Linear model for classification#

In regression, we saw that the target to be predicted was a continuous variable. In classification, this target will be discrete (e.g. categorical).

We will go back to our penguin dataset. However, this time we will try to predict the penguin species using the culmen information. We will also simplify our classification problem by selecting only 2 of the penguin species to solve a binary classification problem.


If you want a deeper overview regarding this dataset, you can refer to the Appendix - Datasets description section at the end of this MOOC.

import pandas as pd

penguins = pd.read_csv("../datasets/penguins_classification.csv")

# only keep the Adelie and Chinstrap classes
penguins = (
    penguins.set_index("Species").loc[["Adelie", "Chinstrap"]].reset_index()
culmen_columns = ["Culmen Length (mm)", "Culmen Depth (mm)"]
target_column = "Species"

We can quickly start by visualizing the feature distribution by class:

import matplotlib.pyplot as plt

for feature_name in culmen_columns:
    # plot the histogram for each specie
    penguins.groupby("Species")[feature_name].plot.hist(alpha=0.5, legend=True)
../_images/7344869f202467c4b9a6a68a1b2117ced331ff3e39038ee39cdc9f50ac7bf893.png ../_images/9277c62360c8e300e90be889277fed86c701bc53f554c4c9c60d5b8925c2e99f.png

We can observe that we have quite a simple problem. When the culmen length increases, the probability that the penguin is a Chinstrap is closer to 1. However, the culmen depth is not helpful for predicting the penguin species.

For model fitting, we will separate the target from the data and we will create a training and a testing set.

from sklearn.model_selection import train_test_split

penguins_train, penguins_test = train_test_split(penguins, random_state=0)

data_train = penguins_train[culmen_columns]
data_test = penguins_test[culmen_columns]

target_train = penguins_train[target_column]
target_test = penguins_test[target_column]

The linear regression that we previously saw will predict a continuous output. When the target is a binary outcome, one can use the logistic function to model the probability. This model is known as logistic regression.

Scikit-learn provides the class LogisticRegression which implements this algorithm.

from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import LogisticRegression

logistic_regression = make_pipeline(
    StandardScaler(), LogisticRegression(penalty=None)
), target_train)
accuracy = logistic_regression.score(data_test, target_test)
print(f"Accuracy on test set: {accuracy:.3f}")
Accuracy on test set: 1.000

Since we are dealing with a classification problem containing only 2 features, it is then possible to observe the decision function boundary. The boundary is the rule used by our predictive model to affect a class label given the feature values of the sample.


Here, we will use the class DecisionBoundaryDisplay. This educational tool allows us to gain some insights by plotting the decision function boundary learned by the classifier in a 2 dimensional feature space.

Notice however that in more realistic machine learning contexts, one would typically fit on more than two features at once and therefore it would not be possible to display such a visualization of the decision boundary in general.

import seaborn as sns
from sklearn.inspection import DecisionBoundaryDisplay

    palette=["tab:red", "tab:blue"],
_ = plt.title("Decision boundary of the trained\n LogisticRegression")

Thus, we see that our decision function is represented by a line separating the 2 classes. We should also note that we did not impose any regularization by setting the parameter penalty to 'none'.

Since the line is oblique, it means that we used a combination of both features:

coefs = logistic_regression[-1].coef_[0]  # the coefficients is a 2d array
weights = pd.Series(coefs, index=culmen_columns)
_ = plt.title("Weights of the logistic regression")

Indeed, both coefficients are non-null. If one of them had been zero, the decision boundary would have been either horizontal or vertical.

Furthermore the intercept is also non-zero, which means that the decision does not go through the point with (0, 0) coordinates.

For the mathematically inclined reader, the equation of the decision boundary is:

coef0 * x0 + coef1 * x1 + intercept = 0

where x0 is "Culmen Length (mm)" and x1 is "Culmen Depth (mm)".

This equation is equivalent to (assuming that coef1 is non-zero):

x1 = coef0 / coef1 * x0 - intercept / coef1

which is the equation of a straight line.