In [1]:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
sns.set_context("talk")
%matplotlib inline
In [2]:
import plotly.offline as py
py.init_notebook_mode(connected=False)

from IPython.core.display import display, HTML
# The polling here is to ensure that plotly.js has already been loaded before
# setting display alignment in order to avoid a race condition.
display(HTML(
    '<script>'
        'var waitForPlotly = setInterval( function() {'
            'if( typeof(window.Plotly) !== "undefined" ){'
                'MathJax.Hub.Config({ SVG: { font: "STIX-Web" }, displayAlign: "center" });'
                'MathJax.Hub.Queue(["setRenderer", MathJax.Hub, "SVG"]);'
                'clearInterval(waitForPlotly);'
            '}}, 5000 );'
    '</script>'
))
In [3]:
import plotly.graph_objs as go
import plotly.figure_factory as ff
import cufflinks as cf

cf.set_config_file(offline=False, world_readable=True, theme='ggplot')

This notebook which accompanies the lecture on the Bias Variance Tradeoff and Regularization.

Notebook created by Joseph E. Gonzalez for DS100.

Toy Dataset

As with the previous lectures we will continue to use an easy to visualize synthetic dataset.

In [4]:
np.random.seed(42)
n = 50
sigma = 10
X = np.linspace(-10, 10, n)
X = np.sort(X)
Y = 2. * X + 10. + sigma * np.random.randn(n) +  20*np.sin(X) + 0.8*(X)**2
X = X/5
data_points = go.Scatter(name="data", x=X, y=Y, mode='markers')
py.iplot([data_points])
In [14]:
rnd_seed = 42
np.random.seed(rnd_seed)
n = 50
sigma = 10
noise = np.random.normal(scale=sigma, size=n)
X = np.random.uniform(-10, 10, n)

# these are three different data models for us to play with, each of higher order
Y1 = np.sin(5*X) + 100 + 20. * X
Y2 = np.sin(5*X) + 100 + 20. * X  - 2 * X**2 + noise
Y5 = 70*np.sin(X) + 100 + 20. * X  - 2 * X**2 - 0.6 * X**3 + 0.004 * X**5 + noise

# Choose one of Y1, Y2 or Y5
Y = Y5

data_points = go.Scatter(name="data", x=X, y=Y, mode='markers')
py.iplot([data_points])

Making a Train-Test Split

It is always a good habit to split data into training and test sets.

In [15]:
## Train Test Split
from sklearn.model_selection import train_test_split 
X_tr, X_te, Y_tr, Y_te = train_test_split(X, Y, test_size=0.25, random_state=42)
train_points = go.Scatter(name="Train Data", 
                          x=X_tr, y=Y_tr, mode='markers',  marker=dict(color="blue", symbol="o"))
test_points = go.Scatter(name="Test Data",
                         x=X_te, y=Y_te, mode='markers', marker=dict(color="red", symbol="x"))
py.iplot([train_points, test_points])

Polynomial Features:

In the following we work through a basic bias variance tradeoff for different numbers of polynomial features. To make things a bit more interesting we will also add an additional sine feature:

$$ \large \Phi_d(x) = \left[\sin(\omega x), x, x^2, \ldots, x^d \right] $$

with the appropriate $\omega = 1$ value determined from prior knowledge (because we created the data).

We use the following Python function to generate a $\Phi$:

In [16]:
def poly_phi(k):
    return lambda X: np.array([np.sin(X)] + 
                              [X ** i for i in range(1, k+1)]).T

Trying different numbers of features

We will consider the following basis sizes:

In [17]:
basis_sizes = [1, 2, 5, 8, 16, 24, 32]

Create a $\Phi$ feature matrix for each number of basis:

In [18]:
Phis_tr = { k: poly_phi(k)(X_tr) for k in basis_sizes }

Train a model for each number of feature

In [19]:
from sklearn import linear_model
models = {}
for k in Phis_tr:
    model = linear_model.LinearRegression()
    model.fit(Phis_tr[k], Y_tr)
    models[k] = model

Visualize the various models

The following code is a bit complicated but essentially makes an interactive visualization of the various models.

