真正的Python散布函数

我写这篇文章是为了在用散点图绘制圆时有一个参考帖子，但用户想要提供分散圆的真实半径，而不是抽象的大小.

我一直在四处寻找，还有其他帖子解释了它的原理，但还没有遇到现成的功能.我已经试过了，希望我已经接近找到解决方案.

这就是我到目前为止所拥有的:

import matplotlib.pyplot as plt
import numpy as np

def true_scatter(x, y, r, ax, **kwargs):
    # Should work for an equal aspect axis
    ax.set_aspect('equal')
    
    # Access the DPI and figure size
    dpi = ax.figure.dpi
    fig_width_inch, _ = ax.figure.get_size_inches()

    # Calculate plot size in data units
    xlim = ax.get_xlim()
    plot_width_data_units = xlim[1] - xlim[0]

    # Calculate the scale factor: pixels per data unit
    plot_width_pixels = fig_width_inch * dpi
    scale = plot_width_pixels / plot_width_data_units

    # Convert radius to pixels, then area to points squared
    radius_pixels = r * scale
    area_pixels_squared = np.pi * (radius_pixels ** 2)
    area_points_squared = area_pixels_squared * (72 / dpi) ** 2

    # Scatter plot with converted area
    scatter = ax.scatter(x, y, s=area_points_squared, **kwargs)
    return scatter

# Example with single scatter
fig, ax = plt.subplots()
ax.set_xlim(-2, 2)
ax.set_ylim(-2, 2)
scatter = true_scatter(0, 0, 1, ax)
ax.set_xlim(-2, 2)
ax.set_ylim(-2, 2)
plt.grid()
plt.show()

Wrong single scatter plot个

不幸的是，这并不完全是答案.我得到一个半径为~1.55的圆，而不是1. 有没有人能发现我的方法有什么问题？

谢谢!