Lasso Regression

# 1 Lasso Regression Basics

Lasso performs a so called L1 regularization (a process of introducing additional information in order to prevent overfitting), i.e. adds penalty equivalent to absolute value of the magnitude of coefficients.

In particular, the minimization objective does not only include the residual sum of squares (RSS) - like in the OLS regression setting - but also the sum of the absolute value of coefficients.

The residual sum of squares (RSS) is calculated as follows:

$RSS = \sum_{i=1}^{n} (y_i - \hat{y_i})^2$

This formula can be stated as:

$RSS = \sum_{i=1}^{n} \bigg(y_i - \big( \beta_{0} + \sum_{j=1}^{p} \beta_{j} x_{ij} \big) \bigg)^2$

• n represents the number of distinct data points, or observations, in our sample.
• p denotes the number of variables that are available in the dataset.
• x_{ij} represents the value of the jth variable for the ith observation, where i = 1, 2, . . ., n and j = 1, 2, . . . , p.

In the lasso regression, the minimization objective becomes:

$\sum_{i=1}^{n} \bigg(y_i - \big( \beta_{0} + \sum_{j=1}^{p} \beta_{j} x_{ij} \big) \bigg)^2 + \lambda \sum_{j=1}^{p} |\beta_j|$

which equals:

$RSS + \lambda \sum_{j=1}^{p} |\beta_j|$

$$\lambda$$ (lambda) provides a trade-off between balancing RSS and magnitude of coefficients.

$$\lambda$$ can take various values:

• $$\lambda$$ = 0: Same coefficients as simple linear regression
• $$\lambda$$ = ∞: All coefficients zero (same logic as before)
• 0 < $$\lambda$$ < ∞: coefficients between 0 and that of simple linear regression

# 2 Implementation of Lasso regression

Python set up:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline
plt.style.use('ggplot')
import warnings; warnings.simplefilter('ignore')

This notebook involves the use of the Lasso regression on the “Auto” dataset. In particular, we only use observations 1 to 200 for our analysis. Furthermore, you can drop the name variable.

Import data:

df = pd.read_csv("https://raw.githubusercontent.com/kirenz/datasets/master/Auto.csv")

Tidying data:

df = df.iloc[0:200]
df = df.drop(['name'], axis=1)
df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 200 entries, 0 to 199
Data columns (total 8 columns):
mpg             200 non-null float64
cylinders       200 non-null int64
displacement    200 non-null float64
horsepower      200 non-null object
weight          200 non-null int64
acceleration    200 non-null float64
year            200 non-null int64
origin          200 non-null int64
dtypes: float64(3), int64(4), object(1)
memory usage: 12.6+ KB
df['origin'] = pd.Categorical(df['origin'])
df['horsepower'] = pd.to_numeric(df['horsepower'], errors='coerce')
print(df.isnull().sum())
mpg             0
cylinders       0
displacement    0
horsepower      2
weight          0
acceleration    0
year            0
origin          0
dtype: int64
# drop missing cases
df = df.dropna()

We use scikit learn to fit a Lasso regression (see documentation) and follow a number of steps (note that scikit-learn uses $$\alpha$$ instead of $$\lambda$$ in their notation):

## 2.1 Standardization

Standardize the features with the module: from sklearn.preprocessing import StandardScaler

It is important to standardize the features by removing the mean and scaling to unit variance. The L1 (Lasso) and L2 (Ridge) regularizers of linear models assume that all features are centered around 0 and have variance in the same order. If a feature has a variance that is orders of magnitude larger that others, it might dominate the objective function and make the estimator unable to learn from other features correctly as expected.

