Matplotlib - Drawing a smooth circle in a polar plot

萝らか妹 提交于 2019-11-28 11:18:31

You can set transform argument of the Circle:

%matplotlib inline
import pylab as pl
import numpy as np

N = 100
theta = np.random.rand(N)*np.pi*2
r = np.cos(theta*2) + np.random.randn(N)*0.1

ax = pl.subplot(111, polar=True)
ax.scatter(theta, r)
circle = pl.Circle((0.5, 0.3), 0.2, transform=ax.transData._b, color="red", alpha=0.4)
ax.add_artist(circle)

output:

or transform=ax.transProjectionAffine + ax.transAxes if you don't like using the private attribute.

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