surface plot on irregular grid in python environment

Question

Is it possible to create surface plot and contour plot of a function for example u(x,y) = x^2 + y^2

on following domain which is bounded by equation

r(t) = 1+(cos(4*t))^2, x = r(t)*cos(t),y = r(t)*sin(t), 0 < t < 2*pi

I want surface plot variant of following scatter plot. I also used scipy griddata as follows

from matplotlib.pyplot  import *
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.interpolate import griddata

data = np.array([[ 2.00000000e+00,  0.00000000e+00],
   [ 1.89614525e+00,  1.51126793e-01],
   [ 1.62613327e+00,  2.60869688e-01],
   [ 1.29639187e+00,  3.15328472e-01],
   [ 1.03183519e+00,  3.39805963e-01],
   [ 9.22330309e-01,  3.87424263e-01],
   [ 9.85968247e-01,  5.09806830e-01],
   [ 1.16496791e+00,  7.25092951e-01],
   [ 1.35481653e+00,  1.00089121e+00],
   [ 1.45215333e+00,  1.26290100e+00],
   [ 1.39897768e+00,  1.42707450e+00],
   [ 1.20351374e+00,  1.44074409e+00],
   [ 9.29854857e-01,  1.31248166e+00],
   [ 6.63453200e-01,  1.11474417e+00],
   [ 4.70396020e-01,  9.55864325e-01],
   [ 3.70174853e-01,  9.32786757e-01],
   [ 3.33831717e-01,  1.08590518e+00],
   [ 3.06157327e-01,  1.37738581e+00],
   [ 2.38732323e-01,  1.70413196e+00],
   [ 1.15900227e-01,  1.94068353e+00],
   [-3.96388724e-02,  1.99329358e+00],
   [-1.83621616e-01,  1.84088591e+00],
   [-2.79344160e-01,  1.54430632e+00],
   [-3.22477316e-01,  1.21965478e+00],
   [-3.47231849e-01,  9.87829012e-01],
   [-4.09520279e-01,  9.23088740e-01],
   [-5.55288737e-01,  1.02352335e+00],
   [-7.90735832e-01,  1.21586141e+00],
   [-1.07111976e+00,  1.39113164e+00],
   [-1.31601256e+00,  1.45381050e+00],
   [-1.44544698e+00,  1.36169834e+00],
   [-1.42013085e+00,  1.13913085e+00],
   [-1.26561597e+00,  8.59376173e-01],
   [-1.06700598e+00,  6.06642746e-01],
   [-9.34540362e-01,  4.37040778e-01],
   [-9.54661396e-01,  3.57053472e-01],
   [-1.14896223e+00,  3.28361061e-01],
   [-1.46070082e+00,  2.94327457e-01],
   [-1.77632959e+00,  2.12929020e-01],
   [-1.97334870e+00,  7.85155418e-02],
   [-1.97334870e+00, -7.85155418e-02],
   [-1.77632959e+00, -2.12929020e-01],
   [-1.46070082e+00, -2.94327457e-01],
   [-1.14896223e+00, -3.28361061e-01],
   [-9.54661396e-01, -3.57053472e-01],
   [-9.34540362e-01, -4.37040778e-01],
   [-1.06700598e+00, -6.06642746e-01],
   [-1.26561597e+00, -8.59376173e-01],
   [-1.42013085e+00, -1.13913085e+00],
   [-1.44544698e+00, -1.36169834e+00],
   [-1.31601256e+00, -1.45381050e+00],
   [-1.07111976e+00, -1.39113164e+00],
   [-7.90735832e-01, -1.21586141e+00],
   [-5.55288737e-01, -1.02352335e+00],
   [-4.09520279e-01, -9.23088740e-01],
   [-3.47231849e-01, -9.87829012e-01],
   [-3.22477316e-01, -1.21965478e+00],
   [-2.79344160e-01, -1.54430632e+00],
   [-1.83621616e-01, -1.84088591e+00],
   [-3.96388724e-02, -1.99329358e+00],
   [ 1.15900227e-01, -1.94068353e+00],
   [ 2.38732323e-01, -1.70413196e+00],
   [ 3.06157327e-01, -1.37738581e+00],
   [ 3.33831717e-01, -1.08590518e+00],
   [ 3.70174853e-01, -9.32786757e-01],
   [ 4.70396020e-01, -9.55864325e-01],
   [ 6.63453200e-01, -1.11474417e+00],
   [ 9.29854857e-01, -1.31248166e+00],
   [ 1.20351374e+00, -1.44074409e+00],
   [ 1.39897768e+00, -1.42707450e+00],
   [ 1.45215333e+00, -1.26290100e+00],
   [ 1.35481653e+00, -1.00089121e+00],
   [ 1.16496791e+00, -7.25092951e-01],
   [ 9.85968247e-01, -5.09806830e-01],
   [ 9.22330309e-01, -3.87424263e-01],
   [ 1.03183519e+00, -3.39805963e-01],
   [ 1.29639187e+00, -3.15328472e-01],
   [ 1.62613327e+00, -2.60869688e-01],
   [ 1.89614525e+00, -1.51126793e-01],
   [ 2.00000000e+00, -4.89858720e-16],
   [ 0.00000000e+00, -1.50000000e+00],
   [-1.00000000e+00, -1.00000000e+00],
   [ 0.00000000e+00, -1.00000000e+00],
   [ 1.00000000e+00, -1.00000000e+00],
   [-5.00000000e-01, -5.00000000e-01],
   [ 0.00000000e+00, -5.00000000e-01],
   [ 5.00000000e-01, -5.00000000e-01],
   [-1.50000000e+00,  0.00000000e+00],
   [-1.00000000e+00,  0.00000000e+00],
   [-5.00000000e-01,  0.00000000e+00],
   [ 0.00000000e+00,  0.00000000e+00],
   [ 5.00000000e-01,  0.00000000e+00],
   [ 1.00000000e+00,  0.00000000e+00],
   [ 1.50000000e+00,  0.00000000e+00],
   [ 2.00000000e+00,  0.00000000e+00],
   [-5.00000000e-01,  5.00000000e-01],
   [ 0.00000000e+00,  5.00000000e-01],
   [ 5.00000000e-01,  5.00000000e-01],
   [-1.00000000e+00,  1.00000000e+00],
   [ 0.00000000e+00,  1.00000000e+00],
   [ 1.00000000e+00,  1.00000000e+00],
   [ 0.00000000e+00,  1.50000000e+00]])   
ua = data[:,0]**2+data[:,1]**2 # u=x^2+y^2