dfs = df.astype('int')
dfs.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 198 entries, 0 to 199
Data columns (total 8 columns):
mpg             198 non-null int64
cylinders       198 non-null int64
displacement    198 non-null int64
horsepower      198 non-null int64
weight          198 non-null int64
acceleration    198 non-null int64
year            198 non-null int64
origin          198 non-null int64
dtypes: int64(8)
memory usage: 13.9 KB
dfs.columns
Index(['mpg', 'cylinders', 'displacement', 'horsepower', 'weight',
'acceleration', 'year', 'origin'],
dtype='object')
from sklearn.preprocessing import StandardScaler

scaler = StandardScaler()
dfs[['cylinders', 'displacement', 'horsepower',
'weight', 'acceleration', 'year', 'origin']] = scaler.fit_transform(dfs[['cylinders',
'displacement',
'horsepower',
'weight',
'acceleration',
'year', 'origin']])
dfs.head(5)
mpg cylinders displacement horsepower weight acceleration year origin
0 18 1.179744 0.726091 0.325216 0.346138 -0.955578 -1.516818 -0.629372
1 15 1.179744 1.100254 1.129264 0.548389 -1.305309 -1.516818 -0.629372
2 18 1.179744 0.821807 0.784672 0.273370 -1.305309 -1.516818 -0.629372
3 16 1.179744 0.699986 0.784672 0.270160 -0.955578 -1.516818 -0.629372
4 17 1.179744 0.682583 0.554944 0.287282 -1.655041 -1.516818 -0.629372

## 2.2 Split data

Split the data set into train and test sets (use X_train, X_test, y_train, y_test), with the first 75% of the data for training and the remaining for testing. (module: from sklearn.model_selection import train_test_split)

X = dfs.drop(['mpg'], axis=1)
y = dfs['mpg']
from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=10)

## 2.3 Lasso regression

Apply Lasso regression on the training set with the regularization parameter lambda = 0.5 (module: from sklearn.linear_model import Lasso) and print the $$R^2$$-score for the training and test set. Comment on your findings.

from sklearn.linear_model import Lasso

reg = Lasso(alpha=0.5)
reg.fit(X_train, y_train)

Lasso(alpha=0.5, copy_X=True, fit_intercept=True, max_iter=1000, normalize=False, positive=False, precompute=False, random_state=None, selection=‘cyclic’, tol=0.0001, warm_start=False)

print('Lasso Regression: R^2 score on training set', reg.score(X_train, y_train)*100)
print('Lasso Regression: R^2 score on test set', reg.score(X_test, y_test)*100)

Lasso Regression: R^2 score on training set 82.49741060950073 Lasso Regression: R^2 score on test set 85.49734440925533

## 2.4 Lasso with different lambdas

Apply the Lasso regression on the training set with the following λ parameters: (0.001, 0.01, 0.1, 0.5, 1, 2, 10). Evaluate the R^2 score for all the models you obtain on both the train and test sets.

lambdas = (0.001, 0.01, 0.1, 0.5, 1, 2, 10)
l_num = 7
pred_num = X.shape

# prepare data for enumerate
coeff_a = np.zeros((l_num, pred_num))
train_r_squared = np.zeros(l_num)
test_r_squared = np.zeros(l_num)
# enumerate through lambdas with index and i
for ind, i in enumerate(lambdas):
reg = Lasso(alpha = i)
reg.fit(X_train, y_train)

coeff_a[ind,:] = reg.coef_
train_r_squared[ind] = reg.score(X_train, y_train)
test_r_squared[ind] = reg.score(X_test, y_test)

## 2.5 Plot values as a function of lambda

Plot all values for both data sets (train and test $$R^2$$-values) as a function of λ. Comment on your findings.

# Plotting
plt.figure(figsize=(18, 8))
plt.plot(train_r_squared, 'bo-', label=r'$R^2$ Training set', color="darkblue", alpha=0.6, linewidth=3)
plt.plot(test_r_squared, 'bo-', label=r'$R^2$ Test set', color="darkred", alpha=0.6, linewidth=3)
plt.xlabel('Lamda index'); plt.ylabel(r'$R^2$')
plt.xlim(0, 6)
plt.title(r'Evaluate lasso regression with lamdas: 0 = 0.001, 1= 0.01, 2 = 0.1, 3 = 0.5, 4= 1, 5= 2, 6 = 10')
plt.legend(loc='best')
plt.grid() ## 2.6 Identify best lambda and coefficients

