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I'm trying to make a 3D surface plot from a 2D array, using the values of my 2D array as the z value:
I want to do a surface plot like in the next pic:
This is my code
import numpy as npimport matplotlib.pyplot as pltimport math as mtfrom mpl_toolkits import mplot3d
ax = plt.axes(projection="3D")
z = [[248, 236,289,300,267,225,266,265,259,279,269,335],[246,253,241,232,276,213,198,201,222,229,193,237],[182,180,200,192,233,211,227,220,174,187,181,197],[124,102,137,130,144,168,149,164,168,156,90,156],[117,124,133,119,155,140,133,120,130,134,138,102],[155,140,137,125,146,102,129,114,119,113,132,122],[104,117,119,138,137,118,117,128,131,133,119,133],[136,115,108,105,133,104,121,135,136,127,135,112],[84,87,93,116,123,110,90,123,112,115,92,107],[118,94,100,83,132,90,111,91,98,116,100,95],[101,76,115,121,108,102,94,80,83,104,101,81],[86,86,109,105,95,75,18,87,92,99,101,128]]
y = np.arange(len(z))x = np.arange(len(z[0]))
(x ,y) = np.meshgrid(x,y)
ax.plot3D(x,y,z)plt.show()
This is what I tried but gives me an error.
you need to use plot_surface:
ax = plt.axes(projection="3d")
z = np.array([[248, 236,289,300,267,225,266,265,259,279,269,335],[246,253,241,232,276,213,198,201,222,229,193,237],[182,180,200,192,233,211,227,220,174,187,181,197],[124,102,137,130,144,168,149,164,168,156,90,156],[117,124,133,119,155,140,133,120,130,134,138,102],[155,140,137,125,146,102,129,114,119,113,132,122],[104,117,119,138,137,118,117,128,131,133,119,133],[136,115,108,105,133,104,121,135,136,127,135,112],[84,87,93,116,123,110,90,123,112,115,92,107],[118,94,100,83,132,90,111,91,98,116,100,95],[101,76,115,121,108,102,94,80,83,104,101,81],[86,86,109,105,95,75,18,87,92,99,101,128]])
y = np.arange(len(z))
x = np.arange(len(z[0]))
(x ,y) = np.meshgrid(x,y)
ax.plot_surface(x,y,z)
plt.show()
I need to calculate the line integral between two points (x1,y1) and (x2,y2) under a surface defined by values on a meshgrid.
I'm not exactly sure on the best tool/approach to use for this process using python.
As I do not have a function which represents the surface, instead values at points on a evenly spaaced meshgrid I am assuming I will need to use one of the following methods
trapz -- Use trapezoidal rule to compute integral from samples.
cumtrapz -- Use trapezoidal rule to cumulatively compute integral.
simps -- Use Simpson's rule to compute integral from samples.
romb -- Use Romberg Integration to compute integral from
(2**k + 1) evenly-spaced samples.
Any help or guidance would be appreciated.
Edit:
import numpy as np
from scipy import interpolate
def f(x, y):
return x**2 + x*y + y*2 + 1
xl = np.linspace(-1.5, 1.5, 101,endpoint = True)
X, Y = np.meshgrid(xl, xl)
Z = f(X, Y)
#And a 2D Line:
arr_2D = np.linspace(start=[-1, 1.2], stop=[0, 1.5], num=101,endpoint =
True) #Creates a 2D line between these two points
#Then we create a multidimensional linear interpolator:
XY = np.stack([X.ravel(), Y.ravel()]).T
S = interpolate.LinearNDInterpolator(XY, Z.ravel())
print(S)
#To interpolate points from 2D curve on the 3D surface:
St = S(arr_2D)
#We also compute the curvilinear coordinates of the 2D curve:
#Using curvilinear coordinates based on cumulative arc length, the integral to solve looks like:
Sd = np.cumsum(np.sqrt(np.sum(np.diff(arr_2D, axis=0)**2, axis=1)))
print(Sd)
I = np.trapz(St[:-1], Sd) # 2.041770932394164
print("Integral: ",I)
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = plt.axes(projection="3d")
x_line = np.linspace(start=[-1], stop=[1.5], num=100,endpoint = True)
y_line = np.linspace(start=[-1.2], stop=[1.5], num=100,endpoint = True)
ax.plot3D(x_line, y_line, 'red') #Line which represents integral
ax.plot_wireframe(X, Y, Z, color='green') #Represents the surface
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('Time')
plt.show()
fig = plt.figure()
ax = plt.axes()
ax.fill_between(Sd, St)
ax.set_xlabel('x')
ax.set_ylabel('Z')
plt.show()
Provided you have surface points (we can even relax the requirement of regular grid) and curve points, then basic analysis provided by numpy and scipy packages should do the trick.
