I'd like to plot implicit equation F(x,y,z) = 0 in 3D. Is it possible in Matplotlib?
You can trick matplotlib into plotting implicit equations in 3D. Just make a one-level contour plot of the equation for each z value within the desired limits. You can repeat the process along the y and z axes as well for a more solid-looking shape.
from mpl_toolkits.mplot3d import axes3d import matplotlib.pyplot as plt import numpy as np def plot_implicit(fn, bbox=(-2.5,2.5)): ''' create a plot of an implicit function fn ...implicit function (plot where fn==0) bbox ..the x,y,and z limits of plotted interval''' xmin, xmax, ymin, ymax, zmin, zmax = bbox*3 fig = plt.figure() ax = fig.add_subplot(111, projection='3d') A = np.linspace(xmin, xmax, 100) # resolution of the contour B = np.linspace(xmin, xmax, 15) # number of slices A1,A2 = np.meshgrid(A,A) # grid on which the contour is plotted for z in B: # plot contours in the XY plane X,Y = A1,A2 Z = fn(X,Y,z) cset = ax.contour(X, Y, Z+z, [z], zdir='z') # [z] defines the only level to plot for this contour for this value of z for y in B: # plot contours in the XZ plane X,Z = A1,A2 Y = fn(X,y,Z) cset = ax.contour(X, Y+y, Z, [y], zdir='y') for x in B: # plot contours in the YZ plane Y,Z = A1,A2 X = fn(x,Y,Z) cset = ax.contour(X+x, Y, Z, [x], zdir='x') # must set plot limits because the contour will likely extend # way beyond the displayed level. Otherwise matplotlib extends the plot limits # to encompass all values in the contour. ax.set_zlim3d(zmin,zmax) ax.set_xlim3d(xmin,xmax) ax.set_ylim3d(ymin,ymax) plt.show()
Here's the plot of the Goursat Tangle:
def goursat_tangle(x,y,z): a,b,c = 0.0,-5.0,11.8 return x**4+y**4+z**4+a*(x**2+y**2+z**2)**2+b*(x**2+y**2+z**2)+c plot_implicit(goursat_tangle)
You can make it easier to visualize by adding depth cues with creative colormapping:
Here's how the OP's plot looks:
def hyp_part1(x,y,z): return -(x**2) - (y**2) + (z**2) - 1 plot_implicit(hyp_part1, bbox=(-100.,100.))
Bonus: You can use python to functionally combine these implicit functions:
def sphere(x,y,z): return x**2 + y**2 + z**2 - 2.0**2 def translate(fn,x,y,z): return lambda a,b,c: fn(x-a,y-b,z-c) def union(*fns): return lambda x,y,z: np.min( [fn(x,y,z) for fn in fns], 0) def intersect(*fns): return lambda x,y,z: np.max( [fn(x,y,z) for fn in fns], 0) def subtract(fn1, fn2): return intersect(fn1, lambda *args:-fn2(*args)) plot_implicit(union(sphere,translate(sphere, 1.,1.,1.)), (-2.,3.))
As far as I know, it is not possible. You have to solve this equation numerically by yourself. Using scipy.optimize is a good idea. The simplest case is that you know the range of the surface that you want to plot, and just make a regular grid in x and y, and try to solve equation F(xi,yi,z)=0 for z, giving a starting point of z. Following is a very dirty code that might help you
from scipy import * from scipy import optimize xrange = (0,1) yrange = (0,1) density = 100 startz = 1 def F(x,y,z): return x**2+y**2+z**2-10 x = linspace(xrange,xrange,density) y = linspace(yrange,yrange,density) points =  for xi in x: for yi in y: g = lambda z:F(xi,yi,z) res = optimize.fsolve(g, startz, full_output=1) if res == 1: zi = res points.append([xi,yi,zi]) points = array(points)