central longitude for NorthPolarStereo

Question

I'd like to plot a polar stereographic plot of the Northern Hemisphere with 180 at the bottom of the plot so I can emphasize the Pacific region. I'm using the latest cartopy from git, and can make a polar stereographic plot no problem, but I can't work out how to change which longitude is at the bottom of the plot. I tried setting the longitude extent to [-180, 180] but this doesn't help, and the NorthPolarStereo() doesn't accept any keyword arguments like central_longitude. Is this possible currently?

Answer 1

This feature has now been implemented in Cartopy (v0.6.x). The following example produces two subplots in Northern Hemisphere polar stereographic projections, one with the default settings and one with the central longitude changed:

"""Stereographic plot with adjusted central longitude."""
import matplotlib.pyplot as plt
import cartopy.crs as ccrs
from cartopy.examples.waves import sample_data


# read sample data
x, y, z = sample_data(shape=(73, 145))

fig = plt.figure(figsize=(8, 4))

# first plot with default settings
ax1 = fig.add_subplot(121, projection=ccrs.NorthPolarStereo())
cs1 = ax1.contourf(x, y, z, 50, transform=ccrs.PlateCarree(),
                   cmap='gist_ncar')
ax1.set_extent([0, 360, 0, 90], crs=ccrs.PlateCarree())
ax1.coastlines()
ax1.set_title('Centred on 0$^\circ$ (default)')

# second plot with 90W at the bottom of the plot
ax2 = fig.add_subplot(
    122, projection=ccrs.NorthPolarStereo(central_longitude=-90))
cs2 = ax2.contourf(x, y, z, 50, transform=ccrs.PlateCarree(),
                   cmap='gist_ncar')
ax2.set_extent([0, 360, 0, 90], crs=ccrs.PlateCarree())
ax2.coastlines()
ax2.set_title('Centred on 90$^\circ$W')

plt.show()

The output of this script is:

Answer 2

For Cartopy 0.17 and matplotlib 3.1.1 (Python 3.7), I got an error in set_extent() with the above solution.

It seems that set_extent() only works this way:

ax1.set_extent([-180, 180, 0, 90], crs=ccrs.PlateCarree())

ax2.set_extent([-179, 179, 0, 90], crs=ccrs.PlateCarree())

So, the rotated image needs some weird longitude boundaries..