This is a practice/laboratory session of SVM tutorial using Python. At the end of this tutorial you will learn the following:
The topics are
Let us start. First, you need to import the necessary modules.
import numpy as np
import math
import pandas as pd
import matplotlib.pyplot as plt
from sklearn import svm
import sklearn.metrics as metric
%matplotlib inline
Our first dataset can be uploaded. Put the CSV files in the same folder as the Jupyter notebook.
filedata='SVM_Dataset1.csv'
data1=pd.read_csv(filedata)
data1
We separate the X training data from the y training data
X1=data1['X1']
X2=data1['X2']
X_training=np.array(list(zip(X1,X2)))
X_training
y_training=data1['y']
y_training
target_names=['-1','+1']
target_names
Let us plot this data. Can you imagine a line separating the two classes?
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
plt.xlabel('X1')
plt.ylabel('X2');
plt.savefig('chart0.png')
To use Vector Support Classification (SVC) algorithm, we need define the model kernel. Let us use linear kernel. Then, we use the fit() function to train the model with our training data.
svc = svm.SVC(kernel='linear').fit(X_training,y_training)
svc
To view the internal model parameters use get_params() method.
svc.get_params(True)
The trained model can be plotted with specifying the decision_function() method.
First, we set the boundary of the plot.
lbX1=math.floor(min(X_training[:,0]))-1
ubX1=math.ceil(max(X_training[:,0]))+1
lbX2=math.floor(min(X_training[:,1]))-1
ubX2=math.ceil(max(X_training[:,1]))+1
[lbX1,ubX1,lbX2,ubX2]
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k'], linestyles=['-'],levels=[0])
plt.title('Linearly Separable')
plt.savefig('chart1.png')
The following plot show the margin and the support vectors
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('Margin and Support Vectors')
plt.savefig('chart2.png')
The number of support vectors for each class can be revealed using `nsupport' attribute
svc.n_support_
To get the indices (= the row numbers in the original dataset) of the support vectors, use support_ attribute
svc.support_
To identify the support vector, use support_vectors_ attribute. The data that become the support vector are
svc.support_vectors_
For linear model, we can reveal the discriminant line that separate the classes using coef_ and intercept_ attributes.
weight=svc.coef_
intercept=svc.intercept_
a = -weight[0,0] / weight[0,1]
print('x2=',a,' * x1 + ',-intercept[0]/weight[0,1])
To get the normalize accuracy, of the training, we can use score(X,y) function.
svc.score(X_training, y_training)
Alternatively, if you have test sample, you can also use the metric from sklearn. To use this on the training sample, we first need to define the y-prediction (which is based on the prediction of the model with X comes from the training sample) and the y-true value (which is based on the y of the training sample).
y_pred=svc.predict(X_training)
y_pred
y_true = y_training
y_true
The absolute accuracy is measured as follow.
metric.accuracy_score(y_true, y_pred, normalize=False)
Confusion matrix is useful to see if there is misclassification. If there is no missclassification, then the corect values would be in the diagonal.
cnf_matrix=metric.confusion_matrix(y_true, y_pred)
cnf_matrix
We can also visualize the confusion matrix through the following function
import itertools
# code from http://scikit-learn.org/stable/auto_examples/model_selection/plot_confusion_matrix.html
def plot_confusion_matrix(cm, classes,
normalize=False,
title='Confusion matrix',
cmap=plt.cm.Blues):
"""
This function prints and plots the confusion matrix.
Normalization can be applied by setting `normalize=True`.