In [20]:
# Make the x values where plot points will be generated
X_plt = np.linspace(np.min(X)-1, np.max(X)+1, 200)

# Generate the Plotly line objects by predicting the value at each X_plt
lines = []
for k in sorted(models.keys()):
    ytmp = models[k].predict(poly_phi(k)(X_plt))
    # Plotting software fails with large numbers
    ytmp[ytmp > 500] = 500
    ytmp[ytmp < -500] = -500
    lines.append(
        go.Scatter(name="degree "+ str(k), x=X_plt,
                   y = ytmp,visible=False))

# Construct steps for the interactive slider
lines[0].visible=True
steps = []
for i in range(len(lines)):
    step = dict(
        label = lines[i]['name'],
        method = 'restyle',
        args = ['visible', [False] * (len(lines)+1)],
    )
    step['args'][1][0] = True
    step['args'][1][i+1] = True # Toggle i'th trace to "visible"
    steps.append(step)

# Build the slider object
sliders = [dict(active = 0, pad = {"t": 20}, steps = steps)]

# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y)+5]),
                   sliders=sliders)
py.iplot(go.Figure(data = [train_points] + lines, layout=layout))
In [21]:
# Make the x values where plot points will be generated
X_plt = np.linspace(np.min(X)-1, np.max(X)+1, 200)

# Generate the Plotly line objects by predicting the value at each X_plt
lines = []
for k in sorted(models.keys()):
    ytmp = models[k].predict(poly_phi(k)(X_plt))
    # Plotting software fails with large numbers
    ytmp[ytmp > 500] = 500
    ytmp[ytmp < -500] = -500
    lines.append(
        go.Scatter(name="degree "+ str(k), x=X_plt,
                   y = ytmp,visible=False))

# Construct steps for the interactive slider
lines[0].visible=True
steps = []
for i in range(len(lines)):
    step = dict(
        label = lines[i]['name'],
        method = 'restyle',
        args = ['visible', [False] * (len(lines)+1)],
    )
    step['args'][1][0] = True
    step['args'][1][i+1] = True # Toggle i'th trace to "visible"
    steps.append(step)

# Build the slider object
sliders = [dict(active = 0, pad = {"t": 20}, steps = steps)]

# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y)+5]),
                   sliders=sliders)
py.iplot(go.Figure(data = [train_points] + lines, layout=layout))

Cross Validation

With the remaining training data:

  1. Split the remaining training dataset k-ways as in the Figure above. The figure uses 5-Fold which is standard. You should split the data in the same way you constructed the test set (this is typically randomly)
  2. For each split train the model on the training fraction and then compute the error (RMSE) on the validation fraction.
  3. Average the error across each validation fraction to estimate the test error.

Define the error Function

We will continue to use the mean squared error but this time rather than define it ourselves we will import it from scikit-learn.

In [22]:
from sklearn.metrics import mean_squared_error

In the following we compute the cross validated RMSE for different numbers of polynomial basis values

In [23]:
from sklearn.model_selection import KFold
kfold_splits = 5
kfold = KFold(kfold_splits, shuffle=True, random_state=42)
In [24]:
mse_scores = []
poly_degrees = np.arange(1,16)
for k in poly_degrees:
    Phi = poly_phi(k)(X_tr)
    
    # One step in k-fold cross validation
    def score_model(train_index, test_index):
        model = linear_model.LinearRegression()
        model.fit(Phi[train_index,], Y_tr[train_index])
        return mean_squared_error(Y_tr[test_index], 
                                  model.predict(Phi[test_index,]))
    

    score = np.mean([score_model(tr_ind, te_ind) 
                     for (tr_ind, te_ind) in kfold.split(Phi)])
    mse_scores.append(score)
    
    
rmse_scores = np.sqrt(np.array(mse_scores))
In [25]:
py.iplot(
    go.Figure(
        data=[go.Scatter(name="CV Curve", x=poly_degrees, y=rmse_scores),
              go.Scatter(name="Optimum", x=[poly_degrees[np.argmin(rmse_scores)]], y=[np.min(rmse_scores)],
                         mode="markers", marker=dict(color="red", size=10))
             ],
        layout=go.Layout(xaxis=dict(title="Degree"), yaxis=dict(title="CV RMSE"))))

Questions?

