Contour line artifacts in matplotlib after interpolating data on a polar plot

Question

I have small sets of irregularly-spaced data taken at various points over a circular area, in polar format. I need to do interpolation to get the data on a regularly-spaced grid, and then I'd like to plot them using a contour plot.

I've managed to do the interpolation and plot the result, but I have to convert from polar to rectangular coordinates to do the interpolation, and I get artifacts on the polar plot when I convert the data back to polar coordinates.

The following code demonstrates what I have so far, and plots the data on a polar and rectangular plot:

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import Rbf

# inputs as 1D arrays
r = np.array([0, 1, 1, 1, 1, 2, 2, 2, 2])
theta = np.radians(np.array([0, 90, 180, 270, 0, 90, 180, 270, 0]))
# z = f(theta, r)
z = np.array([8, 7, 6, 4, 5, 2, 2, 2, 2])

# convert to rect
x = r * np.cos(theta)
y = r * np.sin(theta)

# create RBF for smoothing
rbf = Rbf(x, y, z)

# create grid to smooth over
xi, yi = np.mgrid[-2:2:10j, -2:2:10j]
# smooth
zi = rbf(xi, yi)

# convert back to polar
ri = np.sqrt(xi*xi + yi*yi)
ti = np.arctan2(yi, xi)

# polar plot
fig = plt.figure()
ax = plt.subplot(121, polar=True)
cax = ax.contour(ti, ri, zi, 10, linewidths=0.5, colors='k')
cax = ax.contourf(ti, ri, zi, 10, cmap=plt.cm.Spectral)
ax.set_rmax(2)

# rect plot
ax = plt.subplot(122)
cax = ax.contour(xi, yi, zi, 10, linewidths=0.5, colors='k')
cax = ax.contourf(xi, yi, zi, 10, cmap=plt.cm.Spectral)

plt.show()

The remaining issues are:

Can I fix the contour line artifacts?
Does Scipy provide a more appropriate interpolation algorithm that works for such small datasets containing polar coordinates?

Community · Accepted Answer · 2017-05-23 10:34:55Z

4

You may want to read this as well, but as far as the contour plot in polar coordinates is concerned, matplotlib expects a regularly meshed array in the radius and angle, so you could plot all nicely doing:

# polar plot
ri, ti = np.mgrid[0:2:100j, 0:2*np.pi:100j]
zi = rbf(ri*np.cos(ti), ri*np.sin(ti))

fig = plt.figure()
ax = plt.subplot(121, polar=True)
cax = ax.contour(ti, ri, zi, 10, linewidths=0.5, colors='k')
cax = ax.contourf(ti, ri, zi, 10, cmap=plt.cm.Spectral)
ax.set_rmax(2)

# rect plot
xi, yi = np.mgrid[-2:2:100j, -2:2:100j]
zi = rbf(xi, yi)

ax = plt.subplot(122, aspect='equal')
cax = ax.contour(xi, yi, zi, 10, linewidths=0.5, colors='k')
cax = ax.contourf(xi, yi, zi, 10, cmap=plt.cm.Spectral)
plt.show()

enter image description here

I am a little surprised by your use of Rbf. What exactly are you tring to do and why are you using that interpolator?

edited May 23, 2017 at 10:34

CommunityBot

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answered Jan 16, 2013 at 23:54

Jaime

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3 Comments

Carson Morrow Over a year ago

Well, Rbf was the only interpolator I could get to work. griddata masks off too much (I end up with a diamond shape), and I could never get the BivariateSpline classes to work. I'd love a suggestion for another interpolator, if you know of something more appropriate.

Jaime Over a year ago

@CarsonMorrow The point is that Rbf is not interpolating, it is extrapolating, which unless you know what you are doing, may be the same as "making the data up". The diamond shape you get is actually showing the convex hull of the points you supplied. If you had 6 points in the outer radius, you would get an hexagon. Without knowing in more detail where is your data coming from, it is difficult to make a suggestion for a better interpolator, though.

Carson Morrow Over a year ago

Thanks for the info. I'll have to do some research on the most appropriate interpolator.

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