在Cartopy中创建陆地/海洋面罩？

Question

我有一个网格数据数组以及相关的经/纬度，我想使用Cartopy查找每个特定的网格单元是在海洋还是陆地上。

我可以简单地从Cartopy Feature界面获取土地数据

land_10m = cartopy.feature.NaturalEarthFeature('physical', 'land', '10m')

然后我可以得到几何形状

all_geometries = [geometry for geometry in land_50m.geometries()]

但是我不知道如何使用这些几何图形来测试特定位置是否在陆地上。

这个问题似乎在Mask Ocean or Land from data using Cartopy之前出现了，解决方案几乎是“改为使用底图”，这有点不令人满意。

=========更新=======

感谢bjlittle，我有一个可行的解决方案，可以在下面生成图

from IPython import embed
import numpy as np
import matplotlib.pyplot as plt
import cartopy.crs as ccrs
import cartopy.io.shapereader as shpreader
from shapely.geometry import Point
import cartopy

land_10m = cartopy.feature.NaturalEarthFeature('physical', 'land', '10m')
land_polygons = list(land_10m.geometries())

lats = np.arange(48,58, 0.1)
lons = np.arange(-5,5, 0.1)
lon_grid, lat_grid = np.meshgrid(lons, lats)

points = [Point(point) for point in zip(lon_grid.ravel(), lat_grid.ravel())]

land = []
for land_polygon in land_polygons:
    land.extend([tuple(point.coords)[0] for point in filter(land_polygon.covers, points)])

fig = plt.figure(figsize=(8, 8))
ax = fig.add_axes([0, 0, 1, 1], projection=ccrs.PlateCarree())

ax.add_feature(land_10m, zorder=0, edgecolor='black',     facecolor=sns.xkcd_rgb['black'])

xs, ys = zip(*land)
ax.scatter(xs, ys, transform=ccrs.PlateCarree(),
       s=12, marker='s', c='red', alpha=0.5, zorder=2)

plt.show()

但是这非常慢，最终我将需要以更高的分辨率在全球范围内进行此操作。

有人对提高上述速度有更快的建议吗？

====更新2 ====

准备多边形会带来巨大的不同

将这两行相加可以使示例从40秒加速到0.3秒

from shapely.prepared import prep
land_polygons_prep = [prep(land_polygon) for land_polygon in land_polygons]

我以0.1度跑了一个半小时（即吃午饭）的示例现在运行了1.3秒：-）

Answer 1

为突出显示您可以采用的一种方法，我的回答基于出色的Cartopy Hurricane Katrina (2005)画廊示例，该示例绘制了卡特里娜飓风横扫海湾时的风暴轨迹墨西哥和美国上空。

通过基本上使用cartopy.io.shapereader和某些shapely predicates的简单组合，我们可以完成工作。我的示例有点冗长，但不要拖延……这并不可怕：

import matplotlib.pyplot as plt
import shapely.geometry as sgeom

import cartopy.crs as ccrs
import cartopy.io.shapereader as shpreader


def sample_data():
    """
    Return a list of latitudes and a list of longitudes (lons, lats)
    for Hurricane Katrina (2005).

    The data was originally sourced from the HURDAT2 dataset from AOML/NOAA:
    http://www.aoml.noaa.gov/hrd/hurdat/newhurdat-all.html on 14th Dec 2012.