First, let's create a trial dataset for your problem.
import numpy as np
from scipy import interpolate
Mainly a 3D surface:
def f(x, y):
return x**2 + x*y + y*2 + 1
xl = np.linspace(-1.5, 1.5, 101)
X, Y = np.meshgrid(xl, xl)
Z = f(X, Y)
And a 2D curve:
t = np.linspace(0, 1, 1001)
xt = t**2*np.cos(2*np.pi*t**2)
yt = t**3*np.sin(2*np.pi*t**3)
The complete setup looks like:
axe = plt.axes(projection='3d')
axe.plot_surface(X, Y, Z, cmap='jet', alpha=0.5)
axe.plot(xt, yt, 0)
axe.plot(xt, yt, St)
axe.view_init(elev=25, azim=-45)
Then we create a multidimensional linear interpolator:
XY = np.stack([X.ravel(), Y.ravel()]).T
S = interpolate.LinearNDInterpolator(XY, Z.ravel())
To interpolate points from 2D curve on the 3D surface:
xyt = np.stack([xt, yt]).T
St = S(xyt)
We also compute the curvilinear coordinates of the 2D curve:
Sd = np.cumsum(np.sqrt(np.sum(np.diff(xyt, axis=0)**2, axis=1)))
Using curvilinear coordinates based on cumulative arc length, the integral to solve looks like:
fig, axe = plt.subplots()
axe.plot(Sd, St[:-1])
axe.fill_between(Sd, St[:-1], alpha=0.5)
axe.grid()
Finally we integrate using the method of our choice, here the simplest Trapezoidal Rule from numpy:
I = np.trapz(St[:-1], Sd) # 2.041770932394164
I have a spreadsheet file that I would like to input to create a 3D surface graph using Matplotlib in Python.
I used plot_trisurf and it worked, but I need the projections of the contour profiles onto the graph that I can get with the surface function, like this example.
I'm struggling to arrange my Z data in a 2D array that I can use to input in the plot_surface method. I tried a lot of things, but none seems to work.
Here it is what I have working, using plot_trisurf
import matplotlib
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import pandas as pd
df=pd.read_excel ("/Users/carolethais/Desktop/Dissertação Carol/Códigos/Resultados/res_02_0.5.xlsx")
fig = plt.figure()
ax = fig.gca(projection='3d')
# I got the graph using trisurf
graf=ax.plot_trisurf(df["Diametro"],df["Comprimento"], df["temp_out"], cmap=matplotlib.cm.coolwarm)
ax.set_xlim(0, 0.5)
ax.set_ylim(0, 100)
ax.set_zlim(25,40)
fig.colorbar(graf, shrink=0.5, aspect=15)
ax.set_xlabel('Diâmetro (m)')
ax.set_ylabel('Comprimento (m)')
ax.set_zlabel('Temperatura de Saída (ºC)')
plt.show()
This is a part of my df, dataframe:
Diametro Comprimento temp_out
0 0.334294 0.787092 34.801994
1 0.334294 8.187065 32.465551
2 0.334294 26.155976 29.206090
3 0.334294 43.648591 27.792126
4 0.334294 60.768219 27.163233
... ... ... ...
