How do I change matplotlib's subplot projection of an existing axis?

PythonMatplotlibSubplotCartopy

Python Problem Overview


I'm trying to construct a simple function that takes a subplot instance (matplotlib.axes._subplots.AxesSubplot) and transforms its projection to another projection, for example, to one of the cartopy.crs.CRS projections.

The idea looks something like this

import cartopy.crs as ccrs
import matplotlib.pyplot as plt

def make_ax_map(ax, projection=ccrs.PlateCarree()):
    # set ax projection to the specified projection
    ...
    # other fancy formatting
    ax2.coastlines()
    ...

# Create a grid of plots
fig, (ax1, ax2) = plt.subplots(ncols=2)
# the first subplot remains unchanged
ax1.plot(np.random.rand(10))
# the second one gets another projection
make_ax_map(ax2)

Of course, I can just use fig.add_subplot() function:

fig = plt.figure(figsize=(10,5))
ax1 = fig.add_subplot(121)
ax1.plot(np.random.rand(10))

ax2 = fig.add_subplot(122,projection=ccrs.PlateCarree())
ax2.coastlines()

but I was wondering if there is a proper matplotlib method to change a subplot axis projection after it was defined. Reading matplotlib API didn't help unfortunately.

Python Solutions


Solution 1 - Python

You can't change the projection of an existing axes, the reason is given below. However the solution to your underlying problem is simply to use the subplot_kw argument to plt.subplots() described in the matplotlib documentation here. For example, if you wanted all your subplots to have the cartopy.crs.PlateCarree projection you could do

import matplotlib.pyplot as plt
import cartopy.crs as ccrs

# Create a grid of plots
fig, (ax1, ax2) = plt.subplots(ncols=2, subplot_kw={'projection': ccrs.PlateCarree()})

Regarding the actual question, specifying a projection when you create an axes set determines the axes class you get, which is different for each projection type. For example

import matplotlib.pyplot as plt
import cartopy.crs as ccrs

ax1 = plt.subplot(311)
ax2 = plt.subplot(312, projection='polar')
ax3 = plt.subplot(313, projection=ccrs.PlateCarree())

print(type(ax1))
print(type(ax2))
print(type(ax3))

This code will print the following

<class 'matplotlib.axes._subplots.AxesSubplot'>
<class 'matplotlib.axes._subplots.PolarAxesSubplot'>
<class 'cartopy.mpl.geoaxes.GeoAxesSubplot'>

Notice how each axes is actually an instance of a different class.

Solution 2 - Python

Assuming there are multiple axes being used for 2D plotting, like...

fig = matplotlib.pyplot.Figure()
axs = fig.subplots(3, 4) # prepare for multiple subplots
# (some plotting here)
axs[0,0].plot([1,2,3])

... one can simply destroy one of them and replace it with a new one having the 3D projection:

axs[2,3].remove()
ax = fig.add_subplot(3, 4, 12, projection='3d')
ax.plot_surface(...)

Just note that unlike rest of Python, the add_subplot uses row-column indexing starting from 1 (not from 0).

EDIT: Changed my typo about indexing.

Solution 3 - Python

following the answer to this question :

> https://stackoverflow.com/questions/44090563/in-python-how-can-i-inherit-and-override-a-method-on-a-class-instance-assignin

I found a hack to change the projection of an axe after creating it which seems to work at least in the simple example below, but I have no idea if this solution is the best way

from matplotlib.axes import Axes
from matplotlib.projections import register_projection

class CustomAxe(Axes):
    name = 'customaxe'

    def plotko(self, x):
        self.plot(x, 'ko')
        self.set_title('CustomAxe')

register_projection(CustomAxe)


if __name__ == '__main__':
    import matplotlib.pyplot as plt

    fig = plt.figure()

    ## use this syntax to create a customaxe directly
    # ax = fig.add_subplot(111, projection="customaxe")

    ## change the projection after creation
    ax = plt.gca()
    ax.__class__ = CustomAxe

    ax.plotko(range(10))    
    plt.show()

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionDenis SergeevView Question on Stackoverflow
Solution 1 - PythonajdawsonView Answer on Stackoverflow
Solution 2 - PythondominecfView Answer on Stackoverflow
Solution 3 - Pythonuser2660966View Answer on Stackoverflow