  1. Is 2 a reasonable answer given the data? (Look at the earlier plot.)
  2. Does this depend on the order in which we consider basis?







Variance

  1. What happens if we repeat the training procedure with different versions of our trainin data?

  2. How will the models change as we inrease the degree of our polynomial?






Here we use cross validation to simulate multiple train and test splits by splitting the training data into train and validation datasets. We then visualize each of the corresponding models.

In [26]:
from sklearn.model_selection import KFold

# Construct a KFold object which will define random index splits
kfold_splits = 5
kfold = KFold(kfold_splits, shuffle=True, random_state=42)

# Construct the lines
lines = []

# Make the test lines
X_plt = np.linspace(np.min(X)-1, np.max(X)+1, 200)

# for each train validation split (we ignore the validation data)
for k in basis_sizes:
    for (tr_ind, val_ind) in kfold.split(X_tr):
        # Construct the Phi matrices for each basis size
        Phi = poly_phi(k)(X_tr[tr_ind]) 
        # Fit the model
        model = linear_model.LinearRegression()
        model.fit(Phi, Y_tr[tr_ind])
        # Make predictions at the plotting points
        ytmp = model.predict(poly_phi(k)(X_plt))
        # Plotly has a bug and fails with large numbers
        ytmp[ytmp > 500] = 500
        ytmp[ytmp < -500] = -500
        line = go.Scatter(name="degree "+ str(k), visible=False,
                          x=X_plt, y=ytmp)
        lines.append(line)

for l in lines[0:kfold_splits]:
    l.visible=True

steps = []
for i in range(len(basis_sizes)):
    step = dict(
        label = "Degree " + str(basis_sizes[i]) ,
        method = 'restyle',
        args = ['visible', [False] * (len(lines)+1)]
    )
    step['args'][1][-1] = True
    for u in range(i*kfold_splits, (i+1)*kfold_splits):
        step['args'][1][u] = True # Toggle i'th trace to "visible"
    steps.append(step)
    
sliders = [dict(
    active = 0,
    pad = {"t": 20},
    steps = steps
)]

layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y) + 5]),
                   sliders=sliders,
                   showlegend=False)

py.iplot(go.Figure(data = lines + [train_points], layout=layout))





Regularization

In the previous notebook we adjusted the number of polynomial features to control model complexity and tradeoff bias and variance. However, this approach to managing model complexity has a few critical limitations:

  1. complexity varies discretely
  2. we may only need a few of the higher degree polynomial terms
  3. In general we may not have a natural way to order our basis

Rather than changing the dimension we can instead apply regularization to the weights. More generally, we can adopt the framework of regularized loss minimization.

$$ \large \hat{\theta} = \arg \min_\theta \frac{1}{n} \sum_{i=1}^n \textbf{Loss}\left(y_i, f_\theta(x_i)\right) + \lambda \textbf{R}(\theta) $$

The regularization term $\textbf{R}(\theta)$ penalizes for $\theta$ values that result in more complex and therefore higher variance models. The regularization parameter $\lambda$ determines the degree of regularization to apply and is typically determined through cross validation.






Ridge Regression

There are many forms for $\textbf{R}(\theta)$ but a common form is the squared $L^2$ norm of $\theta$.

$$\large \large \textbf{R}_{L^2}(\theta) = \large||\theta||_2^2 = \theta^T \theta = \sum_{k=1}^p \theta_k^2 $$

In the context of least squares regression this is often referred to as Ridge Regression with the objective:

$$ \large \hat{\theta} = \arg \min_\theta \frac{1}{n} \sum_{i=1}^n \left(y_i - f_\theta(x_i)\right)^2 + \lambda ||\theta||_2^2 $$

This is also sometimes called Tikhonov Regularization.

Deriving the optimal $\hat{\theta}$ with $L^2$ Regularization

We return to our linear model formulation:

$$ \large f_\theta(x) = x^T \theta $$

Using the standard matrix notation:

We can rewrite the objection

\begin{align}\large \hat{\theta}_{\text{L2}} = \arg\min_\theta \frac{1}{n}\left(Y - X \theta \right)^T \left(Y - X \theta \right) + \lambda \theta^T \theta \end{align}

Expanding the objective term:

\begin{align}\large L_\lambda(\theta) = \left(Y - X \theta \right)^T \left(Y - X \theta \right) + \lambda \theta^T \theta = \frac{1}{n} \left( Y^T Y - 2 Y^T X \theta + \theta^T X^T X \theta \right) + \lambda \theta^T \theta \end{align}

Taking the gradient with respect to $\theta$:

\begin{align} \large \nabla_\theta L_\lambda(\theta) & \large = \frac{1}{n} \left( \nabla_\theta Y^T Y - \nabla_\theta 2 Y^T X \theta + \nabla_\theta \theta^T X^T X \theta \right) + \nabla_\theta \lambda \theta^T \theta \\ & \large = \frac{1}{n} \left( 0 - 2 X^T Y + 2 X^T X \theta \right) + 2\lambda \theta \end{align}

The above gradient derivation uses the following identities:

  1. $\large \nabla_\theta \left( A \theta \right) = A^T$
  2. $\large \nabla_\theta \left( \theta^T A \theta \right) = A\theta + A^T \theta$ and $\large A = X^T X$ is symmetric

Setting the gradient equal to zero we get a regularized version of the normal equations:

$$\large (X^T X + n \lambda I) \theta = X^T Y $$

$$\large \theta = \left(X^T X + n \lambda I \right)^{-1} X^T Y $$

Optimal $\theta$ under $L^2$ regularization

Because $\lambda$ is a tuning parameter we often will absorb the $n$ into $\lambda$ and rewrite the above equations as:

$$\large (X^T X + \lambda I) \theta = X^T Y $$

$$\large \theta = \left(X^T X + \lambda I \right)^{-1} X^T Y $$

Notice: The addition of $\lambda I$ ensures that $X^T X + \lambda I$ is full rank. This addresses the earlier issue in least-squares regression when we had co-linear features.






How does $L^2$ Regularization Help

The $L^2$ penalty helps in several ways:

Manages Model Complexity

  1. It ensures that uninformative features weights are relatively small (near zero) mitigating the affect of those features.
  2. It evenly distributes weight over similar features to reduce variance.

Practical Concerns

  1. It removes degeneracy created by co-linear features
  2. It improves the numerical stability of








Visualizing $L^2$ Regularization

In the following we visualize the regularization surface. Notice that it pushes weights towards zero but is relatively smooth around the origin.

In [27]:
theta_range = np.linspace(-2,2,100) 
(u,v) = np.meshgrid(theta_range, theta_range)
w_values = np.vstack((u.flatten(), v.flatten())).T

def l2_sq_reg(w):
    return np.sum(w**2)
reg_values = [l2_sq_reg(w) for w in w_values]
reg_surface = go.Surface(
    x = u, y = v,
    z = np.reshape(reg_values, u.shape),
    contours=dict(z=dict(show=True))
)

# Axis labels
layout = go.Layout(
    scene=go.Scene(
        xaxis=go.XAxis(title='w0'),
        yaxis=go.YAxis(title='w1'),
        aspectratio=dict(x=2.,y=2., z=1.), 
        camera=dict(eye=dict(x=-2, y=-2, z=2))
    )
)
fig = go.Figure(data = [reg_surface], layout = layout)
py.iplot(fig)

Applying $L^2$ Regularization using Scikit Learn

In the following we use the linear_model.Ridge model in scikit learn. To demonstrate the efficacy of regularization we will use the degree 32 polynomials which substantially overfit the data.

In [28]:
maxpoly = 32
Phi = poly_phi(maxpoly)(X_tr)





Normalization and the Intercept

Before we proceed it is important that we appropriately normalize the data. Because the standard $L^2$ regularization methods treat each dimensional equivalently it is important that all dimensions are in the same range of values.

However, if we examine the polynomial features in $\Phi$ we notice that the distribution of values can be quite different for each dimension.

