Is it possible to plot implicit equations using Matplotlib?

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Question :

Is it possible to plot implicit equations using Matplotlib?

I would like to plot implicit equations (of the form f(x, y)=g(x, y) eg. X^y=y^x) in Matplotlib. Is this possible?

Asked By: Geddes


Answer #1:

I don’t believe there’s very good support for this, but you could try something like

import matplotlib.pyplot
from numpy import arange
from numpy import meshgrid

delta = 0.025
xrange = arange(-5.0, 20.0, delta)
yrange = arange(-5.0, 20.0, delta)
X, Y = meshgrid(xrange,yrange)

# F is one side of the equation, G is the other
F = Y**X
G = X**Y

matplotlib.pyplot.contour(X, Y, (F - G), [0])

See the API docs for contour: if the fourth argument is a sequence then it specifies which contour lines to plot. But the plot will only be as good as the resolution of your ranges, and there are certain features it may never get right, often at self-intersection points.

Answered By: Steve

Answer #2:

Since you’ve tagged this question with sympy, I will give such an example.

From the documentation:

from sympy import var, plot_implicit
var('x y')
plot_implicit(x*y**3 - y*x**3)
Answered By: Gary Kerr

Answer #3:

matplotlib does not plot equations; it plots serieses of points. You can use a tool like scipy?.optimize to numerically calculate y points from x values (or vice versa) of implicit equations numerically or any number of other tools as appropriate.

For example, here is an example where I plot the implicit equation x ** 2 + x * y + y ** 2 = 10 in a certain region.

from functools import partial

import numpy
import scipy.optimize
import matplotlib.pyplot as pp

def z(x, y):
    return x ** 2 + x * y + y ** 2 - 10

x_window = 0, 5
y_window = 0, 5

xs = []
ys = []
for x in numpy.linspace(*x_window, num=200):
        # A more efficient technique would use the last-found-y-value as a 
        # starting point
        y = scipy.optimize.brentq(partial(z, x), *y_window)
    except ValueError:
        # Should we not be able to find a solution in this window.

pp.plot(xs, ys)
Answered By: Mike Graham

Answer #4:

There is an implicit equation (and inequality) plotter in sympy. It is created as a part of GSoC and it produces the plots as matplotlib figure instances.

Docs at

Since sympy version 0.7.2 it is available as:

>>> from sympy.plotting import plot_implicit
>>> p = plot_implicit(x < sin(x)) # also creates a window with the plot
>>> the_matplotlib_axes_instance = p._backend._ax
Answered By: Krastanov

Answer #5:

If you’re willing to use something other than matplotlib (but still python), there’s sage:

An example:

Documentation for implicit_plot

The Sage Homepage

Answered By: Alex

Answer #6:

Examples (using the contour approach) for all conic sections are available at

Answered By: soap262

Answer #7:

Many thanks Steve, Mike, Alex. I have gone along with Steve’s solution (please see code below). My only remaining issue is that the contour plot appears behind my gridlines, as opposed to a regular plot, which I can force to the front with zorder. Any more halp greatly appreciated.


import matplotlib.pyplot as plt 
from matplotlib.ticker import MultipleLocator, FormatStrFormatter
import numpy as np 

fig = plt.figure(1) 
ax = fig.add_subplot(111) 

# set up axis 

# setup x and y ranges and precision
x = np.arange(-0.5,5.5,0.01) 
y = np.arange(-0.5,5.5,0.01)

# draw a curve 
line, = ax.plot(x, x**2,zorder=100) 

# draw a contour

#set bounds 

#produce gridlines of different colors/widths

minor_grid_lines = [tick.gridline for tick in ax.xaxis.get_minor_ticks()] 
for idx,loc in enumerate(ax.xaxis.get_minorticklocs()): 
    if loc % 2.0 == 0:
    elif loc % 1.0 == 0:

minor_grid_lines = [tick.gridline for tick in ax.yaxis.get_minor_ticks()] 
for idx,loc in enumerate(ax.yaxis.get_minorticklocs()): 
    if loc % 2.0 == 0:
    elif loc % 1.0 == 0:
Answered By: Geddes

Answer #8:

Edit: If you plot a hyperbola using plt.plot() then you will get the undesired branching effect. plt.scatter() in its place should still work. Then there is no need to reverse the order of negative or positive values, but if you wanted to use this code for some reason (instead of using contour plot from scipy) it will work anyways with plt.scatter()

An implicit function in two dimensions in general can be written as:


Since we cannot write this as f(x) = y, then we cannot compute y from an easily programmable set of discrete x. It is possible, however, to see how close a point generated from a grid is from the true function.

So make a grid of x and y to a custom point density and see how close each point is to satisfying the equation.

In other words, if we can’t get f(x,y) =0, perhaps we can get close to 0. Instead of looking for f(x,y) =0 look for f(x,y) > -epsilon and f(x,y) < epsilon.

epsilon is your tolerance and if this condition fits within your tolerance of 0 and tuning the grid appropriately you can get your function plotted.

The code below does just that for a circle of radius 1 (f(x,y)= x^2 + y^2 -1 = 0). I used the symbol dr for epsilon.

Also, to make sure the plt.plot function connects the lines in the correct order, I use a reversed version of the x values for the negative y values. That way, the evaluation of f(x,y) is done in a clockwise loop so that the nearest values are one after another. Without this, lines from opposite sides of the function would connect and it would appear slightly filled in.

import numpy as np
import matplotlib.pyplot as plt
r = 1 #arbitrary radius to set up the span of points
points = 250 
dr = r/points #epsilon window 

x=list(np.linspace(-5*r,5*r,5*points+1)) #setting up the x,y grid

xreversed = reversed(x) #reversing the array

x_0=[] #placeholder arrays

for i in x:
    for j in y:
        if i**2 + j**2 -1 < dr and i**2+j**2 -1 > -dr  and j >= 0: #positive values of y
for i in xreversed:            
    for j in y:
        if i**2+j**2 -1 < dr and i**2+j**2 -1 > -dr  and j < 0: #negative values of y, using x reversed


Answered By: ??? ???

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