59995 0.437266 14.113660 31.947302
59996 0.437266 25.208851 30.317583
59997 0.437266 33.823035 29.405461
59998 0.437266 57.724209 27.891616
59999 0.437266 62.455890 27.709298
I tried this approach to use the imported data with plot_surface, but what I got was indeed a graph but it didn't work, here it's the way the graph looked with this approach:
Thank you so much
A different approach, based on re-gridding the data, that doesn't require that the original data is specified on a regular grid [deeply inspired by this example;-].
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.tri as tri
from mpl_toolkits.mplot3d import Axes3D
np.random.seed(19880808)
# compute the sombrero over a cloud of random points
npts = 10000
x, y = np.random.uniform(-5, 5, npts), np.random.uniform(-5, 5, npts)
z = np.cos(1.5*np.sqrt(x*x + y*y))/(1+0.33*(x*x+y*y))
# prepare the interpolator
triang = tri.Triangulation(x, y)
interpolator = tri.LinearTriInterpolator(triang, z)
# do the interpolation
xi = yi = np.linspace(-5, 5, 101)
Xi, Yi = np.meshgrid(xi, yi)
Zi = interpolator(Xi, Yi)
# plotting
fig = plt.figure()
ax = fig.gca(projection='3d')
norm = plt.Normalize(-1,1)
ax.plot_surface(Xi, Yi, Zi,
cmap='inferno',
norm=plt.Normalize(-1,1))
plt.show()
plot_trisurf expects x, y, z as 1D arrays while plot_surface expects X, Y, Z as 2D arrays or as x, y, Z with x, y being 1D array and Z a 2D array.
Your data consists of 3 1D arrays, so plotting them with plot_trisurf is immediate but you need to use plot_surface to be able to project the isolines on the coordinate planes... You need to reshape your data.
It seems that you have 60000 data points, in the following I assume that you have a regular grid 300 points in the x direction and 200 points in y — but what is important is the idea of regular grid.
The code below shows
the use of plot_trisurf (with a coarser mesh), similar to your code;
the correct use of reshaping and its application in plot_surface;
note that the number of rows in reshaping corresponds to the number
of points in y and the number of columns to the number of points in x;
and 4. incorrect use of reshaping, the resulting subplots are somehow
similar to the plot you showed, maybe you just need to fix the number
of row and columns.
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
x, y = np.arange(30)/3.-5, np.arange(20)/2.-5
x, y = (arr.flatten() for arr in np.meshgrid(x, y))
z = np.cos(1.5*np.sqrt(x*x + y*y))/(1+0.1*(x*x+y*y))
fig, axes = plt.subplots(2, 2, subplot_kw={"projection" : "3d"})
axes = iter(axes.flatten())
ax = next(axes)
ax.plot_trisurf(x,y,z, cmap='Reds')
ax.set_title('Trisurf')
X, Y, Z = (arr.reshape(20,30) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 20×30')
X, Y, Z = (arr.reshape(30,20) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 30×20')
X, Y, Z = (arr.reshape(40,15) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 40×15')
plt.tight_layout()
plt.show()
Using Matplotlib, I want to plot a 2D heat map. My data is an n-by-n Numpy array, each with a value between 0 and 1. So for the (i, j) element of this array, I want to plot a square at the (i, j) coordinate in my heat map, whose color is proportional to the element's value in the array.
How can I do this?
The imshow() function with parameters interpolation='nearest' and cmap='hot' should do what you want.
