I'd like to plot a certain 3D-function where a parameter changes over time using the sympy.plotting module.
I don't want several plots in one figure, I want one plot that updates, e.g. a parabola getting wider or narrower. How can I do that? I didn't find anything in the docs, but maybe I just missed it...
You can use the link provided by #Lior Cohen. You just need to convert your symbolic expression to a numerical function. Here is how I would do it:
import matplotlib.pyplot as plt
import numpy as np
import sympy as sp
# symbolic expression
x, y, t = sp.symbols("x, y, t")
expr = sp.cos((x**2 + y**2) * t)
# convert to numerical function
func = sp.lambdify([x, y, t], expr)
# numerical discretization
x, y = np.mgrid[-2:2:20j, -2:2:20j]
# evaluate the function at time t=0
z = func(x, y, 0)
plt.ion()
fig = plt.figure()
ax = fig.add_subplot(projection="3d")
surface = ax.plot_surface(x, y, z, color="b")
ax.set_zlim(-1, 1)
for t in np.linspace(0, 1):
# sadly, its not possible to update a surface.
# hence, we remove the previous and add a new one
surface.remove()
# add a new 3d surface
surface = ax.plot_surface(x, y, func(x, y, t), color="b")
fig.canvas.draw()
fig.canvas.flush_events()
My goal is to animate the hyper-specific (canonical) diffusion equation by being able to "tell" Python to increment the "t" variable existing in said equation. I have easily done this in Mathematica but need to use Python for my assigned research project.
The equation is structured/defined as c(x,y,t), and obviously my question applies for any type of function that c(x,y,t) is set to equal. Every answer related to my question ether:
1) Does not include a function that is not a PDE
2) Consists of not incrementing a time variable (t)
Furthermore, I cannot find any method to graph a 3D equation on Python that is for 2 variables.
EDIT: I have figured out a way to do this.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import matplotlib.animation as animation
import random
def fun(x, t):
return x+t #Any equation
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = np.arange(-20.0, 20.0, 0.05)
t = np.arange(0.0,50.0,1)
X, Y = np.meshgrid(x, t)
zs = np.array([fun(x,t) for x,t in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)
ax.plot_surface(X, Y, Z)
ax.set_xlabel('X Position')
ax.set_ylabel('Time')
ax.set_zlabel('Concentration')
plt.show()
Credit: Wim I want to use matplotlib to make a 3d plot given a z function
Any help or simple code of an animation procedure would mean a lot, as my research project deals with 7D mathematics and this is essentially the most basic example of a non-trivial representation of what I am trying to do. So expect more questions to come (regardless of an answer).
Ok so let's take the example from this answer. We can easily modify it to use a function c(x,y,t) instead of f(x,y,sig) (those are just variable names).
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.animation as animation
def update_plot(frame_number, zarray, plot):
plot[0].remove()
plot[0] = ax.plot_surface(x, y, zarray[:,:,frame_number], cmap="magma")
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
N = 14
nmax=20
x = np.linspace(-4,4,N+1)
x, y = np.meshgrid(x, x)
zarray = np.zeros((N+1, N+1, nmax))
sig = lambda t: 1.5+np.sin(t*2*np.pi/nmax)
c = lambda x,y,t : 1/np.sqrt(sig(t))*np.exp(-(x**2+y**2)/sig(t)**2)
for t in range(nmax):
zarray[:,:,t] = c(x,y,t)
plot = [ax.plot_surface(x, y, zarray[:,:,0], color='0.75', rstride=1, cstride=1)]
ax.set_zlim(0,1.5)
animate = animation.FuncAnimation(fig, update_plot, nmax, fargs=(zarray, plot))
plt.show()
Suppose I have a function such as:
Now, I want to make a surface plot of it (matplotlib plot_surface). I've constructed the three arrays using np.arange(stop,end,increment).
And here, I'm stuck. I don't want to use a for-loop, since I think I should be able to solve this with np.sum.
However, I don't know how to construct the function. Ideally, it can be constructed as f(x,y,k) and I'd use f(x,y) = np.sum(f(x,y,k), k).
I am failing to think of the solution.
