你可以繪製羅賓遜預測的流線嗎？

Question

我試圖用羅賓遜投影繪製全球地圖上的流線，但底圖似乎不像預計的座標。當然，對於一個普通的舊圓柱投影，它在x方向上是規則的。

下面是一個例子：

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

u = np.ones((21,21)) 
v = np.ones((21,21)) 
lats = np.arange(-90,91,9) 
lons = np.arange(-180,181,18) 
x,y = np.meshgrid(lons,lats) 

# It works for Cylindrical 
mp = Basemap(projection='cyl') 
xx,yy = mp(x,y) 
mp.streamplot(xx,yy,u,v) 
mp.drawcoastlines() 
plt.show() 

# But not Robinson 
mp = Basemap(projection='robin',lon_0=0) 
xx, yy = mp(x, y) 
mp.streamplot(xx,yy,u,v) 
mp.drawcoastlines() 
plt.show() 

它抱怨的X座標，養：

ValueError: The rows of 'x' must be equal 

所以是有可能的情節簡化羅賓遜預測？

Answer 1

使用命令xx,yy = mp(x,y)根據特定投影的座標轉換應用於您的lon和lats。對於大多數投影來說，這將導致網格點的變形，使得x行不再相等，因此錯誤：ValueError: The rows of 'x' must be equal。要解決這個問題，您需要重新格網化數據，例如像這樣：

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

u = np.ones((21,21)) 
v = np.ones((21,21)) 
lats = np.arange(-90,91,9) 
lons = np.arange(-180,181,18) 
x,y = np.meshgrid(lons,lats) 

mp = Basemap(projection='robin',lon_0=0) 
xx, yy = mp(x, y) 

# generate a grid that is equally spaced in a plot with the current pojection 
lons, lats, xxnew, yynew = mp.makegrid(21,21, returnxy=True) 

# project the data onto the new grid 
unew = plt.mlab.griddata(xx.flatten(), yy.flatten(),u.flatten(), xxnew, yynew ,interp='linear') 
vnew = plt.mlab.griddata(xx.flatten(), yy.flatten(),v.flatten(), xxnew, yynew ,interp='linear') 

mp.streamplot(xxnew,yynew,unew,vnew) 
mp.drawcoastlines() 
plt.show()