Problem

I was trying to make a bar chart with rounded corners, which necessitates using FancyBBoxPatch. However, I would have expected rounded corners to be isotropic by default.

Example showing default behavior and expected behavior (bar on the right):

import matplotlib.pyplot as plt
from matplotlib.patches import FancyBboxPatch, BoxStyle

def add_bar(ax, x, y, w, h, mutation_aspect=1):
    patch = FancyBboxPatch(
        (x, y),
        w,
        h,
        boxstyle=BoxStyle("Round", pad=0, rounding_size=w / 2),
        linewidth=0,
        facecolor="C0",
        mutation_aspect=mutation_aspect,
    )
    ax.add_patch(patch)


fig, ax = plt.subplots(figsize=(10, 5))


x, y, w, h = 0.3, 0, 0.18, 1

# Left: default mutation_aspect=1 (shows issue)
add_bar(ax, x, y, w, h)

# Right: using 1/ax._get_aspect_ratio()
add_bar(ax, x + 0.3, y, w, h, 1 / ax._get_aspect_ratio())

plt.tight_layout()
plt.savefig("reproduce.png", dpi=220)
plt.show()

which gave:

The solution uses private API (ax._get_aspect_ratio()) because ax.get_aspect() returns "auto".

I'm guessing changing the default behavior is not possible for backward compatibility, so my question is: Would an opt-in auto mode for mutation_aspect be acceptable to ensure isotropic rounding?

The same question applies to other BoxStyle types.

Proposed solution

An opt-in auto mode for mutation_aspect (for example passing mutation_aspect="auto") which changes default behavior and sets the mutation_aspect value to 1 / ax._get_aspect_ratio() internally in the FancyBboxPatch class