xx,yy = np.meshgrid(np.linspace(-2,2,100),np.linspace(-2,2,100))
Ua = griddata((data[:,0],data[:,1]),ua,(xx,yy),method='cubic') 

fig = figure(1)
plot (data[:,0], data[:,1], '*'); # 
fig = figure(2)
ax = fig.gca(projection='3d')

ax.plot_wireframe(xx,yy,Ua,rstride=1,cstride=1,linewidth=.5) 

but the result is not good as below

Answer 1

Just calculate the function on a rectangular grid in the first place!

scipy.interpolate.griddata interpolates on a convex hull. You can read that in the docs. That means that it also interpolates between your lobes. That's why you don't get the correct domain. Interpolation is also inherently less accurate than just calculating the function across the grid.

plot_wireframe wants a rectangular grid which you already created. All you need to do is calculate the function values on the rectangular grid. To only plot values on your domain boundary, set everything outside it to np.NaN (not-a-number).

Here's how to do it:

import matplotlib.pyplot as plt
import numpy as np

# cartesian coordinates
xx,yy = np.meshgrid(np.linspace(-2,2,100),np.linspace(-2,2,100))

# function value across square domain
Ua = xx**2 + yy**2

# polar coordinates
tt = np.arctan2(yy, xx)
rr = np.sqrt(Ua) # re-using x^2 + y^2 -- only works for this function

# r coordinate of domain boundary
domain_boundary = 1 + (np.cos(4*tt))**2

# function value across actual domain, with rest set to NaN
Ua[rr > domain_boundary] = np.NaN

# plotting
fig = plt.figure(2)
ax = fig.gca(projection='3d')
ax.plot_wireframe(xx,yy,Ua,rstride=1,cstride=1,linewidth=.5)

That solution is not perfect as you can recognize the rectangular grid in the result. You could try working in polar coordinates as shown in this official matplotlib example.