Note: This is a generated markdown export from the Jupyter notebook file multiple_linear_regression_sklearn.ipynb. You can also view the notebook with the nbviewer from Jupyter.
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
import pandas as pd
import numpy as np
from sklearn import linear_model, datasets, metrics, model_selection
We load the boston house-prices dataset and X
are our features and y
is the target variable medv
(Median value of owner-occupied homes in $1000s).
boston = datasets.load_boston()
print(boston.DESCR)
.. _boston_dataset:
Boston house prices dataset
---------------------------
**Data Set Characteristics:**
:Number of Instances: 506
:Number of Attributes: 13 numeric/categorical predictive. Median Value (attribute 14) is usually the target.
:Attribute Information (in order):
- CRIM per capita crime rate by town
- ZN proportion of residential land zoned for lots over 25,000 sq.ft.
- INDUS proportion of non-retail business acres per town
- CHAS Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
- NOX nitric oxides concentration (parts per 10 million)
- RM average number of rooms per dwelling
- AGE proportion of owner-occupied units built prior to 1940
- DIS weighted distances to five Boston employment centres
- RAD index of accessibility to radial highways
- TAX full-value property-tax rate per $10,000
- PTRATIO pupil-teacher ratio by town
- B 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town
- LSTAT % lower status of the population
- MEDV Median value of owner-occupied homes in $1000's
:Missing Attribute Values: None
:Creator: Harrison, D. and Rubinfeld, D.L.
This is a copy of UCI ML housing dataset.
https://archive.ics.uci.edu/ml/machine-learning-databases/housing/
This dataset was taken from the StatLib library which is maintained at Carnegie Mellon University.
The Boston house-price data of Harrison, D. and Rubinfeld, D.L. 'Hedonic
prices and the demand for clean air', J. Environ. Economics & Management,
vol.5, 81-102, 1978. Used in Belsley, Kuh & Welsch, 'Regression diagnostics
...', Wiley, 1980. N.B. Various transformations are used in the table on
pages 244-261 of the latter.
The Boston house-price data has been used in many machine learning papers that address regression
problems.
.. topic:: References
- Belsley, Kuh & Welsch, 'Regression diagnostics: Identifying Influential Data and Sources of Collinearity', Wiley, 1980. 244-261.
- Quinlan,R. (1993). Combining Instance-Based and Model-Based Learning. In Proceedings on the Tenth International Conference of Machine Learning, 236-243, University of Massachusetts, Amherst. Morgan Kaufmann.
X = pd.DataFrame(boston.data, columns=boston.feature_names)
y = boston.target
Let's describe our features to see whta kind of type we have. Note chas
is a dummy variable.
X.describe()
CRIM | ZN | INDUS | CHAS | NOX | RM | AGE | DIS | RAD | TAX | PTRATIO | B | LSTAT | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
count | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 | 506.000000 |
mean | 3.613524 | 11.363636 | 11.136779 | 0.069170 | 0.554695 | 6.284634 | 68.574901 | 3.795043 | 9.549407 | 408.237154 | 18.455534 | 356.674032 | 12.653063 |
std | 8.601545 | 23.322453 | 6.860353 | 0.253994 | 0.115878 | 0.702617 | 28.148861 | 2.105710 | 8.707259 | 168.537116 | 2.164946 | 91.294864 | 7.141062 |
min | 0.006320 | 0.000000 | 0.460000 | 0.000000 | 0.385000 | 3.561000 | 2.900000 | 1.129600 | 1.000000 | 187.000000 | 12.600000 | 0.320000 | 1.730000 |
25% | 0.082045 | 0.000000 | 5.190000 | 0.000000 | 0.449000 | 5.885500 | 45.025000 | 2.100175 | 4.000000 | 279.000000 | 17.400000 | 375.377500 | 6.950000 |
50% | 0.256510 | 0.000000 | 9.690000 | 0.000000 | 0.538000 | 6.208500 | 77.500000 | 3.207450 | 5.000000 | 330.000000 | 19.050000 | 391.440000 | 11.360000 |
75% | 3.677083 | 12.500000 | 18.100000 | 0.000000 | 0.624000 | 6.623500 | 94.075000 | 5.188425 | 24.000000 | 666.000000 | 20.200000 | 396.225000 | 16.955000 |
max | 88.976200 | 100.000000 | 27.740000 | 1.000000 | 0.871000 | 8.780000 | 100.000000 | 12.126500 | 24.000000 | 711.000000 | 22.000000 | 396.900000 | 37.970000 |
Let's plot the the target variable
fig, axs = plt.subplots(nrows=2)
_ = sns.histplot(x=y, ax=axs[0])
_ = sns.boxplot(x=y, ax=axs[1])
Let's plot the features
_ = sns.pairplot(X);
Split the data into train and test set
X_train, X_test, y_train, y_test = model_selection.train_test_split(X, y, train_size=0.7)
print('train samples:', len(X_train))
print('test samples', len(X_test))
train samples: 354
test samples 152
df_train = pd.DataFrame(y_train, columns=['target'])
df_train['type'] = 'train'
df_test = pd.DataFrame(y_test, columns=['target'])
df_test['type'] = 'test'
df_set = df_train.append(df_test)
_ = sns.displot(df_set, x="target" ,hue="type", kind="kde", log_scale=False)
Fit the model with the training data
lr = linear_model.LinearRegression().fit(X_train, y_train)
Let's print the model parameters (intercept and coefficients)
print('No coef:', len(lr.coef_))
print('Coefficients:
', lr.coef_)
print('Intercept:', lr.intercept_)
No coef: 13
Coefficients:
[-8.17031896e-02 5.36412027e-02 4.94534675e-02 1.95999242e+00
-2.05292222e+01 3.33263437e+00 4.64537600e-03 -1.56096179e+00
3.21585868e-01 -1.28939394e-02 -9.61120854e-01 9.71525278e-03
-5.41283235e-01]
Intercept: 41.085636049592665
Print the predicated values against the the true values. Perfect match should lie be on the red line.
predicted = lr.predict(X_test)
fig, ax = plt.subplots()
ax.scatter(y_test, predicted)
ax.set_xlabel('True Values')
ax.set_ylabel('Predicted')
_ = ax.plot([0, y.max()], [0, y.max()], ls='-', color='red')
Residual plot for our test set
residual = y_test - predicted
fig, ax = plt.subplots()
ax.scatter(y_test, residual)
ax.set_xlabel('y')
ax.set_ylabel('residual')
_ = plt.axhline(0, color='red', ls='--')
_ = sns.displot(residual, kind="kde");
The trainig scores
print("r2 score: {}".format(metrics.r2_score(y_test, predicted)))
print("mse: {}".format(metrics.mean_squared_error(y_test, predicted)))
print("rmse: {}".format(np.sqrt(metrics.mean_squared_error(y_test, predicted))))
print("mae: {}".format(metrics.mean_absolute_error(y_test, predicted)))
r2 score: 0.7876294370225919
mse: 17.99110865100866
rmse: 4.2415927021590205
mae: 3.184767980717786