"""
plt.imshow(cm, interpolation='nearest', cmap=cmap)
plt.title(title)
plt.colorbar()
tick_marks = np.arange(len(classes))
plt.xticks(tick_marks, classes, rotation=45)
plt.yticks(tick_marks, classes)
if normalize:
cm = cm.astype('float') / cm.sum(axis=1)[:, np.newaxis]
print("Normalized confusion matrix")
else:
print('Confusion matrix, without normalization')
print(cm)
thresh = cm.max() / 2.
for i, j in itertools.product(range(cm.shape[0]), range(cm.shape[1])):
plt.text(j, i, cm[i, j],
horizontalalignment="center",
color="white" if cm[i, j] > thresh else "black")
plt.tight_layout()
plt.ylabel('True label')
plt.xlabel('Predicted label')
plt.figure()
plot_confusion_matrix(cnf_matrix, classes=target_names, normalize=False)
plt.show()
Now we can also use the trained SVM to predict something that is outside the training data. Let us predict the class y of the given test data [X1, X2] = [3, 6]
svc.predict([[3,6]])
The test data is now plotted.
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.scatter(3,6,c='r',marker='s',s=90)
plt.legend(['-1','+1','prediction'],loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k'], linestyles=['-'],levels=[0])
plt.title('Prediction')
plt.savefig('chart3.png')
Optionally (in case you have limited memory in our laptop), if you want to clear the memory for the next training data, you can delete the variables with large memory.
del X1, X2, X_training, y_training
filedata='SVM_Dataset2.csv'
data2=pd.read_csv(filedata)
data2
We separate the X training data from the y training data
X1=data2['x1']
X2=data2['x2']
X_training=np.array(list(zip(X1,X2)))
X_training
y_training=data2['y']
y_training
Let us plot this data. Can you imagine a line separating the two classes?
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=3)
plt.xlabel('X1')
plt.ylabel('X2');
plt.savefig('nl-chart0.png')
Since the plot show that no line can separate the two classes. If we still want to use linear kernel, we can define a regularization cost parameter C. We use the fit() function to train the model with our training data. Feel free to change the regularization parameter to make such that the error of classification would be minimum.
lbX1=math.floor(min(X_training[:,0]))-1
ubX1=math.ceil(max(X_training[:,0]))+1
lbX2=math.floor(min(X_training[:,1]))-1
ubX2=math.ceil(max(X_training[:,1]))+1
[lbX1,ubX1,lbX2,ubX2]
svc = svm.SVC(kernel='linear',C=0.001).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=3)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k'], linestyles=['-'],levels=[0])
plt.title('Non-Linearly Separable')
plt.savefig('nl-chart1.png')
svc.score(X_training,y_training)
Now we increase the regularization parameter C=100.
svc = svm.SVC(kernel='linear',C=100).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=3)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('Linear Kernel with regularization')
plt.savefig('nl-chart-regularization.png')
svc.score(X_training,y_training)
The number of support vectors for each class
svc.n_support_
svc.support_vectors_
svc.score(X_training,y_training)
y_pred=svc.predict(X_training)
y_true = y_training
metric.accuracy_score(y_true, y_pred, normalize=False)
Using linear kernel, we found 4 support vectors but one out of 20 data is in the wrong side. Thus the accuracy is 19 / 20 * 100 = 94.99%.
It seems we are stuck with one misclassification.
Now it is the time to change the kernel into non-linear kernel. Let us try to use polynomial kernel.
There is no misclassification if the degree = 2 and above.
You can play with the degree and regularization parameter C.
svc = svm.SVC(kernel='poly',C=1, degree=2, probability=True).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=3)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('Polynomial Kernel')
plt.savefig('nl-poly2-kernel.png')
svc.score(X_training,y_training)
Note that some of the support vectors are not the same point as the earlier support vectors using linear kernel.
svc.support_vectors_
Just to give rough idea of overfitting, now let us use RBF kernel.
svc = svm.SVC(kernel='rbf',C=1, gamma=3).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(['-1','+1'],loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('RBF Kernel gamma 1')
plt.savefig('nl-RBF1-kernel.png')
svc.score(X_training,y_training)
y_pred=svc.predict(X_training)
y_true = y_training
metric.accuracy_score(y_true, y_pred, normalize=False)
svc.n_support_
The accuracy of the training has reached 100% (20/20) due to the RBF kernel. However, look at the support vectors. The whole dataset becomes the support vectors and it is an indication that the RBF model overfit our data. Adding more data may need to change the model.
Thus, in our example the Polynomial model with lower degree (degree 2) works better.
filedata='SVM_Dataset3.csv'
data3=pd.read_csv(filedata)
data3
The actual X data is just one dimension. The scatter plot will not work.
To make it works, we need to add dummy X2 such that we can plot using scatter plot and train using SVM.
X1=data3['x']
X2=np.ones((len(X),1),int)
X_training=np.array(list(zip(X1,X2)))
# X_training=np.array(np.transpose([X1])) # alternative way, but you cannot plot
y_training=data3['y']
y_training
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=3)
plt.xlabel('X1')
plt.ylabel('X2');
plt.savefig('data3-chart0.png')
Now let us do the SVM training to this dataset and plot
lbX1=math.floor(min(X_training[:,0]))-1
ubX1=math.ceil(max(X_training[:,0]))+1
lbX2=math.floor(min(X_training[:,1]))-1
ubX2=math.ceil(max(X_training[:,1]))+1
[lbX1,ubX1,lbX2,ubX2]
svc = svm.SVC(kernel='poly',C=1, degree=2).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('Polynomial Kernel')
plt.savefig('data3-Poly-kernel.png')
svc.score(X_training,y_training)
This model has 9 support vectors.
svc.n_support_
svc = svm.SVC(kernel='poly',C=1, degree=3).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('Polynomial Kernel')
plt.savefig('data3-Poly-kernel.png')
svc.score(X_training,y_training)
We have three support vector for Polynomial degree 3 and training accuracy of 100%. This should be the best model.
svc.n_support_
Just for curiosity, we can also try to use RBF kernel
svc = svm.SVC(kernel='rbf',C=1, gamma=3).fit(X_training,y_training)
idxPlus=y_training[y_training<0].index
idxMin=y_training[y_training>0].index
plt.scatter(X_training[idxPlus,0],X_training[idxPlus,1],c='b',s=50)
plt.scatter(X_training[idxMin,0],X_training[idxMin,1],c='r',s=50)
plt.legend(target_names,loc=2)
X,Y = np.mgrid[lbX1:ubX1:100j,lbX2:ubX2:100j]
Z = svc.decision_function(np.c_[X.ravel(),Y.ravel()])
Z = Z.reshape(X.shape)
plt.contourf(X,Y,Z > 0,alpha=0.4)
plt.contour(X,Y,Z,colors=['k','k','k'], linestyles=['--','-','--'],levels=[-1,0,1])
plt.scatter(svc.support_vectors_[:,0],svc.support_vectors_[:,1],s=120,facecolors='none')
plt.scatter(X_training[:,0],X_training[:,1],c=y_training,s=50,alpha=0.95);
plt.title('RBF Kernel')
plt.savefig('data3-RBF-kernel.png')
svc.score(X_training,y_training)
svc.n_support_
Just as before, the accuracy is 100% but the whole data sets now become support vectors. It is a clear sign of overfitting. When the whole dataset become support vectors, it implies that the model memorizes the data rather than generalizing it. Thus, we should avoid to use RBF for this dataset.
In conclusion, the support vectors in SVM are the quality data that we can use to generate the decision boundary (of the same model). Non-support vector data can be ignored, regardless how many data that you have. This also implies that SVM can overcome with ease the imbalance amount of data between classes. That is one of the strengths of SVM.
When the number of support vectors represent the whole dataset, the model is overfit because it memorize the whole dataset and cannot be generalized to predict new data outside the training set.
last update: March 2018
Cite this tutorial as [Teknomo (2018) SVM in Python] (http://people.revoledu.com/kardi/tutorial/SVM/)
See Also: Python for Data Science
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Copyright © 2017 Kardi Teknomo
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