"""

Part 1 Discussions

"""

#https://medium.com/@jayeshbahire/lasso-ridge-and-elastic-net-regularization-4807897cb722

"""

1.L1 Regularization aka Lasso Regularization– This add regularization terms in the model

which are function of absolute value of the coefficients of parameters.

The coefficient of the paratmeters can be driven to zero as well during the

regularization process. Hence this technique can be used for feature selection and

generating more parsimonious model.

2.L2 Regularization aka Ridge Regularization — This add regularization terms in the

model which are function of square of coefficients of parameters. Coefficient of

parameters can approach to zero but never become zero and hence.

3.Combination of the above two such as Elastic Nets– This add regularization terms in

the model which are combination of both L1 and L2 regularization.

"""

#Import packages

import numpy as np

import pandas as pd

import sklearn

import seaborn as sns

import matplotlib.pyplot as plt

from sklearn.preprocessing import StandardScaler

#from sklearn.cross_validation import train_test_split

from sklearn.model_selection import train_test_split

#Import dataset and create a dataframe

from sklearn import datasets

boston = datasets.load_boston ()

#Find out more about this dataset description

print(boston.DESCR)

#Boston dataset features

boston.keys()

boston.data # ( features )

boston.data.shape

boston.feature_names

boston.target # label

#Create a dataframe

boston_df = pd.DataFrame (boston.data, columns= boston.feature_names )

boston_df.head()

#Add dependent Variable

boston_df['House_Price']= boston.target

boston_df.head ()

boston_df.describe()

features = boston_df.drop('House_Price',axis = 1)

labels = boston_df['House_Price']

features.head()

labels.head()

#Create train and test data with 70o/o and 30°/o split

features_train, features_test, labels_train,labels_test = train_test_split(features, labels, test_size=0.3, random_state=1)

features_train.shape

features_test.shape

labels_train.shape

labels_test.shape

#Let's import the Lasso, Ridge, Elasticnet regression object and define model

# These methods are used to handle overfitting

from sklearn.linear_model import LinearRegression

from sklearn.linear_model import Lasso

from sklearn.linear_model import Ridge # RidgeClassier is also there

from sklearn.linear_model import ElasticNet

lm = LinearRegression ()

lm_lasso = Lasso()

lm_ridge = Ridge()

lm_elastic = ElasticNet()

#Fit a model on the train data

lm.fit(features_train, labels_train)

lm_lasso.fit(features_train, labels_train)

lm_ridge.fit(features_train, labels_train)

lm_elastic.fit(features_train, labels_train)

#Evaluate the model

plt.figure (figsize= (15,10))

ft_importances_lm = pd.Series(lm.coef_, index= features.columns)

ft_importances_lm .plot(kind = 'barh')

plt.show()

#R2 Value

print ("RSquare Value for Simple Regresssion TEST data is-")

print (np.round (lm.score(features_test,labels_test)*100,2))

print ("RSquare Value for Lasso Regresssion TEST data is-")

print (np.round (lm_lasso.score(features_test,labels_test)*100,2))

print ("RSquare Value for Ridge Regresssion TEST data is-")

print (np.round (lm_ridge.score(features_test,labels_test)*100,2))

print ("RSquare Value for Elastic Net Regresssion TEST data is-")

print (np.round (lm_elastic.score(features_test,labels_test)*100,2))

#Predict on test and training data

predict_test_lm = lm.predict(features_test )

predict_test_lasso = lm_lasso.predict (features_test)

predict_test_ridge = lm_ridge.predict (features_test)

predict_test_elastic = lm_elastic.predict(features_test)

#Print the Loss Funtion - MSE & MAE

import numpy as np

from sklearn import metrics

print ("Simple Regression Mean Square Error (MSE) for TEST data is")

print (np.round (metrics .mean_squared_error(labels_test, predict_test_lm),2) )

print ("Lasso Regression Mean Square Error (MSE) for TEST data is")

print (np.round (metrics .mean_squared_error(labels_test, predict_test_lasso),2))

print ("Ridge Regression Mean Square Error (MSE) for TEST data is")

print (np.round (metrics .mean_squared_error(labels_test, predict_test_ridge),2))

print ("ElasticNet Mean Square Error (MSE) for TEST data is")

print (np.round (metrics .mean_squared_error(labels_test, predict_test_elastic),2))

"""

The benchmark of random guessing should get you an RMSE = standard_deviation.

So lower than this, your model is demonstrating some ability to learn;

above that number, you haven't even learned to guess the mean correctly.

"""

"""

Part 2 Discussions

"""

"""

Ridge Regression and Lasso

This notebook explores ridge regression and lasso.

These alternative linear fitting techniques can improve a model's performance

and interpretability.

"""

#Import Libraries

import pandas as pd

import numpy as np

import matplotlib.pyplot as plt

#Data Analysis

data = pd.read_csv("All CSV/Advertising.csv")

data.head()

#Drop the first column

data.drop(['Unnamed: 0'], axis=1, inplace=True)

print (data.head())

print (data.columns)

#lets plot few visuals

def scatter_plot(feature, target):

plt.show()

scatter_plot('TV', 'sales')

scatter_plot('radio', 'sales')

scatter_plot('newspaper', 'sales')

#Lets build the models now

#Multiple Linear Regression

from sklearn.model_selection import cross_val_score

from sklearn.linear_model import LinearRegression

features = data.drop(['sales'], axis=1) #drop the target to get the features

labels = data['sales'].values.reshape(-1,1) #choose the target

lin_reg = LinearRegression()

MSEs = cross_val_score(lin_reg, features, labels, scoring='neg_mean_squared_error', cv=5)

mean_MSE = np.mean(MSEs)

print(mean_MSE)

#Ridge Regression

from sklearn.model_selection import GridSearchCV

from sklearn.linear_model import Ridge

#alpha = [1e-15, 1e-10, 1e-8, 1e-4, 1e-3,1e-2, 1, 5, 10, 20]

ridge = Ridge()

parameters = {'alpha': [1e-15, 1e-10, 1e-8, 1e-4, 1e-3,1e-2, 1, 5, 10, 20]}

ridge_regressor = GridSearchCV(ridge, parameters,scoring='neg_mean_squared_error', cv=5)

ridge_regressor.fit(features, labels)

ridge_regressor.best_params_

ridge_regressor.best_score_

#Lasso

from sklearn.linear_model import Lasso

lasso = Lasso()

"""

For ridge regression, we introduce GridSearchCV.

This will allow us to automatically perform 5-fold cross-validation with a range of different regularization parameters in order to find the optimal value of alpha.

"""

parameters = {'alpha': [1e-15, 1e-10, 1e-8, 1e-4, 1e-3,1e-2, 1, 5, 10, 20]}

lasso_regressor = GridSearchCV(lasso, parameters, scoring='neg_mean_squared_error', cv = 5)

lasso_regressor.fit(features, labels)

lasso_regressor.best_params_

lasso_regressor.best_score_

#Version 2

# Standardize x_train

from sklearn.preprocessing import StandardScaler

scaler = StandardScaler().fit(features)

features = scaler.transform(features)

from sklearn.linear_model import Ridge, Lasso

# Fit x_train and y_train

ridge = Ridge().fit(features, labels)

lasso = Lasso().fit(features, labels)

# Print results for both Ridge and Lasso

print('Ridge: ', ridge.coef_)

print('Lasso:', lasso.coef_)

"""

Code Challenges 02: (House Data) kc_house_data.csv

This is kings house society data.

In particular, we will:

• Use Linear Regression and see the results

• Use Lasso (L1) and see the resuls

• Use Ridge and see the score

"""

# skip from here onwards

# coding: utf-8

# This lab on Ridge Regression and the Lasso is a Python adaptation of p. 251-255 of "Introduction to Statistical Learning with Applications in R" by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani. Adapted by R. Jordan Crouser at Smith College for SDS293: Machine Learning (Spring 2016).

#

# # 6.6: Ridge Regression and the Lasso

import pandas as pd

import numpy as np

import matplotlib.pyplot as plt

from sklearn.preprocessing import scale

from sklearn.model_selection import train_test_split

from sklearn.linear_model import Ridge, RidgeCV, Lasso, LassoCV

from sklearn.metrics import mean_squared_error

# We will use the `sklearn` package in order to perform ridge regression and

# the lasso. The main functions in this package that we care about are `Ridge()`, which can be used

# to fit ridge regression models, and `Lasso()` which will fit lasso models. They also have cross-validated counterparts: `RidgeCV()` and `LassoCV()`. We'll use these a bit later.

#

# Before proceeding, let's first ensure that the missing values have

# been removed from the data, as described in the previous lab.

df = pd.read_csv('Hitters.csv').dropna().drop('Player', axis = 1)

df.info()

dummies = pd.get_dummies(df[['League', 'Division', 'NewLeague']])

# We will now perform ridge regression and the lasso in order to predict `Salary` on

# the `Hitters` data. Let's set up our data:

y = df.Salary

# Drop the column with the independent variable (Salary), and columns for which we created dummy variables

X_ = df.drop(['Salary', 'League', 'Division', 'NewLeague'], axis = 1).astype('float64')

# Define the feature set X.

X = pd.concat([X_, dummies[['League_N', 'Division_W', 'NewLeague_N']]], axis = 1)

X.info()

# Ridge Regression

# The `Ridge()` function has an alpha argument ($\lambda$, but with a different name!) that is used to tune the model. We'll generate an array of alpha values ranging from very big to very small, essentially

# covering the full range of scenarios from the null model containing

# only the intercept, to the least squares fit:

alphas = 10**np.linspace(10,-2,100)*0.5

# Associated with each alpha value is a vector of ridge regression coefficients, which we'll

# store in a matrix `coefs`. In this case, it is a $19 \times 100$

# matrix, with 19 rows (one for each predictor) and 100

# columns (one for each value of alpha). Remember that we'll want to standardize the

# variables so that they are on the same scale. To do this, we can use the

# `normalize = True` parameter:

ridge = Ridge(normalize = True)

coefs = []

for a in alphas:

ridge.set_params(alpha = a)

ridge.fit(X, y)

coefs.append(ridge.coef_)

np.shape(coefs)

# We expect the coefficient estimates to be much smaller, in terms of $l_2$ norm,

# when a large value of alpha is used, as compared to when a small value of alpha is

# used. Let's plot and find out:

ax = plt.gca()

ax.plot(alphas, coefs)

ax.set_xscale('log')

plt.axis('tight')

plt.xlabel('alpha')

plt.ylabel('weights')

# We now split the samples into a training set and a test set in order

# to estimate the test error of ridge regression and the lasso:

# Split data into training and test sets

X_train, X_test , y_train, y_test = train_test_split(X, y, test_size=0.5, random_state=1)

# Next we fit a ridge regression model on the training set, and evaluate

# its MSE on the test set, using $\lambda = 4$:

ridge2 = Ridge(alpha = 4, normalize = True)

ridge2.fit(X_train, y_train) # Fit a ridge regression on the training data

pred2 = ridge2.predict(X_test) # Use this model to predict the test data

print(pd.Series(ridge2.coef_, index = X.columns)) # Print coefficients

print(mean_squared_error(y_test, pred2)) # Calculate the test MSE

# The test MSE when alpha = 4 is 106216. Now let's see what happens if we use a huge value of alpha, say $10^{10}$:

ridge3 = Ridge(alpha = 10**10, normalize = True)

ridge3.fit(X_train, y_train) # Fit a ridge regression on the training data

pred3 = ridge3.predict(X_test) # Use this model to predict the test data

print(pd.Series(ridge3.coef_, index = X.columns)) # Print coefficients

print(mean_squared_error(y_test, pred3)) # Calculate the test MSE

# This big penalty shrinks the coefficients to a very large degree, essentially reducing to a model containing just the intercept. This over-shrinking makes the model more biased, resulting in a higher MSE.

