Python 绘制显示圆的等式
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Plot equation showing a circle
提问by Elyakim
The following formula is used to classify points from a 2-dimensional space:
以下公式用于对二维空间中的点进行分类：
f(x1,x2) = np.sign(x1^2+x2^2-.6)
All points are in space X = [-1,1] x [-1,1]with a uniform probability of picking each x.
所有点都在空间中X = [-1,1] x [-1,1]，以均匀的概率选择每个 x。
Now I would like to visualize the circle that equals:
现在我想形象化等于的圆圈：
0 = x1^2+x2^2-.6
The values of x1 should be on the x-axis and values of x2 on the y-axis.
x1 的值应该在 x 轴上，x2 的值应该在 y 轴上。
It must be possible but I have difficulty transforming the equation to a plot.
这一定是可能的，但我很难将方程转换为绘图。
采纳答案by Bas Jansen
You can use a contour plot, as follows (based on the examples at http://matplotlib.org/examples/pylab_examples/contour_demo.html):
您可以使用等高线图，如下所示（基于http://matplotlib.org/examples/pylab_examples/contour_demo.html上的示例）：
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-1.0, 1.0, 100)
y = np.linspace(-1.0, 1.0, 100)
X, Y = np.meshgrid(x,y)
F = X**2 + Y**2 - 0.6
plt.contour(X,Y,F,[0])
plt.show()
This yields the following graph
这产生了下图
Lastly, some general statements:
最后，一些一般性声明：
x^2does not mean what you thinkit does in python, you have to usex**2.x1andx2are terribly misleading (to me), especially if you state thatx2has to be on the y-axis.- (Thanks to Dux) You can add
plt.gca().set_aspect('equal')to make the figure actually look circular, by making the axis equal.
x^2并不意味着你认为它在 python 中的作用，你必须使用x**2.x1并且x2非常具有误导性（对我而言），特别是如果您声明x2必须在 y 轴上。- （感谢 Dux）您可以
plt.gca().set_aspect('equal')通过使轴相等来添加使图形实际上看起来是圆形的。
回答by Dux
How about drawing x-values and calculating the corresponding y-values?
如何绘制 x 值并计算相应的 y 值？
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-1, 1, 100, endpoint=True)
y = np.sqrt(-x**2. + 0.6)
plt.plot(x, y)
plt.plot(x, -y)
produces
产生
This can obviously be made much nicer, but this is only for demonstration...
这显然可以做得更好，但这仅用于演示......
回答by hitzg
The solution of @BasJansen certainly gets you there, it's either very inefficient (if you use many grid points) or inaccurate (if you use only few grid points).
@BasJansen 的解决方案肯定会让你到达那里，它要么非常低效（如果你使用很多网格点）要么不准确（如果你只使用很少的网格点）。
You can easily draw the circle directly. Given 0 = x1**2 + x**2 - 0.6it follows that x2 = sqrt(0.6 - x1**2)(as Dux stated).
您可以轻松地直接绘制圆圈。鉴于0 = x1**2 + x**2 - 0.6它如下x2 = sqrt(0.6 - x1**2)（如 Dux 所述）。
But what you really want to do is to transform your cartesian coordinates to polar ones.
但是您真正想做的是将笛卡尔坐标转换为极坐标。
x1 = r*cos(theta)
x2 = r*sin(theta)
if you use these substitions in the circle equation you will see that r=sqrt(0.6).
如果你在圆方程中使用这些替换，你会看到r=sqrt(0.6)。
So now you can use that for your plot:
所以现在你可以将它用于你的情节：
import numpy as np
import matplotlib.pyplot as plt
# theta goes from 0 to 2pi
theta = np.linspace(0, 2*np.pi, 100)
# the radius of the circle
r = np.sqrt(0.6)
# compute x1 and x2
x1 = r*np.cos(theta)
x2 = r*np.sin(theta)
# create the figure
fig, ax = plt.subplots(1)
ax.plot(x1, x2)
ax.set_aspect(1)
plt.show()
Result:
结果：
回答by chen_767
# x**2 + y**2 = r**2
r = 6
x = np.linspace(-r,r,1000)
y = np.sqrt(-x**2+r**2)
plt.plot(x, y,'b')
plt.plot(x,-y,'b')
plt.gca().set_aspect('equal')
plt.show()
produces:
产生：
