We are nerds, we are cheesy

%matplotlib inline

import matplotlib
matplotlib.rcParams['figure.figsize'] = (16,13)

# Source: http://stackoverflow.com/a/4687582
# Thanks!

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

def plot_implicit(fn, bbox=(-1.5,1.5)):
    ''' 
    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()

def happy_saint_valentine_ipython(x,y,z):
    return -x**2*z**3 - 9*y**2*z**3/80 + (4*x**2 + 9*y**2 + 4*z**2 - 4)**3/64

plot_implicit(happy_saint_valentine_ipython)