CODEWITHSUNDEEP
pythonnumpymatplotlibplotequation
Elyakim
Elyakim

Reputation: 521

Plot equation showing a circle

The following formula is used to classify points from a 2-dimensional space:

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.

Now I would like to visualize the circle that equals:

0 = x1^2+x2^2-.6

The values of x1 should be on the x-axis and values of x2 on the y-axis.

It must be possible but I have difficulty transforming the equation to a plot.

Upvotes: 16

Views: 54104

Answers (5)

Amjad
Amjad

Reputation: 3678

Plotting a circle using complex numbers

The idea: multiplying a point by complex exponential (enter image description here) rotates the point on a circle

import numpy as np
import matplotlib.pyplot as plt

num_pts=20 # number of points on the circle
ps = np.arange(num_pts+1)
# j = np.sqrt(-1)
pts = np.exp(2j*np.pi/num_pts *(ps))

fig, ax = plt.subplots(1)
ax.plot(pts.real, pts.imag , '-o')
ax.set_aspect(1)
plt.show()

enter image description here

Upvotes: 5

chen_767
chen_767

Reputation: 365

# 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:

enter image description here

Upvotes: 5

hitzg
hitzg

Reputation: 12691

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).

You can easily draw the circle directly. Given 0 = x1**2 + x**2 - 0.6 it follows that x2 = sqrt(0.6 - x1**2) (as Dux stated).

But what you really want to do is to transform your cartesian coordinates to polar ones.

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).

So now you can use that for your plot:

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:

enter image description here

Upvotes: 16

Bas Jansen
Bas Jansen

Reputation: 3343

You can use a contour plot, as follows (based on the examples at http://matplotlib.org/examples/pylab_examples/contour_demo.html):

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

enter image description here

Lastly, some general statements:

  1. x^2 does not mean what you think it does in python, you have to use x**2.
  2. x1 and x2 are terribly misleading (to me), especially if you state that x2 has to be on the y-axis.
  3. (Thanks to Dux) You can add plt.gca().set_aspect('equal') to make the figure actually look circular, by making the axis equal.

Upvotes: 19

Dux
Dux

Reputation: 1246

How about drawing x-values and calculating the corresponding y-values?

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

enter image description here

This can obviously be made much nicer, but this is only for demonstration...

Upvotes: 5

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