This gives the width and height in inches.
bbox = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
width, height = bbox.width, bbox.height
That probably suffices for your purpose, but to get pixels, you can multiply by fig.dpi
:
width *= fig.dpi
height *= fig.dpi
For example,
import matplotlib.pyplot as plt
def get_ax_size(ax):
bbox = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
width, height = bbox.width, bbox.height
width *= fig.dpi
height *= fig.dpi
return width, height
fig, ax = plt.subplots()
print(get_ax_size(ax))
#(496.0, 384.00000000000006)
ax2 = plt.axes([0.3, 0.3, 0.7, 0.7])
print(get_ax_size(ax2))
# (448.0, 336.0)
To make an image of exactly that figure size, you have to remove whitespace between the figure and the axis:
import numpy as np
import matplotlib.pyplot as plt
def get_ax_size(ax):
bbox = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
width, height = bbox.width, bbox.height
width *= fig.dpi
height *= fig.dpi
return width, height
data = np.arange(9).reshape((3, 3))
fig = plt.figure(figsize=(8,6), dpi=80)
ax = plt.Axes(fig, [0., 0., 1., 1.])
ax.set_axis_off()
fig.add_axes(ax)
ax.imshow(data, aspect='equal')
print(get_ax_size(ax))
# (640.0, 480.0)
plt.savefig('/tmp/test.png', dpi=80)
% identify /tmp/test.png
/tmp/test.png PNG 640x480 640x480+0+0 8-bit DirectClass 50.5KB 0.020u 0:00.020
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