我是否应该在scikit-learn的多项式回归中使用特征缩放?

时间:2019-05-07 05:38:04

标签: python scikit-learn linear-regression preprocessor polynomial-approximations

我一直在使用下面的代码来处理多项式函数的套索回归。我的问题是我应该在套索回归中进行特征缩放(尝试拟合多项式函数时)。如我在下面粘贴的代码中概述的那样,R ^ 2结果和图表明不是。感谢有关为什么不是这种情况或我是否从根本上塞了东西的任何建议。预先感谢您的任何建议。

import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split


np.random.seed(0)
n = 15
x = np.linspace(0,10,n) + np.random.randn(n)/5
y = np.sin(x)+x/6 + np.random.randn(n)/10


X_train, X_test, y_train, y_test = train_test_split(x, y, random_state=0)

def answer_regression():
    from sklearn.preprocessing import PolynomialFeatures
    from sklearn.linear_model import Lasso, LinearRegression
    from sklearn.metrics.regression import r2_score
    from sklearn.preprocessing import MinMaxScaler
    import matplotlib.pyplot as plt
    scaler = MinMaxScaler()
    global X_train, X_test, y_train, y_test

    degrees = 12
    poly = PolynomialFeatures(degree=degrees)
    X_train_poly = poly.fit_transform(X_train.reshape(-1,1))
    X_test_poly = poly.fit_transform(X_test.reshape(-1,1))

    #Lasso Regression Model
    X_train_scaled = scaler.fit_transform(X_train_poly)
    X_test_scaled = scaler.transform(X_test_poly)

    #No feature scaling
    linlasso = Lasso(alpha=0.01, max_iter = 10000).fit(X_train_poly, y_train)
    y_test_lassopredict = linlasso.predict(X_test_poly)
    Lasso_R2_test_score = r2_score(y_test, y_test_lassopredict)

    #With feature scaling
    linlasso = Lasso(alpha=0.01, max_iter = 10000).fit(X_train_scaled, y_train)
    y_test_lassopredict_scaled = linlasso.predict(X_test_scaled)
    Lasso_R2_test_score_scaled = r2_score(y_test, y_test_lassopredict_scaled)

    %matplotlib notebook
    plt.figure()

    plt.scatter(X_test, y_test, label='Test data')
    plt.scatter(X_test, y_test_lassopredict, label='Predict data - No Scaling')
    plt.scatter(X_test, y_test_lassopredict_scaled, label='Predict data - With Scaling')

    return (Lasso_R2_test_score, Lasso_R2_test_score_scaled)

answer_regression()```

1 个答案:

答案 0 :(得分:1)

您的 X 范围在 [0,10] 左右,因此多项式特征的范围要大得多。如果不进行缩放,它们的权重已经很小(因为它们的值较大),因此 Lasso 不需要将它们设置为零。如果缩放它们,它们的权重会大得多,而 Lasso 会将它们中的大部分设置为零。这就是为什么它对缩放案例的预测很差(需要这些特征来捕捉 y 的真实趋势)。

您可以通过获取两种情况的权重 (linlasso.coef_) 来确认这一点,您会看到第二种情况(缩放后的情况)的大部分权重都设置为零。

看来您的 alpha 值大于最佳值,应该进行调整。如果你降低 alpha,你会在两种情况下得到相似的结果。