    """
    lons = [-75.1, -75.7, -76.2, -76.5, -76.9, -77.7, -78.4, -79.0,
            -79.6, -80.1, -80.3, -81.3, -82.0, -82.6, -83.3, -84.0,
            -84.7, -85.3, -85.9, -86.7, -87.7, -88.6, -89.2, -89.6,
            -89.6, -89.6, -89.6, -89.6, -89.1, -88.6, -88.0, -87.0,
            -85.3, -82.9]

    lats = [23.1, 23.4, 23.8, 24.5, 25.4, 26.0, 26.1, 26.2, 26.2, 26.0,
            25.9, 25.4, 25.1, 24.9, 24.6, 24.4, 24.4, 24.5, 24.8, 25.2,
            25.7, 26.3, 27.2, 28.2, 29.3, 29.5, 30.2, 31.1, 32.6, 34.1,
            35.6, 37.0, 38.6, 40.1]

    return lons, lats


# create the figure and geoaxes
fig = plt.figure(figsize=(14, 12))
ax = fig.add_axes([0, 0, 1, 1], projection=ccrs.LambertConformal())
ax.set_extent([-125, -66.5, 20, 50], ccrs.Geodetic())

# load the shapefile geometries
shapename = 'admin_1_states_provinces_lakes_shp'
states_shp = shpreader.natural_earth(resolution='110m',
                                     category='cultural', name=shapename)
states = list(shpreader.Reader(states_shp).geometries())

# get the storm track lons and lats
lons, lats = sample_data()

# to get the effect of having just the states without a map "background"
# turn off the outline and background patches
ax.background_patch.set_visible(False)
ax.outline_patch.set_visible(False)

# turn the lons and lats into a shapely LineString
track = sgeom.LineString(zip(lons, lats))

# turn the lons and lats into shapely Points
points = [sgeom.Point(point) for point in zip(lons, lats)]

# determine the storm track points that fall on land
land = []
for state in states:
    land.extend([tuple(point.coords)[0] for point in filter(state.covers, points)])

# determine the storm track points that fall on sea
sea = set([tuple(point.coords)[0] for point in points]) - set(land)

# plot the state polygons
facecolor = [0.9375, 0.9375, 0.859375]    
for state in states:
    ax.add_geometries([state], ccrs.PlateCarree(),
                      facecolor=facecolor, edgecolor='black', zorder=1)

# plot the storm track land points 
xs, ys = zip(*land)
ax.scatter(xs, ys, transform=ccrs.PlateCarree(),
           s=100, marker='s', c='green', alpha=0.5, zorder=2)

# plot the storm track sea points
xs, ys = zip(*sea)
ax.scatter(xs, ys, transform=ccrs.PlateCarree(),
           s=100, marker='x', c='blue', alpha=0.5, zorder=2)

# plot the storm track
ax.add_geometries([track], ccrs.PlateCarree(),
                  facecolor='none', edgecolor='k')

plt.show()

上面的示例应生成以下plot，其中着重说明了如何使用匀称的几何形状来区分陆地和海洋点。

HTH

Answer 2

下面是使用上述解决方案解决此问题的代码示例：

from IPython import embed
import numpy as np
import matplotlib.pyplot as plt
import cartopy.crs as ccrs
import cartopy.io.shapereader as shpreader
from shapely.geometry import Point
import cartopy
from shapely.prepared import prep
import seaborn as sns


land_10m = cartopy.feature.NaturalEarthFeature('physical', 'land', '10m')
land_polygons = list(land_10m.geometries())

lats = np.arange(48,58, 0.1)
lons = np.arange(-5,5, 0.1)
lon_grid, lat_grid = np.meshgrid(lons, lats)

points = [Point(point) for point in zip(lon_grid.ravel(), lat_grid.ravel())]


land_polygons_prep = [prep(land_polygon) for land_polygon in land_polygons]


land = []
for land_polygon in land_polygons_prep:
    land.extend([tuple(point.coords)[0] for point in filter(land_polygon.covers, points)])

fig = plt.figure(figsize=(8, 8))
ax = fig.add_axes([0, 0, 1, 1], projection=ccrs.PlateCarree())

ax.add_feature(land_10m, zorder=0, edgecolor='black', facecolor=sns.xkcd_rgb['black'])

xs, ys = zip(*land)
ax.scatter(xs, ys, transform=ccrs.PlateCarree(),
       s=12, marker='s', c='red', alpha=0.5, zorder=2)