I'd rather not take a wild guess at what you want to do, but here's an illustrative example:
import numpy as np
x = np.random.rand(100)
y = np.random.rand(100)
k = np.arange(1, 11)
# define f_k = (x + y)^k
fk = lambda xx, yy, kk: (xx + yy)**kk
Then,
X, Y, K = np.meshgrid(x, y, k)
# sum over k after evaluating f_k
f = fk(X, Y, K).sum(axis=-1)
f.shape
# (100, 100)
Finally,
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X[...,0], Y[...,0], f)
plt.show()
Alternatively, you could do
X, Y = np.meshgrid(x, y)
ax.plot_surface(X, Y, f)
plt.show()
I am trying to visualize a function of 3 parameters over a cube in R^3 to get an idea of the smoothness of the function. An example of this problem is shown in the sample code below
%pylab
from mpl_toolkits.mplot3d import Axes3D
import itertools
x = np.linspace(0,10,50)
y = np.linspace(0,15,50)
z = np.linspace(0,8,50)
points = []
for element in itertools.product(x, y, z):
points.append(element)
def f(vals):
return np.cos(vals[0]) + np.sin(vals[1]) + vals[2]**0.5
fxyz = map(f, points)
xi, yi, zi = zip(*points)
fig = plt.figure(figsize=(12, 8))
ax = fig.add_subplot(111, projection='3d')
ax.scatter(xi, yi, zi, c=fxyz, alpha=0.5)
plt.show()
The problem with this approach is that the inside of the cube cannot be visualized. Is there a better way to graph a function over some dense subset of R^3?
As #HYRY and #nicoguaro suggested in the comments above, Mayavi is much better suited for this type of work. There is a good set of examples here that I used for reference. Here is what I came up with
import numpy as np
from mayavi import mlab
x = np.linspace(0,10,50)
y = np.linspace(0,15,50)
z = np.linspace(0,8,50)
X, Y, Z = np.meshgrid(x, y, z)
s = np.cos(X) + np.sin(Y) + Z**0.5
b1 = np.percentile(s, 20)
b2 = np.percentile(s, 80)
mlab.pipeline.volume(mlab.pipeline.scalar_field(s), vmin=b1, vmax=b2)
mlab.axes()
mlab.show()
After which I rotated the figure to desired angles with the GUI and saved desired views
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.))
Update: I finally have found an easy way to render 3D implicit surface with matplotlib and scikit-image, see my other answer. I left this one for whom is interested in plotting parametric 3D surfaces.
Motivation
Late answer, I just needed to do the same and I found another way to do it at some extent. So I am sharing this another perspective.
This post does not answer: (1) How to plot any implicit function F(x,y,z)=0? But does answer: (2) How to plot parametric surfaces (not all implicit functions, but some of them) using mesh with matplotlib?
#Paul's method has the advantage to be non parametric, therefore we can plot almost anything we want using contour method on each axe, it fully addresses (1). But matplotlib cannot easily build a mesh from this method, so we cannot directly get a surface from it, instead we get plane curves in all directions. This is what motivated my answer, I wanted to address (2).
Rendering mesh
If we are able to parametrize (this may be hard or impossible), with at most 2 parameters, the surface we want to plot then we can plot it with matplotlib.plot_trisurf method.
That is, from an implicit equation F(x,y,z)=0, if we are able to get a parametric system S={x=f(u,v), y=g(u,v), z=h(u,v)} then we can plot it easily with matplotlib without having to resort to contour.
Then, rendering such a 3D surface boils down to:
# Render:
ax = plt.axes(projection='3d')
ax.plot_trisurf(x, y, z, triangles=tri.triangles, cmap='jet', antialiased=True)
Where (x, y, z) are vectors (not meshgrid, see ravel) functionally computed from parameters (u, v) and triangles parameter is a Triangulation derived from (u,v) parameters to shoulder the mesh construction.
Imports
Required imports are:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits import mplot3d
from matplotlib.tri import Triangulation
Some surfaces
Lets parametrize some surfaces...
