SVC and OneVsOneClassifier decision_function inconsistent on sub-sample

Hi,

I’m seeing inconsistent numerical results with SVC’s decision_function. When estimated over an entire batch of samples ( (n_samples, n_features) matrix ) compared to analyzing sample-by-sample, the results are not the same. This is true for both the individual numerical values per sample and the overall distribution of the results.

The model is SVC with RBF kernel, for a 3-class classification:

SVC(C=1.0, gamma=0.007, class_weight = new_class_weight, probability = True, random_state = 30, 
decision_function_shape = 'ovr')

The models are loaded from file:

ML = joblib.load("model.pkl")

Option A, analyze a matrix:

distances = ML.decision_function(X)

Option B, analyze individual samples:

distances = numpy.zeros([X.shape[0], 3])
for i in range(X.shape[0]):     
    distances[i,:]` = ML.decision_function(X[i,:].reshape(1,-1))

Output for first two samples: Option A: sample 1: [ 0.90835588, -0.17305875, 2.26470288] sample 2: [ 1.10437313, -0.2371539 , 2.13278077]

Option B: sample 1: [ 0.82689247, -0.32689247, 2.5 ] sample 2: [ 1.22005359, -0.5 , 2.27994641]

I couldn’t find any indication for this behavior in the documentation.

Windows-10-10.0.15063-SP0 Python 3.5.2 |Anaconda 4.2.0 (64-bit)| (default, Jul 5 2016, 11:41:13) [MSC v.1900 64 bit (AMD64)] NumPy 1.12.1 SciPy 0.18.1 Scikit-Learn 0.18.1

Thanks!

[DanaAverbuch]

Here you go:

import numpy as np
from sklearn.datasets import load_iris
data = load_iris()
y=data.target
X = data.data
from sklearn.svm import SVC
from sklearn.multiclass import OneVsRestClassifier
clf = SVC(decision_function_shape='ovr').fit(X, y)
dfall = clf.decision_function(X)
same = list()
for i in range(X.shape[0]):
    dfone = clf.decision_function(X[i, :].reshape(1, -1))
    same.append(np.all(dfone == dfall[i, :]))

print(same)

[jnothman]

yes, I suppose it checks the main properties on a small sample. It assumes you can’t control the ovo output, while testing the helper directly would allow us to do so and hence be more rigorous.