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Introduction
In my previous article, I discussed the theory of hard-margin SVM.
This time, we will implement it using Python.
Also, the following code works with Google Colab.
\begin{align*} \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\f}[2]{\frac{#1}{#2}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\d}[2]{\frac{{\rm d}#1}{{\rm d}#2}} \newcommand{\T}{\mathsf{T}} \newcommand{\(}{\left(} \newcommand{\)}{\right)} \newcommand{\{}{\left\{} \newcommand{\}}{\right\}} \newcommand{\[}{\left[} \newcommand{\]}{\right]} \newcommand{\dis}{\displaystyle} \newcommand{\eq}[1]{{\rm Eq}(\ref{#1})} \newcommand{\n}{\notag\\} \newcommand{\t}{\ \ \ \ } \newcommand{\tt}{\t\t\t\t} \newcommand{\argmax}{\mathop{\rm arg\, max}\limits} \newcommand{\argmin}{\mathop{\rm arg\, min}\limits} \def\l<#1>{\left\langle #1 \right\rangle} \def\us#1_#2{\underset{#2}{#1}} \def\os#1^#2{\overset{#2}{#1}} \newcommand{\case}[1]{\{ \begin{array}{ll} #1 \end{array} \right.} \newcommand{\s}[1]{{\scriptstyle #1}} \definecolor{myblack}{rgb}{0.27,0.27,0.27} \definecolor{myred}{rgb}{0.78,0.24,0.18} \definecolor{myblue}{rgb}{0.0,0.443,0.737} \definecolor{myyellow}{rgb}{1.0,0.82,0.165} \definecolor{mygreen}{rgb}{0.24,0.47,0.44} \newcommand{\c}[2]{\textcolor{#1}{#2}} \newcommand{\ub}[2]{\underbrace{#1}_{#2}} \end{align*}
Theoretical Overview of Hard Margin SVM
The $n$ $p$-dimensional data observed are denoted by $X$ and the $n$ label variable pairs are denoted by $\bm{y}$, respectively, as follows.
\us X_{[n \times p]} = \mat{x_1^{(1)} & x_2^{(1)} & \cdots & x_p^{(1)} \\
x_1^{(2)} & x_2^{(2)} & \cdots & x_p^{(2)} \\
\vdots & \vdots & \ddots & \vdots \\
x_1^{(n)} & x_2^{(2)} & \cdots & x_p^{(n)}}
=
\mat{ – & \bm{x}^{(1)\T} & – \\
– & \bm{x}^{(2)\T} & – \\
& \vdots & \\
– & \bm{x}^{(n)\T} & – \\},
\t
\us \bm{y}_{[n \times 1]} = \mat{y^{(1)} \\ y^{(2)} \\ \vdots \\ y^{(n)}}.
\end{align*}
From the previous consequence, the parameter that determines the separating hyperplane could be calculated as follows.
\hat{\bm{w}} &= \sum_{\bm{x}^{(i)} \in S} \hat{\alpha}_i y^{(i)} \bm{x}^{(i)}, \\
\hat{b} &= \f{1}{|S|} \sum_{\bm{x}^{(i)} \in S} (y^{(i)} – \hat{\bm{w}}^\T \bm{x}^{(i)}).
\end{align}
(where $S$ is the set of support vectors)
Also, $\bm{\alpha} = (\alpha_1, \alpha_2, \dots, \alpha_n)^\T$ is a pair of Lagrangian undetermined multipliers, and we use the steepest descent method to find its optimal solution $\hat{\bm{\alpha}}$.
\bm{\alpha}^{[t+1]} = \bm{\alpha}^{[t]} + \eta \pd{\tilde{L}(\bm{\alpha})}{\bm{\alpha}}.
\end{align*}
The value of the gradient vector $\pd{\tilde{L}(\bm{\alpha})}{\bm{\alpha}}$ is
\us H_{[n \times n]} \equiv \us \bm{y}_{[n \times 1]} \us \bm{y}^{\T}_{[1 \times n]} \odot \us X_{[n \times p]} \us X^{\T}_{[p \times n]}
\end{align}
\begin{align} \pd{\tilde{L}(\bm{\alpha})}{\bm{\alpha}} = \bm{1}\, – H \bm{\alpha}. \end{align}
Full-scratch implementation of hard-margin SVM
From the above, a hard-margin SVM will be implemented in full scratch.
import numpy as np
class HardMarginSVM:
"""
Attributes
----------
eta : float
epoch : int
random_state : int
is_trained : bool
num_samples : int
num_features : int
w : NDArray[float]
b : float
alpha : NDArray[float]
Methods
-------
fit -> None
Fitting parameter vectors for training data
predict -> NDArray[int]
Return predicted value
"""
def __init__(self, eta=0.001, epoch=1000, random_state=42):
self.eta = eta
self.epoch = epoch
self.random_state = random_state
self.is_trained = False
def fit(self, X, y):
"""
Fitting parameter vectors for training data
Parameters
----------
X : NDArray[NDArray[float]]
y : NDArray[float]
"""
self.num_samples = X.shape[0]
self.num_features = X.shape[1]
self.w = np.zeros(self.num_features)
self.b = 0
rgen = np.random.RandomState(self.random_state)
self.alpha = rgen.normal(loc=0.0, scale=0.01, size=self.num_samples)
for _ in range(self.epoch):
self._cycle(X, y)
indexes_sv = [i for i in range(self.num_samples) if self.alpha[i] != 0]
for i in indexes_sv:
self.w += self.alpha[i] * y[i] * X[i]
for i in indexes_sv:
self.b += y[i] - (self.w @ X[i])
self.b /= len(indexes_sv)
self.is_trained = True
def predict(self, X):
"""
Return predicted value
Parameters
----------
X : NDArray[NDArray[float]]
Returns
-------
result : NDArray[int]
"""
if not self.is_trained:
raise Exception('This model is not trained.')