Store your test data results in a DataFrame and indentify the lambda where the $$R^2$$ has it’s maximum value in the test data. Fit a Lasso model with this lambda parameter (use the training data) and obtain the corresponding regression coefficients. Furthermore, obtain the mean squared error for the test data of this model (module: from sklearn.metrics import mean_squared_error)

df_lam = pd.DataFrame(test_r_squared*100, columns=['R_squared'])
df_lam['lambda'] = (lambdas)
# returns the index of the row where column has maximum value.
df_lam.loc[df_lam['R_squared'].idxmax()]

R_squared 88.105773 lambda 0.001000 Name: 0, dtype: float64

# Coefficients of best model
reg_best = Lasso(alpha = 0.1)
reg_best.fit(X_train, y_train)
reg_best.coef_

array([-0.35554113, -1.13104696, -0.00596296, -3.31741775, -0. , 0.37914648, 0.74902885])

from sklearn.metrics import mean_squared_error
mean_squared_error(y_test, reg_best.predict(X_test))

3.586249592807347

## 2.7 Cross Validation

Evaluate the performance of a Lasso regression for different regularization parameters λ using 5-fold cross validation on the training set (module: from sklearn.model_selection import cross_val_score) and plot the cross-validation (CV) $$R^2$$ scores of the training and test data as a function of λ.

Use the following lambda parameters: l_min = 0.05 l_max = 0.2 l_num = 20 lambdas = np.linspace(l_min,l_max, l_num)

l_min = 0.05
l_max = 0.2
l_num = 20
lambdas = np.linspace(l_min,l_max, l_num)

train_r_squared = np.zeros(l_num)
test_r_squared = np.zeros(l_num)

pred_num = X.shape
coeff_a = np.zeros((l_num, pred_num))
from sklearn.model_selection import cross_val_score

for ind, i in enumerate(lambdas):
reg = Lasso(alpha = i)
reg.fit(X_train, y_train)
results = cross_val_score(reg, X, y, cv=5, scoring="r2")

train_r_squared[ind] = reg.score(X_train, y_train)
test_r_squared[ind] = reg.score(X_test, y_test)
# Plotting
plt.figure(figsize=(18, 8))
plt.plot(train_r_squared, 'bo-', label=r'$R^2$ Training set', color="darkblue", alpha=0.6, linewidth=3)
plt.plot(test_r_squared, 'bo-', label=r'$R^2$ Test set', color="darkred", alpha=0.6, linewidth=3)
plt.xlabel('Lamda value'); plt.ylabel(r'$R^2$')
plt.xlim(0, 19)
plt.title(r'Evaluate 5-fold cv with different lamdas')
plt.legend(loc='best')
plt.grid() ## 2.8 Best Model

Finally, store your test data results in a DataFrame and identify the lambda where the $$R^2$$ has it’s maximum value in the test data. Fit a Lasso model with this lambda parameter (use the training data) and obtain the corresponding regression coefficients. Furthermore, obtain the mean squared error for the test data of this model (module: from sklearn.metrics import mean_squared_error)

df_lam = pd.DataFrame(test_r_squared*100, columns=['R_squared'])
df_lam['lambda'] = (lambdas)
# returns the index of the row where column has maximum value.
df_lam.loc[df_lam['R_squared'].idxmax()]

R_squared 87.897525 lambda 0.050000 Name: 0, dtype: float64

# Best Model
reg_best = Lasso(alpha = 0.144737)
reg_best.fit(X_train, y_train)

Lasso(alpha=0.144737, copy_X=True, fit_intercept=True, max_iter=1000, normalize=False, positive=False, precompute=False, random_state=None, selection=‘cyclic’, tol=0.0001, warm_start=False)

from sklearn.metrics import mean_squared_error

mean_squared_error(y_test, reg_best.predict(X_test))

3.635187490993961

reg_best.coef_

array([-0.34136411, -1.18223273, -0. , -3.27132984, 0. , 0.33262331, 0.71385488]) ##### Jan Kirenz
###### Professor

I’m a data scientist educator and consultant.