真正的Python散点函数

Question

我写这篇文章是为了在用散点图绘制圆时提供参考帖子，但用户希望为散点圆提供真实的半径而不是抽象的尺寸。

我一直在四处寻找，还有其他帖子解释了它的理论，但还没有遇到现成的功能。我已经尝试过自己，希望我很快就能找到解决方案。

这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()

不幸的是，这并不是答案。我得到一个半径约为 1.55 而不是 1 的圆。 有人能发现我的方法有什么问题吗？

谢谢！

Answer 1

使用这个答案，我能够更新您的功能以正常工作。请注意以下更改：

• 您应该在获取尺寸之前绘制轴，因为这些值尚未初始化（我不确定该解释是否完全正确，但至少知道在绘制之前，这些值不是图形之后的值已绘制）。
• 您应该使用轴宽度而不是图形宽度，因为您是在轴上绘图。
• 您需要将数据单位的绘图宽度加 1。我不太清楚为什么......
• 圆的面积为

  pi/2*r^2

  （你忘了 1/2）。
• 边缘颜色会使标记稍大。

import matplotlib.pyplot as plt
import numpy as np

plt.close("all")

def true_scatter(x, y, r, ax, **kwargs):
    # Should work for an equal aspect axis
    ax.set_aspect("equal")
    fig = ax.figure
    fig.canvas.draw()   # need to draw before getting the dimensions
    
    # Access the DPI and axes size
    dpi = fig.dpi
    ax_width = ax.get_window_extent().width

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

    # Calculate the scale factor: pixels per data unit
    # 72/dpi because that is the conversion factor from matplotlib
    scale = ax_width/plot_width_data_units*(72/dpi)

    # Convert radius to pixels, then area to points squared
    radius_pixels = r*scale
    s = 0.5*np.pi*radius_pixels**2
    
    # the edge color makes the point look slightly larger than it is
    kwargs["edgecolor"] = kwargs.get("edgecolor", "none")

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

正如链接答案中提到的，调整图形大小或放大会破坏缩放比例。