在源代码的圆等级中，用于半径的测量单元是什么:matplotlib.patches

Question

在源代码中查看circle类：https：//matplotlib.org/3.1.1/_modules/matplotlib/patches.html#circle.set_radius

我试图识别半径是英里，m还是公里。我该如何检查。

圆圈已经被绘制了，我试图手动估计查看圆圈，但不要认为这是验证的好方法

class Circle(Ellipse):
    """
    A circle patch.
    """
    def __str__(self):
        pars = self.center[0], self.center[1], self.radius
        fmt = "Circle(xy=(%g, %g), radius=%g)"
        return fmt % pars
    @docstring.dedent_interpd
    def __init__(self, xy, radius=5, **kwargs):
        """
        Create true circle at center *xy* = (*x*, *y*) with given
        *radius*.  Unlike :class:`~matplotlib.patches.CirclePolygon`
        which is a polygonal approximation, this uses Bezier splines
        and is much closer to a scale-free circle.
        Valid kwargs are:
        %(Patch)s
        """
        Ellipse.__init__(self, xy, radius * 2, radius * 2, **kwargs)
        self.radius = radius
[docs]    def set_radius(self, radius):
        """
        Set the radius of the circle
        Parameters
        ----------
        radius : float
        """
        self.width = self.height = 2 * radius
        self.stale = True

[docs]    def get_radius(self):
        """
        Return the radius of the circle
        """
        return self.width / 2.

    radius = property(get_radius, set_radius)

预期：半径应为英里实际：尚不清楚用于半径的单位测量

Answer 1

如果您塑造了实现的塑造，例如从两个点获得距离，它将为您提供两个分之间的距离，而您可以使用半径。

import math
p1 = [-118.22191509999999, 34.0431494]
p2 = [-118.13458169630128, 34.04311852494086]
distance = math.sqrt( ((p1[0]-p2[0])**2)+((p1[1]-p2[1])**2) )
self.ax1.add_patch(Circle((-118.22191509999999, 34.0431494), 0.08733340915635142, fill=False))

0.0873340915635142是您的距离和半径。