(pacificWest)

    target = [(lon, lat)]

    result1 = p1.contains_points(target)
    result2 = p2.contains_points(target)
    result3 = p3.contains_points(target)
    result4 = p4.contains_points(target)

    # if target is in one of the polygons, it is in ocean
    if result1==True or result2==True or result3==True or result4==True: 
        print("In Ocean")     
    else:
        print("Land")

对于我来说，lon和lat变量位于我在问题程序中计算出的ISS的范围内。

Answer 3

Karin Todd的

global-land-mask非常易于使用和高效：

from global_land_mask import globe
print(globe.is_land(49.22, -2.23))
# → True
print(globe.is_land(49.22, -2.25))
# → False

可以通过pip使用它，它的唯一依赖项是numpy。

确定一个位置（给定坐标）是在陆地上还是在海洋上

3 个答案: