# 16.4. Regularization¶

Cross-validation helps you find a dimension for a fitted model that balances under- and over-fitting. Rather than selecting the dimension of the model, we can instead, build a large model but restrict the size of the coefficients. To do this, we penalize the average square error from over-fitting by adding a penalty term on the size of the coefficients. We fit the model by minimizing the combination of mean squared error with a regularization term:

\begin{aligned} \frac{1}{n} \sum_{i=1}^{n}(y_i - \mathbf{x}_i \boldsymbol{\theta})^2 ~+~ \lambda \sum_{j = 1}^{p} \theta_j^2 \end{aligned}

The regularization term $$\lambda \sum_{j = 1}^{p} \theta_j^2$$ penalizes large coefficients. The regularization parameter $$\lambda$$ determines the size of the penalty to apply, and we typically choose it through cross validation.

Penalizing the square of the coefficients, is called $$L_2$$ regularization, and an $$L_2$$ regularized linear model is called ridge regression. Another popular regularization penalizes the absolute size of the coefficients. The $$L_1$$ regularized linear model is called Lasso regression. (Lasso stands for Least Absolute Shrinkage and Selection Operator.)

\begin{aligned} \frac{1}{n} \sum_{i=1}^{n}(y_i - \mathbf{x}_i \boldsymbol{\theta})^2 ~+~ \lambda \sum_{j = 1}^{p} |\theta_j| \end{aligned}

We can think about what happens to this minimization when $$\lambda$$ is really large. The coefficients are heavily penalized so they will shrink to 0. On the other hand, when $$\lambda$$ is tiny, then the coefficients can take on any value. In fact, when $$\lambda$$, we’re back in the world on ordinary least squares. A couple of issue crop up when we think about controlling the size of the coefficients.

• We do not want to regularize the intercept term. This way, a large $$\lambda$$ fits a constant model.

• When features have very different scales, the penalty can impact them differently, with large-valued features being penalized more than others. To avoid this, we standardize all of the features to have mean 0 and variance 1.

## 16.4.1. Regularizing a Market Analysis¶

A market research project for a pharmaceutical company wanted to model consumer interest in purchasing a cold sore healthcare product. They gathered data from 1023 consumers. Each consumer was asked to rate on a 10-point Likert scale 35 factors according to whether the factor matters to them when considering purchasing a cold sore treatment. They also rated their interest in purchasing the product.1

ma_df = pd.read_csv('data/market-analysis.csv')


The table below contains the 35 factors and their correlation to their interest in purchasing the product.

Corr

Description

Corr

Description

x1

0.70

provides soothing relief

x19

0.54

has a non-messy application

x2

0.58

moisturizes cold sore blister

x20

0.70

good for any stage of a cold

x3

0.69

provides long-lasting relief

x21

0.49

easy to apply/take

x4

0.70

provides fast-acting relief

x22

0.52

package keeps from contamination

x5

0.72

shortens duration of a cold

x23

0.57

easy to dispense a right amount

x6

0.68

x24

0.63

worth the price it costs

x7

0.67

dries up cold sore

x25

0.57

recommended most by pharamacists

x8

0.72

heals fast

x26

0.54

recommended by doctors

x9

0.72

penetrates deep

x27

0.54

FDA approved

x10

0.65

relieves pain

x28

0.64

a brand I trust

x11

0.61

prevents cold

x29

0.60

clinically proven

x12

0.73

prevents from getting worse

x30

0.68

a brand I would recommend

x13

0.57

medicated

x31

0.74

an effective treatment

x14

0.61

prescription strength

x32

0.37

portable

x15

0.63

repairs damaged skin

x33

0.37

discreet packaging

x16

0.67

x34

0.55

helps conceal cold sores

x17

0.42

contains SPF

x35

0.63

absorbs quickly

x18

0.57

non-irritating

Some of these 35 features appear to be measuring similar aspects of desirability, based on their labels. We can compute the correlations between the explanatory variables to confirm this observation.

ma_df.corr()

y x1 x2 x3 ... x32 x33 x34 x35
y 1.00 0.70 0.58 0.69 ... 0.37 0.37 0.55 0.63
x1 0.70 1.00 0.69 0.74 ... 0.42 0.33 0.57 0.71
x2 0.58 0.69 1.00 0.63 ... 0.43 0.36 0.55 0.63
... ... ... ... ... ... ... ... ... ...
x33 0.37 0.33 0.36 0.34 ... 0.34 1.00 0.35 0.36
x34 0.55 0.57 0.55 0.67 ... 0.16 0.35 1.00 0.54
x35 0.63 0.71 0.63 0.70 ... 0.46 0.36 0.54 1.00

36 rows × 36 columns

We see, for example, the last feature, “absorbs quickly”, is highly correlated with the first three: “provides soothing relief”, “moisturizes”, and “provides long-lasting relief”.

Before we start to fit regularized models, we set up the design matrix and output vector and split the data into train and test sets. We set aside 200 observations for the train set.

y = ma_df[['y']]
X = ma_df.drop(['y'], axis=1)

test_size = 200

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=test_size, random_state=42)


As mentioned earlier, we need to standardize the features. We standardize the train set, and then, when we go to test the model, we use the train set standardization on the test set. The StandardScaler method helps us with this process.

from sklearn.preprocessing import StandardScaler

scalerX = StandardScaler().fit(X_train)
X_train_scaled = scalerX.transform(X_train)
X_test_scaled = scalerX.transform(X_test)


The means of each of the 35 features in the train set are all 0 and the SDs are all 1.

np.mean(X_train_scaled, axis=0)

array([ 0.,  0., -0., ..., -0.,  0.,  0.])

np.std(X_train_scaled, axis=0)

array([1., 1., 1., ..., 1., 1., 1.])


This is not the case for the test set, because we have used the averages and SDs from train set to standardize them.

np.mean(X_test_scaled, axis=0)

array([-0.17, -0.12, -0.15, ..., -0.12, -0.1 , -0.24])


We perform a LASSO regression on the train set, setting $$\lambda$$ (represented as the parameter alpha in the call)

from sklearn.linear_model import Lasso

lasso_model1 = Lasso(alpha=0.1)
lasso_model1.fit(X_train_scaled, y_train)

lasso_model1.coef_[0:20]

array([0.06, 0.  , 0.01, 0.  , 0.17, 0.  , 0.12, 0.02, 0.01, 0.  , 0.  ,
0.24, 0.  , 0.  , 0.16, 0.  , 0.  , 0.  , 0.07, 0.02])


We see that several of the coefficients are 0. This is a property of LASSO regression that many find appealing. The nature of $$L_1$$ distance leads to coefficients being zeroed out, which can be viewed as a model selection technique.

sum(lasso_model1.coef_ > 0)

15


Twenty of the 35 coefficients are 0 in this model.

Let’s see how the coefficients change with $$\lambda$$.

lasso_model = Lasso()
coefs = []
mses = []
alphas = np.arange(0.01, 2, 0.01)

for a in alphas:
lasso_model.set_params(alpha = a)
lasso_model.fit(X_train_scaled,y_train)
coefs.append(lasso_model.coef_)
mses.append(mean_squared_error(y_test,
lasso_model.predict(X_test_scaled)))


As $$\lambda$$ increases, the model fitting is penalized more heavily and many coefficients shrink to 0. The left hand side of the plot shows greater model complexity, corresponds to small $$\lambda$$. The model that we first fit was for $$\lambda = 0.1$$, and we can see from this plot which coefficients are non-zero. The largest coefficients are for “x12”, “x30”, “x5”, “x15”, “x31”, “x28”, and “x7”.

fig = px.line(x = alphas, y = mses,
labels={"x": "lambda",
"y": "MSE"
})

fig.update_layout(width=350, height=250)
fig.show()


