pythonscikit-learnpipelinelasso-regressionregularized

Updating Python sklearn Lasso(normalize=True) to Use Pipeline


I am new to Python. I am trying to practice basic regularization by following along with a DataCamp exercise using this CSV: https://assets.datacamp.com/production/repositories/628/datasets/a7e65287ebb197b1267b5042955f27502ec65f31/gm_2008_region.csv

# Import numpy and pandas
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Read the CSV file into a DataFrame: df
df = pd.read_csv('gm_2008_region.csv')

# Create arrays for features and target variable
X = df.drop(['life','Region'], axis=1)
y = df['life'].values.reshape(-1,1)
df_columns = df.drop(['life','Region'], axis=1).columns

The code that I use for the DataCamp exercise is as follows:

# Import Lasso
from sklearn.linear_model import Lasso

# Instantiate a lasso regressor: lasso
lasso = Lasso(alpha=0.4, normalize=True)

# Fit the regressor to the data
lasso.fit(X, y)

# Compute and print the coefficients
lasso_coef = lasso.coef_
print(lasso_coef)

# Plot the coefficients
plt.plot(range(len(df_columns)), lasso_coef)
plt.xticks(range(len(df_columns)), df_columns.values, rotation=60)
plt.margins(0.02)
plt.show()

Original

I get the output above, indicating that child_mortality is the most important feature in predicting life expectancy, but this code also results in a deprecation warning due to the use of "normalize."

I'd like to update this code using the current best practice. I have tried the following, but I get a different output. I am hoping someone can help identify what I need to modify in the updated code in order to produce the same output.

# Modified based on https://scikit-learn.org/stable/modules/preprocessing.html#preprocessing-scaler
# and https://stackoverflow.com/questions/28822756/getting-model-attributes-from-pipeline
# Import Lasso
from sklearn.linear_model import Lasso
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import StandardScaler

# Instantiate a lasso regressor: lasso
#lasso = Lasso(alpha=0.4, normalize=True)
pipe = Pipeline(steps=[    
('scaler',StandardScaler()),
('lasso',Lasso(alpha=0.4))
])

# Fit the regressor to the data
#lasso.fit(X, y)
pipe.fit(X, y)

# Compute and print the coefficients
#lasso_coef = lasso.coef_
#print(lasso_coef)
lasso_coef = pipe.named_steps['lasso'].coef_
print(lasso_coef)

# Plot the coefficients
plt.plot(range(len(df_columns)), lasso_coef)
plt.xticks(range(len(df_columns)), df_columns.values, rotation=60)
plt.margins(0.02)
plt.show()

Updated

As you can see, I draw the same conclusion, but I'd be more comfortable that I was doing this correctly if the output images were more similar. What am I doing wrong with the Pipeline?


Solution

  • When you set Lasso(..normalize=True) the normalization is different from that in StandardScaler(). It divides by the l2-norm instead of the standard deviation. If you read the help page:

    normalize bool, default=False This parameter is ignored when fit_intercept is set to False. If True, the regressors X will be normalized before regression by subtracting the mean and dividing by the l2-norm. If you wish to standardize, please use StandardScaler before calling fit on an estimator with normalize=False.

    Deprecated since version 1.0: normalize was deprecated in version 1.0 and will be removed in 1.2.

    It is also touched upon in this post. Since it will be deprecated, I think it's better to just use the StandardScaler normalization. You can see it's reproducible as long as you scale it in the same way:

    lasso = Lasso(alpha=0.4,random_state=99)
    lasso.fit(StandardScaler().fit_transform(X),y)
    print(lasso.coef_)
    
    [-0.         -0.30409556 -2.33203165 -0.          0.51040194  1.45942351
     -1.02516505 -4.57678764]
    
    pipe = Pipeline(steps=[    
    ('scaler',StandardScaler()),
    ('lasso',Lasso(alpha=0.4,random_state=99))
    ])
    
    pipe.fit(X, y)
    lasso_coef = pipe.named_steps['lasso'].coef_
    print(lasso_coef)
    
    [-0.         -0.30409556 -2.33203165 -0.          0.51040194  1.45942351
     -1.02516505 -4.57678764]