Linear Regression¶

This example shows how to use the pylops.LinearRegression operator to perform Linear regression analysis.

In short, linear regression is the problem of finding the best fitting coefficients, namely intercept $$\mathbf{x_0}$$ and gradient $$\mathbf{x_1}$$, for this equation:

$y_i = x_0 + x_1 t_i \qquad \forall i=0,1,\ldots,N-1$

As we can express this problem in a matrix form:

$\mathbf{y}= \mathbf{A} \mathbf{x}$

our solution can be obtained by solving the following optimization problem:

$J= \|\mathbf{y} - \mathbf{A} \mathbf{x}\|_2$

See documentation of pylops.LinearRegression for more detailed definition of the forward problem.

import matplotlib.pyplot as plt
import numpy as np

import pylops

plt.close("all")
np.random.seed(10)


Define the input parameters: number of samples along the t-axis (N), linear regression coefficients (x), and standard deviation of noise to be added to data (sigma).

N = 30
x = np.array([1.0, 2.0])
sigma = 1


Let’s create the time axis and initialize the pylops.LinearRegression operator

t = np.arange(N, dtype="float64")
LRop = pylops.LinearRegression(t, dtype="float64")


We can then apply the operator in forward mode to compute our data points along the x-axis (y). We will also generate some random gaussian noise and create a noisy version of the data (yn).

y = LRop * x
yn = y + np.random.normal(0, sigma, N)


We are now ready to solve our problem. As we are using an operator from the pylops.LinearOperator family, we can simply use /, which in this case will solve the system by means of an iterative solver (i.e., scipy.sparse.linalg.lsqr).

xest = LRop / y
xnest = LRop / yn


Let’s plot the best fitting line for the case of noise free and noisy data

plt.figure(figsize=(5, 7))
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * x[1] + x[0],
"k",
lw=4,
label=r"true: $x_0$ = %.2f, $x_1$ = %.2f" % (x[0], x[1]),
)
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * xest[1] + xest[0],
"--r",
lw=4,
label=r"est noise-free: $x_0$ = %.2f, $x_1$ = %.2f" % (xest[0], xest[1]),
)
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * xnest[1] + xnest[0],
"--g",
lw=4,
label=r"est noisy: $x_0$ = %.2f, $x_1$ = %.2f" % (xnest[0], xnest[1]),
)
plt.scatter(t, y, c="r", s=70)
plt.scatter(t, yn, c="g", s=70)
plt.legend()


Out:

<matplotlib.legend.Legend object at 0x7f770be0ea58>


Once that we have estimated the best fitting coefficients $$\mathbf{x}$$ we can now use them to compute the y values for a different set of values along the t-axis.

t1 = np.linspace(-N, N, 2 * N, dtype="float64")
y1 = LRop.apply(t1, xest)

plt.figure(figsize=(5, 7))
plt.plot(t, y, "k", label="Original axis")
plt.plot(t1, y1, "r", label="New axis")
plt.scatter(t, y, c="k", s=70)
plt.scatter(t1, y1, c="r", s=40)
plt.legend()


Out:

<matplotlib.legend.Legend object at 0x7f770bcd7320>


We consider now the case where some of the observations have large errors. Such elements are generally referred to as outliers and can affect the quality of the least-squares solution if not treated with care. In this example we will see how using a L1 solver such as pylops.optimization.sparsity.IRLS can drammatically improve the quality of the estimation of intercept and gradient.

# Add outliers
yn[1] += 40
yn[N - 2] -= 20

# IRLS
nouter = 20
epsR = 1e-2
epsI = 0
tolIRLS = 1e-2

xnest = LRop / yn
xirls, nouter, xirls_hist, rw_hist = pylops.optimization.sparsity.IRLS(
LRop,
yn,
nouter,
threshR=False,
epsR=epsR,
epsI=epsI,
tolIRLS=tolIRLS,
returnhistory=True,
)
print("IRLS converged at %d iterations..." % nouter)

plt.figure(figsize=(5, 7))
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * x[1] + x[0],
"k",
lw=4,
label=r"true: $x_0$ = %.2f, $x_1$ = %.2f" % (x[0], x[1]),
)
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * xnest[1] + xnest[0],
"--r",
lw=4,
label=r"L2: $x_0$ = %.2f, $x_1$ = %.2f" % (xnest[0], xnest[1]),
)
plt.plot(
np.array([t.min(), t.max()]),
np.array([t.min(), t.max()]) * xirls[1] + xirls[0],
"--g",
lw=4,
label=r"L1 - IRSL: $x_0$ = %.2f, $x_1$ = %.2f" % (xirls[0], xirls[1]),
)
plt.scatter(t, y, c="r", s=70)
plt.scatter(t, yn, c="g", s=70)
plt.legend()


Out:

IRLS converged at 14 iterations...

<matplotlib.legend.Legend object at 0x7f770bc0cbe0>


Let’s finally take a look at the convergence of IRLS. First we visualize the evolution of intercept and gradient

fig, axs = plt.subplots(2, 1, figsize=(8, 10))
fig.suptitle("IRLS evolution", fontsize=14, fontweight="bold", y=0.95)
axs[0].plot(xirls_hist[:, 0], xirls_hist[:, 1], ".-k", lw=2, ms=20)
axs[0].scatter(x[0], x[1], c="r", s=70)
axs[0].set_title("Intercept and gradient")
axs[0].grid()
for iiter in range(nouter):
axs[1].semilogy(
rw_hist[iiter],
color=(iiter / nouter, iiter / nouter, iiter / nouter),
label="iter%d" % iiter,
)
axs[1].set_title("Weights")
axs[1].legend(loc=5, fontsize="small")
plt.tight_layout()
plt.subplots_adjust(top=0.8)


Total running time of the script: ( 0 minutes 1.067 seconds)

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