# Okay, so fitting a ridge regression model with alpha = 4 leads to a much lower test

# MSE than fitting a model with just an intercept. We now check whether

# there is any benefit to performing ridge regression with alpha = 4 instead of

# just performing least squares regression. Recall that least squares is simply

# ridge regression with alpha = 0.

ridge2 = Ridge(alpha = 0, normalize = True)

ridge2.fit(X_train, y_train) # Fit a ridge regression on the training data

pred = ridge2.predict(X_test) # Use this model to predict the test data

print(pd.Series(ridge2.coef_, index = X.columns)) # Print coefficients

print(mean_squared_error(y_test, pred)) # Calculate the test MSE

# It looks like we are indeed improving over regular least-squares!

#

# Instead of arbitrarily choosing alpha $ = 4$, it would be better to

# use cross-validation to choose the tuning parameter alpha. We can do this using

# the cross-validated ridge regression function, `RidgeCV()`. By default, the function

# performs generalized cross-validation (an efficient form of LOOCV), though this can be changed using the

# argument `cv`.

ridgecv = RidgeCV(alphas = alphas, scoring = 'neg_mean_squared_error', normalize = True)

ridgecv.fit(X_train, y_train)

ridgecv.alpha_

# Therefore, we see that the value of alpha that results in the smallest cross-validation

# error is 0.57. What is the test MSE associated with this value of

# alpha?

ridge4 = Ridge(alpha = ridgecv.alpha_, normalize = True)

ridge4.fit(X_train, y_train)

mean_squared_error(y_test, ridge4.predict(X_test))

# This represents a further improvement over the test MSE that we got using

# alpha $ = 4$. Finally, we refit our ridge regression model on the full data set,

# using the value of alpha chosen by cross-validation, and examine the coefficient

# estimates.

#This is the final model with alpha = 0.57

# L2 Regularization

ridge4.fit(X, y)

pd.Series(ridge4.coef_, index = X.columns)

# As expected, none of the coefficients are exactly zero - ridge regression does not

# perform variable selection!

# # The Lasso

# We saw that ridge regression with a wise choice of alpha can outperform least

# squares as well as the null model on the Hitters data set. We now ask

# whether the lasso can yield either a more accurate or a more interpretable

# model than ridge regression. In order to fit a lasso model, we'll

# use the `Lasso()` function; however, this time we'll need to include the argument `max_iter = 10000`.

# Other than that change, we proceed just as we did in fitting a ridge model:

lasso = Lasso(max_iter = 10000, normalize = True)

coefs = []

for a in alphas:

lasso.set_params(alpha=a)

lasso.fit(scale(X_train), y_train)

coefs.append(lasso.coef_)

ax = plt.gca()

ax.plot(alphas*2, coefs)

ax.set_xscale('log')

plt.axis('tight')

plt.xlabel('alpha')

plt.ylabel('weights')

# Notice that in the coefficient plot that depending on the choice of tuning

# parameter, some of the coefficients are exactly equal to zero. We now

# perform 10-fold cross-validation to choose the best alpha, refit the model, and compute the associated test error:

lassocv = LassoCV(alphas = None, cv = 10, max_iter = 100000, normalize = True)

lassocv.fit(X_train, y_train)

#As we get the final value of alpha from above (LassoCV), we make the final model

# L1 Regularization

lasso.set_params(alpha=lassocv.alpha_)

lasso.fit(X_train, y_train)

mean_squared_error(y_test, lasso.predict(X_test))

# This is substantially lower than the test set MSE of the null model and of

# least squares, and only a little worse than the test MSE of ridge regression with alpha

# chosen by cross-validation.

#

# However, the lasso has a substantial advantage over ridge regression in

# that the resulting coefficient estimates are sparse. Here we see that 13 of

# the 19 coefficient estimates are exactly zero:

# Some of the coefficients are now reduced to exactly zero.

pd.Series(lasso.coef_, index=X.columns)