我正在使用matplotlib的底图绘制地图。这些数据遍布全球,但我只想保留非洲大陆上的所有数据并将其丢弃在海洋上。有没有办法可以过滤数据,还是有办法再次绘制海洋以覆盖数据?
答案 0 :(得分:6)
matplotlib.basemap中的方法:is_land(xpt, ypt)
如果给定的x,y点(在投影坐标中)超过陆地,则返回True
,否则返回False
。土地的定义基于与类实例相关的GSHHS海岸线多边形。陆地内湖泊点不计入陆地点。
有关详细信息,请参阅here。
答案 1 :(得分:6)
is_land()
将循环所有多边形以检查它是否为陆地。对于大数据量,它非常慢。您可以使用matplotlib中的points_inside_poly()
快速检查点阵列。这是代码。它不会检查lakepolygons
,如果你想在湖泊中删除点,你可以添加你自己。
在我的电脑上检查100000点需要2.7秒。如果您想要更快的速度,可以将多边形转换为位图,但这样做有点困难。如果您的数据集没有快速提供以下代码,请告诉我。
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()
答案 2 :(得分:4)
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
答案 3 :(得分:2)
最简单的方法是使用底图' s maskoceans。
如果对于每个lat,lon你有一个数据,你想要 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):
如果对于每个纬度而言,您没有任何数据,而您只想将这些点分散到陆地上:
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')
答案 4 :(得分:2)
当我被告知在这里发布我的答案会更好时,我正在回答this question。基本上,我的解决方案提取用于绘制Basemap
实例的海岸线的多边形,并将这些多边形与地图轮廓组合以生成覆盖地图海洋区域的matplotlib.PathPatch
。
如果数据很粗糙并且不需要数据插值,这尤其有用。在这种情况下,使用maskoceans
会产生一个非常粗糙的海岸线轮廓,看起来不太好。
以下是我发布的另一个问题的答案:
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()
这会产生以下情节:
希望这对某人有帮助:))