Probabilistic SVM, how to get the support vectors?












-1















Currently I'm trying to implement a probabilistic SVM, but I've run into issues, this is my code:



import numpy as np
import random
from sklearn.svm import SVC
import math
from scipy.optimize import minimize
import matplotlib.pyplot as plt
from sklearn import decomposition
from sklearn import svm
from sklearn.model_selection import GridSearchCV
from sklearn.model_selection import train_test_split
import warnings
warnings.filterwarnings("ignore")

def training_banana(name):
inputs =
file = open(name, "r")
for line in file:
vector = line.split()
coordinate =
for i in range(len(vector)):
coordinate.append(float(vector[i]))
inputs.append(coordinate)
file.close()

return np.array(inputs)


def define_inputs(name, name_targets):
inputs = training_banana(name)
targets_array = training_banana(name_targets)
N = targets_array.shape[0]
targets = np.zeros(N)
for i in range(N):
targets[i] = targets_array[i][0]
return inputs, targets, N

#training set
inputs_train, targets_train, N = define_inputs('banana_train_data_10.asc', 'banana_train_labels_10.asc') # banana_train, banana_train_label
permute = list(range(N))
random.shuffle(permute)
inputs_train = inputs_train[permute, :]
targets_train = targets_train[permute]

#test set
inputs_test, targets_test, N = define_inputs('banana_test_data_10.asc', 'banana_test_labels_10.asc') #banana_test.txt, banana_test_label
permute = list(range(N))
random.shuffle(permute)
inputs_test = inputs_test[permute, :]
targets_test = targets_test[permute]

#print(inputs_train[0])


def plotting():
param_C = [2048]
param_grid = {'C': param_C, 'kernel': ['rbf'], 'gamma': [0.5]}
clf = GridSearchCV(svm.SVC(), param_grid)
clf.fit(inputs_train, targets_train)
clf = SVC(C=clf.best_params_['C'], cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape='ovr', degree=5, gamma=clf.best_params_['gamma'],
kernel=clf.best_params_['kernel'],
max_iter=-1, probability=True, random_state=None, shrinking=True,
tol=0.001, verbose=False)
clf.fit(inputs_train, targets_train)
index = clf.support_vectors_
print(len(index))
list_to_compare =
for i in range(inputs_test.shape[0]):
if clf.predict([inputs_test[i]]) == targets_test[i]:
list_to_compare.append(1)
return_value = np.sum(list_to_compare)/inputs_test.shape[0]
print(return_value)









#print(predicting_classes_pos_inputs)
#print(inputs_test)
#print(predicting_classes_pos_targets)
#print(inputs_test.shape[0])




#print(predicting_classes)
#colors = [int(i % 23) for i in xy[0]]
#for i in range(inputs_test.shape[0]):
# if clf.predict([targets_test[i]]) == 1:
# predict_pos_inputs.append(inputs_test[i])
#print(predict_pos_inputs)
#plt.scatter(inputs_test[:, 0], predictions[:, 0], s=30, cmap=plt.cm.Paired)
#plt.show()

#print(inputs_test.shape[0])


#support_vectors = get_supp_vec()
#print(support_vectors[:,0])
plotting()


My problem is that I seem to get the same number of support vectors no matter if I choose the probabilistic version or the "standard" SVM. By the probabilistic version, I mean Platt's implementation of it. As you see, I'm using:



index = clf.support_vectors_
print(len(index))


After running it on 10 datasets, I get approximately 84.7 support vectors, if I instead try using:



index2 = clf.n_support
print(len(index2))


I get only two support vectors. From what I've understood, probabilstic SVM is supposed to be more sparse than the regular, but when I use the second variant (index2), it doesn't really matter how I change gamma or C (the slack), I still get the same result, that is, complete insensitivty to the parameters, which also seems wrong.