For example in the following we plot the degree 3 and degree 6 dimensions:

In [29]:
py.iplot(ff.create_distplot([Phi[:,2], Phi[:,3]], group_labels=['x^', 'x^2']))

Notice:

  1. difference in spread
  2. asymmetry






Standardizing the Data

A common transformation is to center and scale the features to zero mean and unit variance:

$$\large z = \frac{x - \mu}{\sigma} $$

This an be accomplished by applying the StandardScalar scikit learn preprocessor.

In [30]:
from sklearn.preprocessing import StandardScaler

normalizer = StandardScaler()
normalizer.fit(poly_phi(32)(X_tr))
Out[30]:
StandardScaler(copy=True, with_mean=True, with_std=True)
In [31]:
# we define the phi function for reuse in the future
def phi_fun(X):
    return normalizer.transform(poly_phi(32)(X))

Phi = phi_fun(X_tr)

Notice in the above code snippet we define a new $\Phi$ function that applies the pre-trained normalization. This procedure learns something about the training data in the formulation of the normalizer.

  1. Will this be an issue when we cross validate on the training data?

This process of transformations: feature construction, rescaling, and then subsequently model fitting form pipelines. Scikit learn actually has a pipeline framework to aid with this process.






In the following we plot the spread of the transformed dimensions. They are still not the same but are at least on the same scale.

In [32]:
py.iplot(ff.create_distplot([Phi[:,2], Phi[:,3]], group_labels=['x^2', 'x^3'], bin_size=0.3), filename="phi_dist_plot2")

Fitting the Ridge Regression Model

We are now finally ready to fit the ridge regression model. However, we haven't yet decided on a value for the regularization parameter $\lambda$. Therefore, we will try a range of values.

In [33]:
import sklearn.linear_model as linear_model
lam_values = np.hstack((np.logspace(-8,-1,10), 
                        np.logspace(-1,2,10),
                        np.logspace(2,10,10)))

models = [
    linear_model.Ridge(alpha = lam).fit(Phi, Y_tr)
    for lam in lam_values
]

Move the slider in the following plot to see the fit for various $\lambda$ values.

In [35]:
# Make the x values where plot points will be generated
X_plt = np.linspace(np.min(X)-1, np.max(X)+1, 200)

# Generate the Plotly line objects by predicting the value at each X_plt
lines = []
for k in range(len(models)):
    ytmp = models[k].predict(phi_fun(X_plt))
    # Plotting software fails with large numbers
    ytmp[ytmp > 500] = 500
    ytmp[ytmp < -500] = -500
    lines.append(
        go.Scatter(name="Lambda "+ str(lam_values[k]), 
                   x=X_plt, y = ytmp, visible=False))

# Construct steps for the interactive slider
lines[0].visible=True
steps = []
for i in range(len(lines)):
    step = dict(
        label = lines[i]['name'],
        method = 'restyle',
        args = ['visible', [False] * (len(lines)+1)],
    )
    step['args'][1][0] = True
    step['args'][1][i+1] = True # Toggle i'th trace to "visible"
    steps.append(step)

# Build the slider object
sliders = [dict(active = 0, pad = {"t": 20}, steps = steps)]

# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y) + 5]),
                   sliders=sliders,
                   showlegend=False)
py.iplot(go.Figure(data = [train_points] + lines, layout=layout), filename="ridge_regression_lines")

For large $\lambda$ values we see a smoother fit. Notice however that the model does not appear to perform well outside of the input data range. This is a common problem with polynomial feature transformations.

Using Cross Validation in the RidgeCV Model

Because cross validation is essential to determining the optimal regularization parameter there is built-in support for cross validation in linear_model.RidgeCV. Here we call the built-in cross validation routine passing the lambda values we wish to consider.