hyperplane = X @ self.w + self.b
result = np.where(hyperplane > 0, 1, -1)
return result
def _cycle(self, X, y):
"""
One cycle of gradient descent method
Parameters
----------
X : NDArray[NDArray[float]]
y : NDArray[float]
"""
y = y.reshape([-1, 1])
H = (y @ y.T) * (X @ X.T)
grad = np.ones(self.num_samples) - H @ self.alpha
self.alpha += self.eta * grad
self.alpha = np.where(self.alpha < 0, 0, self.alpha)Confirmation of SVM operation using iris dataset
The data used as an example is the iris dataset. The iris dataset consists of petal and sepal lengths for three varieties: Versicolour, Virginica, and Setosa.
Let’s read the iris dataset using the scikit-learn library.
import pandas as pd
from sklearn.datasets import load_iris
iris = load_iris()
df_iris = pd.DataFrame(iris.data, columns=iris.feature_names)
df_iris['class'] = iris.target
df_iris
This time we will perform a binary logistic regression classification, focusing only on data with class = 0, 1. For simplicity, we assume that the two features are petal length and petal width.
df_iris = df_iris[df_iris['class'] != 2]
df_iris = df_iris[['petal length (cm)', 'petal width (cm)', 'class']]
X = df_iris.iloc[:, :-1].values
y = df_iris.iloc[:, -1].values
y = np.where(y==0, -1, 1)The data set is standardized to have a mean of 0 and a standard deviation of 1.
from sklearn.preprocessing import StandardScaler
sc = StandardScaler()
X_std = sc.fit_transform(X)To evaluate the generalization performance of the model, the data set is split into a training data set and a test data set. In this case, we split the training data at a ratio of 80% and the test data at a ratio of 20%.
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X_std, y, test_size=0.2, random_state=42, stratify=y)The plot class should also be defined here.
import matplotlib.pyplot as plt
from matplotlib.colors import ListedColormap
class DecisionPlotter:
def __init__(self, X, y, classifier, test_idx=None):
self.X = X
self.y = y
self.classifier = classifier
self.test_idx = test_idx
self.colors = ['#de3838', '#007bc3', '#ffd12a']
self.markers = ['o', 'x', ',']
self.labels = ['setosa', 'versicolor', 'virginica']
def plot(self):
cmap = ListedColormap(self.colors[:len(np.unique(self.y))])
xx1, xx2 = np.meshgrid(
np.arange(self.X[:,0].min()-1, self.X[:,0].max()+1, 0.01),
np.arange(self.X[:,1].min()-1, self.X[:,1].max()+1, 0.01))
Z = self.classifier.predict(np.array([xx1.ravel(), xx2.ravel()]).T)
Z = Z.reshape(xx1.shape)
plt.contourf(xx1, xx2, Z, alpha=0.2, cmap=cmap)
plt.xlim(xx1.min(), xx1.max())
plt.ylim(xx2.min(), xx2.max())
for idx, cl, in enumerate(np.unique(self.y)):
plt.scatter(
x=self.X[self.y==cl, 0], y=self.X[self.y==cl, 1],
alpha=0.8,
c=self.colors[idx],
marker=self.markers[idx],
label=self.labels[idx])
if self.test_idx is not None:
X_test, y_test = self.X[self.test_idx, :], self.y[self.test_idx]
plt.scatter(
X_test[:, 0], X_test[:, 1],
alpha=0.9,
c='None',
edgecolor='gray',
marker='o',
s=100,
label='test set')
plt.legend()We will now check the operation of SVM using the iris dataset.
hard_margin_svm = HardMarginSVM()
hard_margin_svm.fit(X_train, y_train)
X_comb = np.vstack((X_train, X_test))
y_comb = np.hstack((y_train, y_test))
dp = DecisionPlotter(X=X_comb, y=y_comb, classifier=hard_margin_svm, test_idx=range(len(y_train), len(y_comb)))
dp.plot()
plt.xlabel('petal length [standardized]')
plt.ylabel('petal width [standardized]')
plt.show()
The decision curve could be plotted in this way.
SVM implementation using scikit-learn
You can run SVM using scikit-learn as follows
from sklearn import svm
sk_svm = svm.LinearSVC(C=1e10, random_state=42)
sk_svm.fit(X_train, y_train)
X_comb = np.vstack((X_train, X_test))
y_comb = np.hstack((y_train, y_test))
dp = DecisionPlotter(X=X_comb, y=y_comb, classifier=sk_svm, test_idx=range(len(y_train), len(y_comb)))
dp.plot()
plt.xlabel('petal length [standardized]')
plt.ylabel('petal width [standardized]')
plt.show()
We were also able to plot the decision curve this way in scikit-learn.
You can try the above code here▼.
