LibSVM incluye un contenedor de python que funciona a través de SWIG.
Ejemplo svm-test.py de su distribución:
#!/usr/bin/env python
from svm import *
# a three-class problem
labels = [0, 1, 1, 2]
samples = [[0, 0], [0, 1], [1, 0], [1, 1]]
problem = svm_problem(labels, samples);
size = len(samples)
kernels = [LINEAR, POLY, RBF]
kname = ['linear','polynomial','rbf']
param = svm_parameter(C = 10,nr_weight = 2,weight_label = [1,0],weight = [10,1])
for k in kernels:
param.kernel_type = k;
model = svm_model(problem,param)
errors = 0
for i in range(size):
prediction = model.predict(samples[i])
probability = model.predict_probability
if (labels[i] != prediction):
errors = errors + 1
print "##########################################"
print " kernel %s: error rate = %d/%d" % (kname[param.kernel_type], errors, size)
print "##########################################"
param = svm_parameter(kernel_type = RBF, C=10)
model = svm_model(problem, param)
print "##########################################"
print " Decision values of predicting %s" % (samples[0])
print "##########################################"
print "Numer of Classes:", model.get_nr_class()
d = model.predict_values(samples[0])
for i in model.get_labels():
for j in model.get_labels():
if j>i:
print "{%d, %d} = %9.5f" % (i, j, d[i,j])
param = svm_parameter(kernel_type = RBF, C=10, probability = 1)
model = svm_model(problem, param)
pred_label, pred_probability = model.predict_probability(samples[1])
print "##########################################"
print " Probability estimate of predicting %s" % (samples[1])
print "##########################################"
print "predicted class: %d" % (pred_label)
for i in model.get_labels():
print "prob(label=%d) = %f" % (i, pred_probability[i])
print "##########################################"
print " Precomputed kernels"
print "##########################################"
samples = [[1, 0, 0, 0, 0], [2, 0, 1, 0, 1], [3, 0, 0, 1, 1], [4, 0, 1, 1, 2]]
problem = svm_problem(labels, samples);
param = svm_parameter(kernel_type=PRECOMPUTED,C = 10,nr_weight = 2,weight_label = [1,0],weight = [10,1])
model = svm_model(problem, param)
pred_label = model.predict(samples[0])
Dado que el sitio web o los documentos de LibSVM no lo mencionan explícitamente, envié un correo electrónico a Chih-Jen Lin, preguntándole sobre el soporte de aprendizaje incremental/en línea. Su respuesta fue: "Desafortunadamente no. La razón es que todavía no vemos una configuración estándar para el aprendizaje incremental/decremental". – Cerin
Lo que muestra el código de ejemplo no es aprendizaje en línea. – mrgloom