Sphere
# Parameters:
theta = np.linspace(0, 2*np.pi, 20)
phi = np.linspace(0, np.pi, 20)
theta, phi = np.meshgrid(theta, phi)
rho = 1
# Parametrization:
x = np.ravel(rho*np.cos(theta)*np.sin(phi))
y = np.ravel(rho*np.sin(theta)*np.sin(phi))
z = np.ravel(rho*np.cos(phi))
# Triangulation:
tri = Triangulation(np.ravel(theta), np.ravel(phi))
Cone
theta = np.linspace(0, 2*np.pi, 20)
rho = np.linspace(-2, 2, 20)
theta, rho = np.meshgrid(theta, rho)
x = np.ravel(rho*np.cos(theta))
y = np.ravel(rho*np.sin(theta))
z = np.ravel(rho)
tri = Triangulation(np.ravel(theta), np.ravel(rho))
Torus
a, c = 1, 4
u = np.linspace(0, 2*np.pi, 20)
v = u.copy()
u, v = np.meshgrid(u, v)
x = np.ravel((c + a*np.cos(v))*np.cos(u))
y = np.ravel((c + a*np.cos(v))*np.sin(u))
z = np.ravel(a*np.sin(v))
tri = Triangulation(np.ravel(u), np.ravel(v))
Möbius Strip
u = np.linspace(0, 2*np.pi, 20)
v = np.linspace(-1, 1, 20)
u, v = np.meshgrid(u, v)
x = np.ravel((2 + (v/2)*np.cos(u/2))*np.cos(u))
y = np.ravel((2 + (v/2)*np.cos(u/2))*np.sin(u))
z = np.ravel(v/2*np.sin(u/2))
tri = Triangulation(np.ravel(u), np.ravel(v))
Limitation
Most of the time, Triangulation is required in order to coordinate mesh construction of plot_trisurf method, and this object only accepts two parameters, so we are limited to 2D parametric surfaces. It is unlikely we could represent the Goursat Tangle with this method.
Matplotlib expects a series of points; it will do the plotting if you can figure out how to render your equation.
Referring to Is it possible to plot implicit equations using Matplotlib? Mike Graham's answer suggests using scipy.optimize to numerically explore the implicit function.
There is an interesting gallery at http://xrt.wikidot.com/gallery:implicit showing a variety of raytraced implicit functions - if your equation matches one of these, it might give you a better idea what you are looking at.
Failing that, if you care to share the actual equation, maybe someone can suggest an easier approach.
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[0],xrange[1],density)
y = linspace(yrange[0],yrange[1],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[2] == 1:
zi = res[0]
points.append([xi,yi,zi])
points = array(points)
Actually there is an easy way to plot implicit 3D surface with the scikit-image package. The key is the marching_cubes method.
import numpy as np
from skimage import measure
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
Then we compute the function over a 3D meshgrid, in this example we use the goursat_tangle method #Paul defined in its answer:
xl = np.linspace(-3, 3, 50)
X, Y, Z = np.meshgrid(xl, xl, xl)
F = goursat_tangle(X, Y, Z)
The magic is happening here with marching_cubes:
verts, faces, normals, values = measure.marching_cubes(F, 0, spacing=[np.diff(xl)[0]]*3)
verts -= 3
We just need to correct vertices coordinates as they are expressed in Voxel coordinates (hence scaling using spacing switch and the subsequent origin shift).
Finally it is just about rendering the iso-surface using tri_surface:
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_trisurf(verts[:, 0], verts[:, 1], faces, verts[:, 2], cmap='jet', lw=0)
Which returns:
Have you looked at mplot3d on matplotlib?
Finally, I did it (I updated my matplotlib to 1.0.1).
Here is code:
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
def hyp_part1(x,y,z):
return -(x**2) - (y**2) + (z**2) - 1
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x_range = np.arange(-100,100,10)
y_range = np.arange(-100,100,10)
X,Y = np.meshgrid(x_range,y_range)
A = np.linspace(-100, 100, 15)
A1,A2 = np.meshgrid(A,A)
for z in A:
X,Y = A1, A2
Z = hyp_part1(X,Y,z)
ax.contour(X, Y, Z+z, [z], zdir='z')
for y in A:
X,Z= A1, A2
Y = hyp_part1(X,y,Z)
ax.contour(X, Y+y, Z, [y], zdir='y')
for x in A:
Y,Z = A1, A2
X = hyp_part1(x,Y,Z)
ax.contour(X+x, Y, Z, [x], zdir='x')
ax.set_zlim3d(-100,100)
ax.set_xlim3d(-100,100)
ax.set_ylim3d(-100,100)
Here is result:
Thank You, Paul!
MathGL (GPL plotting library) can plot it easily. Just create a data mesh with function values f[i,j,k] and use Surf3() function to make isosurface at value f[i,j,k]=0. See this sample.