在底图中填充海洋 [英] Fill oceans in basemap

问题描述

我正在尝试在matplotlib.Basemap上绘制1x1度数据，并且我想用白色填充海洋.但是，为了使海洋的边界遵循matplotlib绘制的海岸线，白海面具的分辨率应该比我的数据分辨率高得多.

I am trying to plot 1x1 degree data on a matplotlib.Basemap, and I want to fill the ocean with white. However, in order for the boundaries of the ocean to follow the coastlines drawn by matplotlib, the resolution of the white ocean mask should be much higher than the resolution of my data.

长时间搜索后，我尝试了两种可能的解决方案:

After searching around for a long time I tried the two possible solutions:

(1)maskoceans()和is_land()函数，但是由于我的数据分辨率低于底图绘制的地图，因此边缘看起来不佳.我也不想将数据插值到更高的分辨率.

(1) maskoceans() and is_land() functions, but since my data is lower resolution than the map drawn by basemap it does not look good on the edges. I do not want to interpolate my data to higher resolution either.

(2)m.drawlsmask()，但是由于无法指定zorder，因此pcolormesh图始终覆盖蒙版.

(2) m.drawlsmask(), but since zorder cannot be assigned the pcolormesh plot always overlays the mask.

此代码

import numpy as np
import matplotlib.pyplot as plt
import mpl_toolkits.basemap as bm

#Make data
lon = np.arange(0,360,1)
lat = np.arange(-90,91,1)
data = np.random.rand(len(lat),len(lon))

#Draw map
plt.figure()
m = bm.Basemap(resolution='i',projection='laea', width=1500000, height=2900000, lat_ts=60, lat_0=72, lon_0=319)
m.drawcoastlines(linewidth=1, color='white')
data, lon = bm.addcyclic(data,lon)
x,y = m(*np.meshgrid(lon,lat))
plt.pcolormesh(x,y,data)
plt.savefig('1.png',dpi=300)

产生此图像:

添加m.fillcontinents(color='white')会产生以下图像，这是我需要的，但要填充海洋而不是土地.

Adding m.fillcontinents(color='white') produces the following image, which is what I need but to fill the ocean and not the land.

修改:

m.drawmapboundary(fill_color='lightblue')也会填满土地，因此无法使用.

m.drawmapboundary(fill_color='lightblue') also fills over land and can therefore not be used.

理想的结果是海洋是白色的，而我用plt.pcolormesh(x,y,data)绘制的图则显示在陆地上.

The desired outcome is that the oceans are white, while what I plotted with plt.pcolormesh(x,y,data) shows up over the lands.

推荐答案

我找到了一个更好的解决方案，该问题使用地图中海岸线定义的多边形生成覆盖海洋区域的matplotlib.PathPatch.该解决方案具有更好的分辨率，并且速度更快:

I found a much nicer solution to the problem which uses the polygons defined by the coastlines in the map to produce a matplotlib.PathPatch that overlays the ocean areas. This solution has a much better resolution and is much faster:

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()

输出看起来像这样:

原始解决方案:

您可以在basemap.maskoceans中使用分辨率更高的数组，以使分辨率适合大陆轮廓.然后，您只需反转遮罩并将遮罩的阵列绘制在数据之上即可.

You can use an array with greater resolution in basemap.maskoceans, such that the resolution fits the continent outlines. Afterwards, you can just invert the mask and plot the masked array on top of your data.

以某种方式，当我使用整个地图范围时(例如，经度从-180到180，纬度从-90到90)，只有basemap.maskoceans可以工作.鉴于需要相当高的分辨率才能使其看起来不错，因此计算需要一段时间:

Somehow I only got basemap.maskoceans to work when I used the full range of the map (e.g. longitudes from -180 to 180 and latitudes from -90 to 90). Given that one needs quite a high resolution to make it look nice, the computation takes a while:

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

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)


##producing a mask -- seems to only work with full coordinate limits
lons2 = np.linspace(-180,180,10000)
lats2 = np.linspace(-90,90,5000)
lon2, lat2 = np.meshgrid(lons2,lats2)
x2,y2 = m(lon2,lat2)
pseudo_data = np.ones_like(lon2)
masked = bm.maskoceans(lon2,lat2,pseudo_data)
masked.mask = ~masked.mask

##plotting the mask
cmap = colors.ListedColormap(['w'])
pcol = ax.pcolormesh(x2,y2,masked, cmap=cmap)

plt.show()

结果如下:

这篇关于在底图中填充海洋的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！