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我正在使用 matplotlib 中的底图绘制地图。数据遍布全球,但我只想把大陆上的数据全部保留下来,丢到海里。有没有办法可以过滤数据,或者有没有办法再次绘制海洋来覆盖数据?

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5 回答 5

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matplotlib.basemap 中有方法: is_land(xpt, ypt)

True如果给定的 x,y 点(在投影坐标中)在陆地上,则返回,False否则返回。土地的定义基于与类实例关联的 GSHHS 海岸线多边形。陆地区域内湖泊上方的点不计为陆地点。

有关详细信息,请参阅此处

于 2012-12-10T07:43:14.447 回答
6

is_land()将循环所有多边形以检查它是否着陆。对于大数据量,它非常慢。您可以使用points_inside_poly()matplotlib 快速检查点数组。这是代码。它不检查lakepolygons,如果你想删除湖中的点,你可以添加你自己。

在我的电脑上检查 100000 个点需要 2.7 秒。如果您想要更快的速度,您可以将多边形转换为位图,但这样做有点困难。请告诉我以下代码是否对您的数据集不够快。

from mpl_toolkits.basemap import Basemap
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.nxutils as nx

def points_in_polys(points, polys):
    result = []
    for poly in polys:
        mask = nx.points_inside_poly(points, poly)
        result.extend(points[mask])
        points = points[~mask]
    return np.array(result)

points = np.random.randint(0, 90, size=(100000, 2))
m = Basemap(projection='moll',lon_0=0,resolution='c')
m.drawcoastlines()
m.fillcontinents(color='coral',lake_color='aqua')
x, y = m(points[:,0], points[:,1])
loc = np.c_[x, y]
polys = [p.boundary for p in m.landpolygons]
land_loc = points_in_polys(loc, polys)
m.plot(land_loc[:, 0], land_loc[:, 1],'ro')
plt.show()
于 2012-12-11T00:37:41.073 回答
5

HYRY 的答案不适用于新版本的 matplotlib(不推荐使用 nxutils)。我制作了一个有效的新版本:

from mpl_toolkits.basemap import Basemap
import matplotlib.pyplot as plt
from matplotlib.path import Path
import numpy as np

map = Basemap(projection='cyl', resolution='c')

lons = [0., 0., 16., 76.]
lats = [0., 41., 19., 51.]

x, y = map(lons, lats)

locations = np.c_[x, y]

polygons = [Path(p.boundary) for p in map.landpolygons]

result = np.zeros(len(locations), dtype=bool) 

for polygon in polygons:

    result += np.array(polygon.contains_points(locations))

print result
于 2014-11-06T12:44:26.620 回答
5

最简单的方法是使用底图的maskoceans

如果对于每个 lat, lon 你有一个数据并且你想使用轮廓:在 meshgrid 和插值之后:

from scipy.interpolate import griddata as gd
from mpl_toolkits.basemap import Basemap, cm, maskoceans
xi, yi = np.meshgrid(xi, yi)
zi = gd((mlon, mlat),
            scores,
            (xi, yi),
            method=grid_interpolation_method)
#mask points on ocean
data = maskoceans(xi, yi, zi)
con = m.contourf(xi, yi, data, cmap=cm.GMT_red2green)
#note instead of zi we have data now.

更新(比 in_land 或 in_polygon 解决方案快得多):

如果对于每个 lat, lon 您没有任何数据,并且您只想将点分散在陆地上:

x, y = m(lons, lats)
samples = len(lons)
ocean = maskoceans(lons, lats, datain=np.arange(samples),
                   resolution='i')
ocean_samples = np.ma.count_masked(ocean)
print('{0} of {1} points in ocean'.format(ocean_samples, samples))
m.scatter(x[~ocean.mask], y[~ocean.mask], marker='.', color=colors[~ocean.mask], s=1)
m.drawcountries()
m.drawcoastlines(linewidth=0.7)
plt.savefig('a.png')
于 2017-03-05T05:19:30.613 回答
4

我正在回答这个问题,当我被告知最好在这里发布我的答案时。基本上,我的解决方案提取了用于绘制Basemap实例海岸线的多边形,并将这些多边形与地图的轮廓相结合,以生成matplotlib.PathPatch覆盖地图海洋区域的多边形。

如果数据是粗略的并且不需要数据的插值,这尤其有用。在这种情况下,使用maskoceans会产生非常粗糙的海岸线轮廓,看起来不太好。

这是我作为另一个问题的答案发布的相同示例:

from matplotlib import pyplot as plt
from mpl_toolkits import basemap as bm
from matplotlib import colors
import numpy as np
import numpy.ma as ma
from matplotlib.patches import Path, PathPatch

fig, ax = plt.subplots()

lon_0 = 319
lat_0 = 72

##some fake data
lons = np.linspace(lon_0-60,lon_0+60,10)
lats = np.linspace(lat_0-15,lat_0+15,5)
lon, lat = np.meshgrid(lons,lats)
TOPO = np.sin(np.pi*lon/180)*np.exp(lat/90)

m = bm.Basemap(resolution='i',projection='laea', width=1500000, height=2900000, lat_ts=60, lat_0=lat_0, lon_0=lon_0, ax = ax)
m.drawcoastlines(linewidth=0.5)

x,y = m(lon,lat)
pcol = ax.pcolormesh(x,y,TOPO)

##getting the limits of the map:
x0,x1 = ax.get_xlim()
y0,y1 = ax.get_ylim()
map_edges = np.array([[x0,y0],[x1,y0],[x1,y1],[x0,y1]])

##getting all polygons used to draw the coastlines of the map
polys = [p.boundary for p in m.landpolygons]

##combining with map edges
polys = [map_edges]+polys[:]

##creating a PathPatch
codes = [
    [Path.MOVETO] + [Path.LINETO for p in p[1:]]
    for p in polys
]
polys_lin = [v for p in polys for v in p]
codes_lin = [c for cs in codes for c in cs]
path = Path(polys_lin, codes_lin)
patch = PathPatch(path,facecolor='white', lw=0)

##masking the data:
ax.add_patch(patch)

plt.show()

这将产生以下图:

在此处输入图像描述

希望这对某人有帮助:)

于 2018-02-05T15:36:23.627 回答