Support vectors are data points that maximize the margin around a hyperplane that separates positive and negative instances in dataset.
Basic idea:
Another method for building a classifier where we view the data “spatially”, and predict a new instance’s class based on where it is “located in space”
For simplicity, we’ll assume decision is binary (positive/negative, yes/no, etc.)
Assume numeric attributes (easier to work with as “points in space” and perform arithmetic calculations); however, not restricted to 2D
Hyperplane: plane (or line if 2D space) that separates positive and negative instances in dataset
Support vectors: data points (actual instances from training dataset) that maximize the margin around the hyperplane; they “anchor” the hyperplane
Linear separability
One equation for defining a line is ax + by = c
Find a, b, c such that ax + by ≥ c for red points and ax + by ≤ c for green points In general, many possible solutions for a, b, c
Defining the hyperplane (assuming you knew the support vectors)
Ex: In the dataset below, assume decision attribute is z
Easy to see that support vectors are s1 = (1 0), s2 = (3 1), and s3 = (3 -1)
Augment vectors with 1 more dimension; assign value of 1 for this dimension in each support vector s1 = (1 0 1), s2 = (3 1 1), s3 = (3 -1 1)
Set up equations we need to solve:
φ is a function that maps instance data to a “feature space”
α‘s are parameters which, when “combined” with support vectors and respective decisions, will help us define the hyperplane
α1φ(s1) • φ(s1) + α2φ(s2) • φ(s1) + α3φ(s3) • φ(s1) = -1 s1’s decision
α1φ(s1) • φ(s2) + α2φ(s2) • φ(s2) + α3φ(s3) • φ(s2) = 1 s2’s decision
α1φ(s1) • φ(s3) + α2φ(s2) • φ(s3) + α3φ(s3) • φ(s3) = 1 s3’s decision
Let φ( ) be I (identity vector):
Since data in this example are linearly separable, we’re going to be able to use a linear SVM; so we’re just going to use identify function for φ
α1s1 • s1 + α2s2 • s1 + α3s3 • s1 = -1
α1s1 • s2 + α2s2 • s2 + α3s3 • s2 = 1
α1s1 • s3 + α2s2 • s3 + α3s3 • s3 = 1
Compute dot product in each equation:
2α1 + 4α2 + 4α3 = -1 e.g., s2 • s1 = (3 1 1) • (1 0 1) = (3*1) + (1*0) + (1*1) = 4
4α1 + 11α2 + 9α3 = 1
4α1 + 9α2 + 11α3 = 1
Solve for α1, α2, and α3:
α1 = -3.5, α2 = 0.75, α3 = 0.75
Define the discriminating hyperplane:
w’ = α1s1 + α2s2 + α3s3
= -3.5 (1 0 1) + 0.75 (3 1 1) + 0.75 (3 -1 1)
= (1 0 -2)
For hyperplane in 2D, (a b c) translates to ax + by + c = 0
For hyperplane in 3D, (a b c d) translates to ax + by + cz + d = 0 etc.
So in this case (1 0 -2) translates to 1x + 0y + -2 = 0 or simply x = 2
Doing Linear SVM in Python
Import Libraries
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.svm import SVC Read Data
df = pd.read_csv('svm_exampleData.csv')Change non numeric value into numeric value
# Just for graphing purposes, change non-decision attr's to have numeric values
df= df.replace({'z': r'positive'}, {'z':1}, regex=True)
df= df.replace({'z': r'negative'}, {'z':-1}, regex=True)X = df.iloc[:, 0:2].values
y = df.iloc[:, 2].values
r,_ = df.shape # Display original data points
plt.scatter(X[:, 0], X[:, 1], c=y, s=r, cmap='winter');
plt.show() Make Linear SVM
# Make a linear SVM
model = SVC(kernel='linear')
model.fit(X, y)
print(model.support_vectors_) # The support vectors
print(model.coef_) # The weights (x and y)
print(model.intercept_) # The intercept # Function to plot SVM boundary lines - cool!
def plot_svc_decision_function(model, ax=None, plot_support=True):
if ax is None:
ax = plt.gca()
xlim = ax.get_xlim()
ylim = ax.get_ylim()
# Create grid to evaluate model
x = np.linspace(xlim[0], xlim[1], 30)
y = np.linspace(ylim[0], ylim[1], 30)
Y, X = np.meshgrid(y, x)
xy = np.vstack([X.ravel(), Y.ravel()]).T
P = model.decision_function(xy).reshape(X.shape)
# Plot decision boundary and margins
ax.contour(X, Y, P, colors='k',
levels=[-1, 0, 1], alpha=0.5,
linestyles=['--', '-', '--'])
# Plot support vectors
if plot_support:
ax.scatter(model.support_vectors_[:, 0],
model.support_vectors_[:, 1],
s=300, linewidth=1, facecolors='none');
ax.set_xlim(xlim)
ax.set_ylim(ylim)
# Call the function to show points and SV boundaries
plt.scatter(X[:, 0], X[:, 1], c=y, s=50, cmap='winter')
plot_svc_decision_function(model);
plt.show()
# Make predictions (will be array([-1]) or array([1]))
model.predict([[-3, 4]])
model.predict([[7, 5]])
Thanks for your support!
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