By the time $$\lambda$$ is about 1.25, the penalty is so large that the LASSO regression is fitting a constant model to the data, and so the MSE remains the same. We are again faced with a model selection question, but this time it’s in the form of deciding which $$\lambda$$. We can use cross-validation to help us.

from sklearn.linear_model import LassoCV

lasso_cv_model = LassoCV(alphas = np.arange(0.01, 1, 0.01),
cv = 5, max_iter = 100000)

lasso_best = lasso_cv_model.fit(X_train_scaled, np.squeeze(y_train))


The penalty for the cross-validated model is:

lasso_best.alpha_

0.04


And we can compute the MSE for predictions in the test set using the cross-validated model.

y_test_pred = lasso_best.predict(X_test_scaled)
mean_squared_error(y_test,y_test_pred)

1.0957444655340487

sum(lasso_best.coef_ > 0)

16


Similar to the model that we fitted first, there are 19 coefficients that are 0 in this model.

Next, let’s try using an $$L_2$$ penalty and perform Ridge regression. We’ll make a plot of the coefficients as $$\lambda$$ varies.

from sklearn.linear_model import Ridge

ridge = Ridge( )
coefsR = []
alphasR = np.arange(1, 1001, 25)

for a in alphasR:
ridge.set_params(alpha = a)
ridge.fit(X_train_scaled, y_train)
coefsR.append(ridge.coef_)


This plot shows a very different shape compared to the LASSO coefficient plot. The coefficients do not disappear entirely. The plots do have an important similarity in that as the penalty increases the coefficients shrink. In a way, with Ridge regression, we are using all of the variables a little bit. Again, we are faced with the question of which $$\lambda$$ to use, and we turn to cross-validation to help.

from sklearn.linear_model import RidgeCV

ridgecv = RidgeCV(alphas = alphasR, cv=10)
ridgecv.fit(X_train_scaled, y_train)

RidgeCV(alphas=array([  1,  26,  51, ..., 926, 951, 976]), cv=10)


We use the cross-validated choice for $$\lambda$$ to fit the Ridge regression.

ridge_best = Ridge(alpha = ridgecv.alpha_)
ridge_best.fit(X_train_scaled, y_train)
mean_squared_error(y_test, ridge_best.predict(X_test_scaled))

1.1257716762997172


The MSE is larger than the MSE for LASSO. Many coefficients, while not 0, are quite small. We can count the number of coefficients larger than 0.05.

sum(np.squeeze(ridge_best.coef_ > 0.05))

14


Similar to LASSO, roughly 20 of the features have tiny coefficients. We can compare the coefficients from the two regularized models.

ridge_best.coef_

array([[0.06, 0.02, 0.05, ..., 0.04, 0.03, 0.02]])

lasso_best.coef_[: 20]

array([ 0.07, -0.  ,  0.01,  0.  ,  0.18,  0.  ,  0.12,  0.03,  0.  ,
0.  ,  0.02,  0.23,  0.  ,  0.  ,  0.17,  0.  ,  0.  ,  0.  ,
0.08,  0.02])


Using $$L_1$$ or $$L_2$$ regularization, allows us to avoid model over-fitting by penalizing large coefficients. $$L_1$$ regularization has the advantage of zeroing out coefficients. That is, Lasso regression performs feature selection—it discards a subset of the original features when fitting model parameters. This is particularly useful when working with high-dimensional data with many features. A model that only uses a few features to make a prediction will run much faster than a model that requires many calculations. Since unneeded features tend to increase model variance without decreasing bias, we can sometimes increase the accuracy of other models by using lasso regression to select a subset of features to use.

Sometimes we prefer one type of regularization over the other because it maps more closely to the domain we are working with. For example, if know that the phenomenon we are trying to model results from many small factors, we might prefer ridge regression because it won’t discard these factors. On the other hand, some outcomes result from a few highly influential features. We prefer lasso regression in these situations because it will discard unneeded features.

1

Lipovetsky & Conklin (2015) Predictor relative importance and matching regression parameters, Journal of Applied Statistics, 42:5, 1017-1031, https://doi.org/10.1080/02664763.2014.994480