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%matplotlib inline
%load_ext autoreload
%autoreload 2
import numpy as np
import matplotlib.pyplot as plt
Here we construct a filter, $F$, such that
$$F\left(x\right) = e^{-|x|} \cos{\left(2\pi x\right)} $$We want to show that, if $F$ is used to generate sample calibration data for the MKS, then the calculated influence coefficients are in fact just $F$.
In [2]:
x0 = -10.
x1 = 10.
x = np.linspace(x0, x1, 1000)
def F(x):
return np.exp(-abs(x)) * np.cos(2 * np.pi * x)
p = plt.plot(x, F(x), color='#1a9850')
Next we generate the sample data (X, y) using
scipy.ndimage.convolve. This performs the convolution
for each sample.
In [3]:
import scipy.ndimage
n_space = 101
n_sample = 50
np.random.seed(201)
x = np.linspace(x0, x1, n_space)
X = np.random.random((n_sample, n_space))
y = np.array([scipy.ndimage.convolve(xx, F(x), mode='wrap') for xx in X])
For this problem, a basis is unnecessary, as no discretization is
required in order to reproduce the convolution with the MKS localization. Using
the ContinuousIndicatorBasis with n_states=2 is the equivalent of a
non-discretized convolution in space.
In [4]:
from pymks import MKSLocalizationModel
from pymks import PrimitiveBasis
prim_basis = PrimitiveBasis(n_states=2, domain=[0, 1])
model = MKSLocalizationModel(basis=prim_basis)
Fit the model using the data generated by $F$.
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model.fit(X, y)
To check for internal consistency, we can compare the predicted output with the original for a few values
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y_pred = model.predict(X)
print y[0, :4]
print y_pred[0, :4]
With a slight linear manipulation of the coefficients, they agree perfectly with the shape of the filter, $F$.
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plt.plot(x, F(x), label=r'$F$', color='#1a9850')
plt.plot(x, -model.coeff[:,0] + model.coeff[:, 1],
'k--', label=r'$\alpha$')
l = plt.legend()
Some manipulation of the coefficients is required to reproduce the filter. Remember the convolution for the MKS is
$$ p \left[s\right] = \sum_{l=0}^{L-1} \sum_{r=0}^{S - 1} \alpha[l, r] m[l, s - r] $$However, when the primitive basis is selected, the MKSLocalizationModel solves a modified form of this. There are always redundant coefficients since
Thus, the regression in Fourier space must be done with categorical variables, and the regression takes the following form:
$$ \begin{split} p [s] & = \sum_{l=0}^{L - 1} \sum_{r=0}^{S - 1} \alpha[l, r] m[l, s -r] \\ P [k] & = \sum_{l=0}^{L - 1} \beta[l, k] M[l, k] \\ &= \beta[0, k] M[0, k] + \beta[1, k] M[1, k] \end{split} $$where
$$\beta[0, k] = \begin{cases} \langle F(x) \rangle ,& \text{if } k = 0\\ 0, & \text{otherwise} \end{cases} $$This removes the redundancies from the regression, and we can reproduce the filter.
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