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
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.
In [3]:
xb = np.array([-5,-4,-3,-2,-1,0,1,2,3,4,5])
yb = np.array([-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5])
yt = np.array([5]*11)
yc = np.array(0)
x = np.hstack((xb,xb,yb[1:10],yt[1:10],yc))
y = np.hstack((yb,yt,xb[1:10],xb[1:10],yc))
f1 = np.array([0]*40)
f2 = [1]
f = np.hstack((f1,f2))
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);
In [5]:
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
).
In [6]:
xnew = np.array[-5,5,100]
ynew = np.array[-5,5,100]
Xnew,Ynew = np.meshgrid(xnew,ynew)
Fnew = griddata((x,y),f,(Xnew,Ynew), method = 'cubic')
In [7]:
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 [8]:
In [ ]:
assert True # leave this to grade the plot