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%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
sns.set_style('white')
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from scipy.interpolate import griddata
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:
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.
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# left, top, right, bottom
x = np.hstack((np.array([-5]*10), np.linspace(-5, 5, 10), np.array([5]*10), np.linspace(5, -5, 10), 0))
y = np.hstack((np.linspace(-5, 5, 10), np.array([5]*10), np.linspace(5, -5, 10), np.array([-5]*10), 0))
f = np.hstack(([0]*40, 1))
#raise NotImplementedError()
The following plot should show the points on the boundary and the single point in the interior:
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plt.scatter(x, y);
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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).
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xnew = np.linspace(-5.0, 5.0, 100)
ynew = np.linspace(-5.0, 5.0, 100)
Xnew, Ynew = np.meshgrid(xnew, ynew)
Fnew = griddata((x,y), f, (Xnew, Ynew), method='cubic', fill_value=0.0)
#raise NotImplementedError()
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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.
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plt.contour(Xnew, Ynew, Fnew)
#raise NotImplementedError()
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assert True # leave this to grade the plot