Please review the interpolation parameter details, and see Interpolations for imshow and Image antialiasing.
import matplotlib.pyplot as plt
import numpy as np
a = np.random.random((16, 16))
plt.imshow(a, cmap='hot', interpolation='nearest')
plt.show()
Seaborn is a high-level API for matplotlib, which takes care of a lot of the manual work.
seaborn.heatmap automatically plots a gradient at the side of the chart etc.
import numpy as np
import seaborn as sns
import matplotlib.pylab as plt
uniform_data = np.random.rand(10, 12)
ax = sns.heatmap(uniform_data, linewidth=0.5)
plt.show()
You can even plot upper / lower left / right triangles of square matrices. For example, a correlation matrix, which is square and is symmetric, so plotting all values would be redundant.
corr = np.corrcoef(np.random.randn(10, 200))
mask = np.zeros_like(corr)
mask[np.triu_indices_from(mask)] = True
with sns.axes_style("white"):
ax = sns.heatmap(corr, mask=mask, vmax=.3, square=True, cmap="YlGnBu")
plt.show()
I would use matplotlib's pcolor/pcolormesh function since it allows nonuniform spacing of the data.
Example taken from matplotlib:
import matplotlib.pyplot as plt
import numpy as np
# generate 2 2d grids for the x & y bounds
y, x = np.meshgrid(np.linspace(-3, 3, 100), np.linspace(-3, 3, 100))
z = (1 - x / 2. + x ** 5 + y ** 3) * np.exp(-x ** 2 - y ** 2)
# x and y are bounds, so z should be the value *inside* those bounds.
# Therefore, remove the last value from the z array.
z = z[:-1, :-1]
z_min, z_max = -np.abs(z).max(), np.abs(z).max()
fig, ax = plt.subplots()
c = ax.pcolormesh(x, y, z, cmap='RdBu', vmin=z_min, vmax=z_max)
ax.set_title('pcolormesh')
# set the limits of the plot to the limits of the data
ax.axis([x.min(), x.max(), y.min(), y.max()])
fig.colorbar(c, ax=ax)
plt.show()
For a 2d numpy array, simply use imshow() may help you:
import matplotlib.pyplot as plt
import numpy as np
def heatmap2d(arr: np.ndarray):
plt.imshow(arr, cmap='viridis')
plt.colorbar()
plt.show()
test_array = np.arange(100 * 100).reshape(100, 100)
heatmap2d(test_array)
This code produces a continuous heatmap.
You can choose another built-in colormap from here.
Here's how to do it from a csv:
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import griddata
# Load data from CSV
dat = np.genfromtxt('dat.xyz', delimiter=' ',skip_header=0)
X_dat = dat[:,0]
Y_dat = dat[:,1]
Z_dat = dat[:,2]
# Convert from pandas dataframes to numpy arrays
X, Y, Z, = np.array([]), np.array([]), np.array([])
for i in range(len(X_dat)):
X = np.append(X, X_dat[i])
Y = np.append(Y, Y_dat[i])
Z = np.append(Z, Z_dat[i])
# create x-y points to be used in heatmap
xi = np.linspace(X.min(), X.max(), 1000)
yi = np.linspace(Y.min(), Y.max(), 1000)
# Interpolate for plotting
zi = griddata((X, Y), Z, (xi[None,:], yi[:,None]), method='cubic')
# I control the range of my colorbar by removing data
# outside of my range of interest
zmin = 3
zmax = 12
zi[(zi<zmin) | (zi>zmax)] = None
# Create the contour plot
CS = plt.contourf(xi, yi, zi, 15, cmap=plt.cm.rainbow,
vmax=zmax, vmin=zmin)
plt.colorbar()
plt.show()
where dat.xyz is in the form
x1 y1 z1
x2 y2 z2
...
Use matshow() which is a wrapper around imshow to set useful defaults for displaying a matrix.
a = np.diag(range(15))
plt.matshow(a)
https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.matshow.html
This is just a convenience function wrapping imshow to set useful defaults for displaying a matrix. In particular:
Set origin='upper'.
Set interpolation='nearest'.
Set aspect='equal'.
Ticks are placed to the left and above.
Ticks are formatted to show integer indices.
Here is a new python package to plot complex heatmaps with different kinds of row/columns annotations in Python: https://github.com/DingWB/PyComplexHeatmap
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.