In [36]:
ridge_cv_model = linear_model.RidgeCV(alphas=lam_values, 
                                      store_cv_values=True)
# Fit the model to our training data
ridge_cv_model.fit(Phi, Y_tr)

# Plot the predicted model
ridge_cv_line = go.Scatter(name = "Ridge CV Curve",
                           x = X_plt,
                           y = ridge_cv_model.predict(phi_fun(X_plt)))
# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y)+5]))
py.iplot(go.Figure(data = [train_points, ridge_cv_line], layout=layout))
In [37]:
ridge_cv_loss = np.sqrt(np.mean(ridge_cv_model.cv_values_,axis=0))
py.iplot(
    go.Figure(
        data=[go.Scatter(name="CV Curve", x=lam_values, y=ridge_cv_loss),
              go.Scatter(name="Optimum", x=[lam_values[np.argmin(ridge_cv_loss)]], y=[np.min(ridge_cv_loss)],
                         mode="markers", marker=dict(color="red", size=10))
             ],
        layout=go.Layout(xaxis=dict(title="Lambda", type="log"), 
                         yaxis=dict(title="CV RMSE", range=[10,70]))),
    filename="ridge_cv_model_curve")

Question: What is going on with small $\lambda$ values?

$L^1$ Regularization (Lasso)

Another common regularization function is the sum of the absolute values:

$$\large \large \textbf{R}_{L^1}(\theta) = \sum_{k=1}^p |\theta_k| $$

This is called $L^1$ regularization as it corresponds to the $L^1$ norm. Least squares linear regression in conjunction with the $L^1$ norm is often called the Lasso (Least Absolute Shrinkage and Selection Operator).

In contrast to the $L^2$ norm the $L^1$ norm encourages $\theta_i$ values to be exactly zero in less informative dimensions thereby reducing model complexity. To see how the $L^1$ encourages sparsity consider the following illustration:

In the above figures we plot the loss for settings of a two dimensional ($\theta_1$ and $\theta_2$) model as the elliptical contours. Without regularization the solution would be at the center of the contours. By imposing regularization we constrain the solution to living in the "norm ball" centered at the origin (all zero theta vector). As we increase $\lambda$ we actually shrink the ball. Unlike the $L^2$ solutions in the $L^1$ will often "slide to the corners" which are aligned with axis causing subsets of the $\theta$ vector to be exactly zero.

In some settings a compromise can be achieved by using both the $L^2$ and $L^1$ norms to encourage sparsity while ensuring relatively co-linear features are given equal weight (to improve robustness).

In [38]:
theta_range = np.linspace(-2,2,100) 
(u,v) = np.meshgrid(theta_range, theta_range)
w_values = np.vstack((u.flatten(), v.flatten())).T

def l1_reg(w):
    return np.sum(np.abs(w))
reg_values = [l1_reg(w) for w in w_values]
reg_surface = go.Surface(
    x = u, y = v,
    z = np.reshape(reg_values, u.shape),
    contours=dict(z=dict(show=True))
)

# Axis labels
layout = go.Layout(
    scene=go.Scene(
        xaxis=go.XAxis(title='w0'),
        yaxis=go.YAxis(title='w1'),
        aspectratio=dict(x=2.,y=2., z=1.), 
        camera=dict(eye=dict(x=-2, y=-2, z=2))
    )
)
fig = go.Figure(data = [reg_surface], layout = layout)
py.iplot(fig)

$L^1$ regularized regression in scikit-learn

In the following we use the scikit-learn Lasso package. As before we will try a range of values for the regularization parameter.

In [39]:
lam_values = np.logspace(-1.3,2.5,20)
models = []
for lam in lam_values:
    model = linear_model.Lasso(alpha = lam, max_iter=100000)
    model.fit(Phi, Y_tr)
    models.append(model)

Again we can plot the fit for different regularization penalties.

In [40]:
# Make the x values where plot points will be generated
X_plt = np.linspace(np.min(X)-1, np.max(X)+1, 200)

# Generate the Plotly line objects by predicting the value at each X_plt
lines = []
# Make the full polynomial
poly = np.array([r"\theta_0 \sin(x)"] + [r" \theta_{" + str(d) + "} x^{"+str(d)+"} " for d in range(1, 33)])


for k in range(len(models)):
    ytmp = models[k].predict(phi_fun(X_plt))
    # Plotting software fails with large numbers
    ytmp[ytmp > 500] = 500
    ytmp[ytmp < -500] = -500
    num_features = np.sum(~np.isclose(models[k].coef_, 0.))
    # get all the nonzer terms
    #     non_zero_terms = ~np.isclose(models[k].coef_,0)
    #     poly_str = "$" +("+".join(poly[non_zero_terms])) + "$"
    lines.append(
        go.Scatter(name=(
            "Lambda "+ str(lam_values[k]) + 
            " num features = " + str(num_features)
            + " out of " + str(len(models[k].coef_))), 
                   x=X_plt, y = ytmp, visible=False))