Can someone tell me, what is the correct way of plotting the support vectors for a probabilistic SVM using the sklearn package? Thanks










share|improve this question









New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

    – Konrad S
    yesterday











  • SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

    – Vivek Kumar
    yesterday
















-1















Currently I'm trying to implement a probabilistic SVM, but I've run into issues, this is my code:



import numpy as np
import random
from sklearn.svm import SVC
import math
from scipy.optimize import minimize
import matplotlib.pyplot as plt
from sklearn import decomposition
from sklearn import svm
from sklearn.model_selection import GridSearchCV
from sklearn.model_selection import train_test_split
import warnings
warnings.filterwarnings("ignore")

def training_banana(name):
inputs =
file = open(name, "r")
for line in file:
vector = line.split()
coordinate =
for i in range(len(vector)):
coordinate.append(float(vector[i]))
inputs.append(coordinate)
file.close()

return np.array(inputs)


def define_inputs(name, name_targets):
inputs = training_banana(name)
targets_array = training_banana(name_targets)
N = targets_array.shape[0]
targets = np.zeros(N)
for i in range(N):
targets[i] = targets_array[i][0]
return inputs, targets, N

#training set
inputs_train, targets_train, N = define_inputs('banana_train_data_10.asc', 'banana_train_labels_10.asc') # banana_train, banana_train_label
permute = list(range(N))
random.shuffle(permute)
inputs_train = inputs_train[permute, :]
targets_train = targets_train[permute]

#test set
inputs_test, targets_test, N = define_inputs('banana_test_data_10.asc', 'banana_test_labels_10.asc') #banana_test.txt, banana_test_label
permute = list(range(N))
random.shuffle(permute)
inputs_test = inputs_test[permute, :]
targets_test = targets_test[permute]

#print(inputs_train[0])


def plotting():
param_C = [2048]
param_grid = {'C': param_C, 'kernel': ['rbf'], 'gamma': [0.5]}
clf = GridSearchCV(svm.SVC(), param_grid)
clf.fit(inputs_train, targets_train)
clf = SVC(C=clf.best_params_['C'], cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape='ovr', degree=5, gamma=clf.best_params_['gamma'],
kernel=clf.best_params_['kernel'],
max_iter=-1, probability=True, random_state=None, shrinking=True,
tol=0.001, verbose=False)
clf.fit(inputs_train, targets_train)
index = clf.support_vectors_
print(len(index))
list_to_compare =
for i in range(inputs_test.shape[0]):
if clf.predict([inputs_test[i]]) == targets_test[i]:
list_to_compare.append(1)
return_value = np.sum(list_to_compare)/inputs_test.shape[0]
print(return_value)









#print(predicting_classes_pos_inputs)
#print(inputs_test)
#print(predicting_classes_pos_targets)
#print(inputs_test.shape[0])




#print(predicting_classes)
#colors = [int(i % 23) for i in xy[0]]
#for i in range(inputs_test.shape[0]):
# if clf.predict([targets_test[i]]) == 1:
# predict_pos_inputs.append(inputs_test[i])
#print(predict_pos_inputs)
#plt.scatter(inputs_test[:, 0], predictions[:, 0], s=30, cmap=plt.cm.Paired)
#plt.show()

#print(inputs_test.shape[0])


#support_vectors = get_supp_vec()
#print(support_vectors[:,0])
plotting()


My problem is that I seem to get the same number of support vectors no matter if I choose the probabilistic version or the "standard" SVM. By the probabilistic version, I mean Platt's implementation of it. As you see, I'm using:



index = clf.support_vectors_
print(len(index))


After running it on 10 datasets, I get approximately 84.7 support vectors, if I instead try using:



index2 = clf.n_support
print(len(index2))


I get only two support vectors. From what I've understood, probabilstic SVM is supposed to be more sparse than the regular, but when I use the second variant (index2), it doesn't really matter how I change gamma or C (the slack), I still get the same result, that is, complete insensitivty to the parameters, which also seems wrong.



Can someone tell me, what is the correct way of plotting the support vectors for a probabilistic SVM using the sklearn package? Thanks










share|improve this question









New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

    – Konrad S
    yesterday











  • SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

    – Vivek Kumar
    yesterday














-1












-1








-1








Currently I'm trying to implement a probabilistic SVM, but I've run into issues, this is my code:



import numpy as np
import random
from sklearn.svm import SVC
import math
from scipy.optimize import minimize
import matplotlib.pyplot as plt
from sklearn import decomposition
from sklearn import svm
from sklearn.model_selection import GridSearchCV
from sklearn.model_selection import train_test_split
import warnings
warnings.filterwarnings("ignore")

def training_banana(name):
inputs =
file = open(name, "r")
for line in file:
vector = line.split()
coordinate =
for i in range(len(vector)):
coordinate.append(float(vector[i]))
inputs.append(coordinate)
file.close()

return np.array(inputs)


def define_inputs(name, name_targets):
inputs = training_banana(name)
targets_array = training_banana(name_targets)
N = targets_array.shape[0]
targets = np.zeros(N)
for i in range(N):
targets[i] = targets_array[i][0]
return inputs, targets, N

#training set
inputs_train, targets_train, N = define_inputs('banana_train_data_10.asc', 'banana_train_labels_10.asc') # banana_train, banana_train_label
permute = list(range(N))
random.shuffle(permute)
inputs_train = inputs_train[permute, :]
targets_train = targets_train[permute]

#test set
inputs_test, targets_test, N = define_inputs('banana_test_data_10.asc', 'banana_test_labels_10.asc') #banana_test.txt, banana_test_label
permute = list(range(N))
random.shuffle(permute)
inputs_test = inputs_test[permute, :]
targets_test = targets_test[permute]

#print(inputs_train[0])


def plotting():
param_C = [2048]
param_grid = {'C': param_C, 'kernel': ['rbf'], 'gamma': [0.5]}
clf = GridSearchCV(svm.SVC(), param_grid)
clf.fit(inputs_train, targets_train)
clf = SVC(C=clf.best_params_['C'], cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape='ovr', degree=5, gamma=clf.best_params_['gamma'],
kernel=clf.best_params_['kernel'],
max_iter=-1, probability=True, random_state=None, shrinking=True,
tol=0.001, verbose=False)
clf.fit(inputs_train, targets_train)
index = clf.support_vectors_
print(len(index))
list_to_compare =
for i in range(inputs_test.shape[0]):
if clf.predict([inputs_test[i]]) == targets_test[i]:
list_to_compare.append(1)
return_value = np.sum(list_to_compare)/inputs_test.shape[0]
print(return_value)









#print(predicting_classes_pos_inputs)
#print(inputs_test)
#print(predicting_classes_pos_targets)
#print(inputs_test.shape[0])




#print(predicting_classes)
#colors = [int(i % 23) for i in xy[0]]
#for i in range(inputs_test.shape[0]):
# if clf.predict([targets_test[i]]) == 1:
# predict_pos_inputs.append(inputs_test[i])
#print(predict_pos_inputs)
#plt.scatter(inputs_test[:, 0], predictions[:, 0], s=30, cmap=plt.cm.Paired)
#plt.show()

#print(inputs_test.shape[0])


#support_vectors = get_supp_vec()
#print(support_vectors[:,0])
plotting()


My problem is that I seem to get the same number of support vectors no matter if I choose the probabilistic version or the "standard" SVM. By the probabilistic version, I mean Platt's implementation of it. As you see, I'm using:



index = clf.support_vectors_
print(len(index))


After running it on 10 datasets, I get approximately 84.7 support vectors, if I instead try using:



index2 = clf.n_support
print(len(index2))


I get only two support vectors. From what I've understood, probabilstic SVM is supposed to be more sparse than the regular, but when I use the second variant (index2), it doesn't really matter how I change gamma or C (the slack), I still get the same result, that is, complete insensitivty to the parameters, which also seems wrong.



Can someone tell me, what is the correct way of plotting the support vectors for a probabilistic SVM using the sklearn package? Thanks










share|improve this question









New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Currently I'm trying to implement a probabilistic SVM, but I've run into issues, this is my code:



import numpy as np
import random
from sklearn.svm import SVC
import math
from scipy.optimize import minimize
import matplotlib.pyplot as plt
from sklearn import decomposition
from sklearn import svm
from sklearn.model_selection import GridSearchCV
from sklearn.model_selection import train_test_split
import warnings
warnings.filterwarnings("ignore")

def training_banana(name):
inputs =
file = open(name, "r")
for line in file:
vector = line.split()
coordinate =
for i in range(len(vector)):
coordinate.append(float(vector[i]))
inputs.append(coordinate)
file.close()

return np.array(inputs)


def define_inputs(name, name_targets):
inputs = training_banana(name)
targets_array = training_banana(name_targets)
N = targets_array.shape[0]
targets = np.zeros(N)
for i in range(N):
targets[i] = targets_array[i][0]
return inputs, targets, N

#training set
inputs_train, targets_train, N = define_inputs('banana_train_data_10.asc', 'banana_train_labels_10.asc') # banana_train, banana_train_label
permute = list(range(N))
random.shuffle(permute)
inputs_train = inputs_train[permute, :]
targets_train = targets_train[permute]

#test set
inputs_test, targets_test, N = define_inputs('banana_test_data_10.asc', 'banana_test_labels_10.asc') #banana_test.txt, banana_test_label
permute = list(range(N))
random.shuffle(permute)
inputs_test = inputs_test[permute, :]
targets_test = targets_test[permute]

#print(inputs_train[0])


def plotting():
param_C = [2048]
param_grid = {'C': param_C, 'kernel': ['rbf'], 'gamma': [0.5]}
clf = GridSearchCV(svm.SVC(), param_grid)
clf.fit(inputs_train, targets_train)
clf = SVC(C=clf.best_params_['C'], cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape='ovr', degree=5, gamma=clf.best_params_['gamma'],
kernel=clf.best_params_['kernel'],
max_iter=-1, probability=True, random_state=None, shrinking=True,
tol=0.001, verbose=False)
clf.fit(inputs_train, targets_train)
index = clf.support_vectors_
print(len(index))
list_to_compare =
for i in range(inputs_test.shape[0]):
if clf.predict([inputs_test[i]]) == targets_test[i]:
list_to_compare.append(1)
return_value = np.sum(list_to_compare)/inputs_test.shape[0]
print(return_value)









#print(predicting_classes_pos_inputs)
#print(inputs_test)
#print(predicting_classes_pos_targets)
#print(inputs_test.shape[0])




#print(predicting_classes)
#colors = [int(i % 23) for i in xy[0]]
#for i in range(inputs_test.shape[0]):
# if clf.predict([targets_test[i]]) == 1:
# predict_pos_inputs.append(inputs_test[i])
#print(predict_pos_inputs)
#plt.scatter(inputs_test[:, 0], predictions[:, 0], s=30, cmap=plt.cm.Paired)
#plt.show()

#print(inputs_test.shape[0])


#support_vectors = get_supp_vec()
#print(support_vectors[:,0])
plotting()


My problem is that I seem to get the same number of support vectors no matter if I choose the probabilistic version or the "standard" SVM. By the probabilistic version, I mean Platt's implementation of it. As you see, I'm using:



index = clf.support_vectors_
print(len(index))


After running it on 10 datasets, I get approximately 84.7 support vectors, if I instead try using:



index2 = clf.n_support
print(len(index2))


I get only two support vectors. From what I've understood, probabilstic SVM is supposed to be more sparse than the regular, but when I use the second variant (index2), it doesn't really matter how I change gamma or C (the slack), I still get the same result, that is, complete insensitivty to the parameters, which also seems wrong.



Can someone tell me, what is the correct way of plotting the support vectors for a probabilistic SVM using the sklearn package? Thanks







python scikit-learn






share|improve this question









New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









Vinay Bharadhwaj

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11111






New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Konrad SKonrad S

1




1




New contributor




Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Konrad S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

    – Konrad S
    yesterday











  • SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

    – Vivek Kumar
    yesterday



















  • I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

    – Konrad S
    yesterday











  • SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

    – Vivek Kumar
    yesterday

















I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

– Konrad S
yesterday





I suppose I've done wrong and I should've went for a SMO if I really wanted a probabilistic output?

– Konrad S
yesterday













SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

– Vivek Kumar
yesterday





SVC(probability=True) does not change the fitting of SVC in any way. It just calculates the probabilities for the samples predicted after the fitting from the learned model. So that will not change the number of support vectors.

– Vivek Kumar
yesterday












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