问题
The following python code should plot r(theta) = theta on the range [-pi/2, pi/2].
import matplotlib.pyplot as plt
import numpy
theta = numpy.linspace(-numpy.pi / 2, numpy.pi / 2, 64 + 1)
r = theta
plt.polar(theta, r)
plt.savefig('polar.png')
This produces the plot:
However, I would expect it to produce:
The negative values of r(theta) seem to be clipped. How do I make it so that matplotlib plots the negative values of r(theta)?
回答1:
The first plot seems correct. It just doesn't show the negative values. This can be overcome by explicitely setting the limits of the r axes.
import matplotlib.pyplot as plt
import numpy
theta = numpy.linspace(-numpy.pi / 2, numpy.pi / 2, 64 + 1)
r = theta
plt.polar(theta, r)
plt.ylim(theta.min(),theta.max())
plt.yticks([-1, 0,1])
plt.show()
This behaviour is based on the assumption that any quantity should be plottable on a polar graph, which might be beneficial for technical questions on relative quantities. E.g. one might ask about the deviation of a quantity in a periodic system from its mean value. In this case the convention used by matplotlib is ideally suited.
From a more mathematical (theoretical) perspective one might argue that negative radii are a point reflection on the origin. In order to replicate this behaviour, one needs to rotate the points of negative r values by π. The expected graph from the question can thus be reproduced by the following code
import matplotlib.pyplot as plt
import numpy as np
theta = np.linspace(-np.pi / 2, np.pi / 2, 64 + 1)
r = theta
plt.polar(theta+(r<0)*np.pi, np.abs(r))
plt.show()
来源：https://stackoverflow.com/questions/42982290/polar-plot-of-a-function-with-negative-radii-using-matplotlib