# Construct steps for the interactive slider
lines[0].visible=True
steps = []
for i in range(len(lines)):
    step = dict(
        label = lines[i]['name'],
        method = 'restyle',
        args = ['visible', [False] * (len(lines)+1)],
    )
    step['args'][1][0] = True
    step['args'][1][i+1] = True # Toggle i'th trace to "visible"
    steps.append(step)

# Build the slider object
sliders = [dict(active = 0, pad = {"t": 20}, steps = steps)]

# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y) + 5]),
                   sliders=sliders,
                   showlegend=False)
py.iplot(go.Figure(data = [train_points] + lines, layout=layout), filename="lasso_regression_lines")

Notice:

What happens in the above plot for larger values of $\lambda$?







Cross Validated Solution

As with Ridge regression, scikit-learn provides support for cross validation directly in the Lasso model training procedure. In the following we LassoCV to determine the best regularization parameter.

In [41]:
lasso_cv_model = linear_model.LassoCV(alphas=lam_values, max_iter=1000000)
# Fit the model to our training data
lasso_cv_model.fit(Phi, Y_tr)

# Plot the predicted model
lasso_cv_line = go.Scatter(name = "Lasso CV Curve",
                           x = X_plt,
                           y = lasso_cv_model.predict(phi_fun(X_plt)))
# render the plot
layout = go.Layout(xaxis=dict(range=[np.min(X_plt), np.max(X_plt)]), 
                   yaxis=dict(range=[np.min(Y) -5 , np.max(Y)+5]))
py.iplot(go.Figure(data = [train_points, lasso_cv_line], layout=layout), filename="lasso_cv_line")

Let's look at the polynomial terms with non-zero $\theta$

In [47]:
from IPython.display import display, Markdown 

# get all the nonzer terms
non_zero_terms = ~np.isclose(lasso_cv_model.coef_,0)

# Make the full polynomial
poly = np.array([r"\theta_0 \sin(x)"] + 
                [r" \theta_{" + str(d) + "} x^{"+str(d)+"} " for d in range(1, 33)])

# Print only the nonzero terms
display(Markdown(r"$\large" + ("+".join(poly[non_zero_terms])) + "$"))

$\large\theta_0 \sin(x)+ \theta_{1} x^{1} + \theta_{2} x^{2} + \theta_{3} x^{3} + \theta_{4} x^{4} + \theta_{7} x^{7} + \theta_{9} x^{9} + \theta_{14} x^{14} + \theta_{25} x^{25} + \theta_{27} x^{27} $

In [48]:
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import KFold


kfold_splits = 5
kfold = KFold(kfold_splits, shuffle=True, random_state=42)

mse_scores = []
for lam in lam_values:
    # One step in k-fold cross validation
    def score_model(train_index, test_index):
        model = linear_model.Lasso(alpha=lam, max_iter=1000000)
        model.fit(Phi[train_index,:], Y_tr[train_index])
        return mean_squared_error(Y_tr[test_index], model.predict(Phi[test_index,]))
    
    mse_score = np.mean([score_model(tr_ind, te_ind) 
                     for (tr_ind, te_ind) in kfold.split(Phi)])
    mse_scores.append(mse_score)
rmse_scores = np.sqrt(np.array(mse_scores))


py.iplot(
    go.Figure(
        data=[go.Scatter(name="CV Curve", x=lam_values, y=rmse_scores),
              go.Scatter(name="Optimum", x=[lam_values[np.argmin(rmse_scores)]], y=[np.min(rmse_scores)],
                         mode="markers", marker=dict(color="red", size=10))
             ],
        layout=go.Layout(xaxis=dict(title="Lambda",type="log",range=[-1.2,1.5]), 
                         yaxis=dict(title="CV RMSE", range=[5,30]))),
    filename="lasso_cv_model_curve")