Interpolation Exercise 2


In [1]:
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
sns.set_style('white')

In [2]:
from scipy.interpolate import griddata

Sparse 2d interpolation

In this example the values of a scalar field $f(x,y)$ are known at a very limited set of points in a square domain:

  • The square domain covers the region $x\in[-5,5]$ and $y\in[-5,5]$.
  • The values of $f(x,y)$ are zero on the boundary of the square at integer spaced points.
  • The value of $f$ is known at a single interior point: $f(0,0)=1.0$.
  • The function $f$ is not known at any other points.

Create arrays x, y, f:

  • x should be a 1d array of the x coordinates on the boundary and the 1 interior point.
  • y should be a 1d array of the y coordinates on the boundary and the 1 interior point.
  • f should be a 1d array of the values of f at the corresponding x and y coordinates.

You might find that np.hstack is helpful.


In [87]:
x = np.hstack((np.linspace(-4,4,9), np.full(11, -5), np.linspace(-4,4,9), np.full(11, 5), [0]))
y = np.hstack((np.full(9,-5), np.linspace(-5, 5,11), np.full(9,5), np.linspace(-5,5,11), [0]))
f = np.hstack((np.zeros(20), np.zeros(20),[1.0]))
print(f)


[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.
  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.
  0.  0.  0.  0.  1.]

The following plot should show the points on the boundary and the single point in the interior:


In [88]:
plt.scatter(x, y);



In [89]:
assert x.shape==(41,)
assert y.shape==(41,)
assert f.shape==(41,)
assert np.count_nonzero(f)==1

Use meshgrid and griddata to interpolate the function $f(x,y)$ on the entire square domain:

  • xnew and ynew should be 1d arrays with 100 points between $[-5,5]$.
  • Xnew and Ynew should be 2d versions of xnew and ynew created by meshgrid.
  • Fnew should be a 2d array with the interpolated values of $f(x,y)$ at the points (Xnew,Ynew).
  • Use cubic spline interpolation.

In [109]:
xnew = np.linspace(-5, 5, 100)
ynew = np.linspace(-5, 5, 100)
Xnew, Ynew = np.meshgrid(xnew, ynew)
Fnew = griddata((x, y), f , (Xnew, Ynew),  method='cubic')
plt.imshow(Fnew, extent=(-5,5,-5,5))


Out[109]:
<matplotlib.image.AxesImage at 0x7fb7bd933128>

In [ ]:


In [110]:
assert xnew.shape==(100,)
assert ynew.shape==(100,)
assert Xnew.shape==(100,100)
assert Ynew.shape==(100,100)
assert Fnew.shape==(100,100)

Plot the values of the interpolated scalar field using a contour plot. Customize your plot to make it effective and beautiful.


In [117]:
plt.contourf(Xnew, Ynew, Fnew, cmap='hot')
plt.colorbar(label='Z')
plt.box(False)
plt.title("The interpolated 2d grid of our data.")
plt.xlabel('X')
plt.ylabel('Y');



In [108]:
assert True # leave this to grade the plot

In [ ]: