OK, here's a final version of my script demonstrating that there's a
bug somewhere in the computation of coef_. If I compute coef_ myself
from dual_coef_ and support_vectors_ I am able to match the predict
function with the dot product method, and the new coef_ has a dot
product of about -1.5 with the coef_ provided by the SVC class,
meaning they are not very similar at all.






import numpy as np
from sklearn.svm import SVC
import time

rng = np.random.RandomState([1,2,3])

m = 1000
n = 1000

X = rng.randn(m,n)
w = rng.randn(n)
b = rng.randn(1)
y = (np.dot(X,w) + b ) > 0

t1 = time.time()
svm = SVC(kernel = 'linear', C = 1.0).fit(X,y)
t2 = time.time()
print 'train time ',t2 - t1

X2 = X

t1 = time.time()
y1 = svm.predict(X2)
t2 = time.time()
print 'predict time ',t2 - t1
print '# support vectors:',svm.n_support_
print 'predict time per support vector:',(t2-t1)/float(svm.n_support_.sum())

t1 = time.time()
y2 = ( np.dot(X2, svm.coef_.T) - svm.intercept_ ) > 0
t2 = time.time()
print 'dot product time',t2 -t1

print 'class 1 prevalence ',y.mean()
print 'predict accuracy ',(y1 == y).mean()
print 'dot product accuracy ',(y2 == y).mean()
print 'predict and dot agreement rate',(y1 == y2).mean()

coefs = svm.dual_coef_
assert len(coefs.shape) == 2
assert coefs.shape[0] == 1
coefs = coefs[0,:]
w = np.dot(svm.support_vectors_.T, coefs)
f = np.dot(X,w) + b
y3 = (f < 0)
print 'agreement rate with my method: ',(y3 == y1).mean()

print 'dot prod between sklearn coef_ and my coef_: ',np.dot(w,svm.coef_[0,:])







On Thu, May 24, 2012 at 11:17 AM, Ian Goodfellow
<[email protected]> wrote:
> On Thu, May 24, 2012 at 10:55 AM, Alexandre Gramfort
> <[email protected]> wrote:
>> is this test buggy:
>
> Yes. The test passes for me, but if I replace X and Y from the binary
> section of that test with the X and y from my script, the test fails.
>
>>
>> https://github.com/scikit-learn/scikit-learn/blob/master/sklearn/svm/tests/test_svm.py#L264
>>
>> ?
>>
>> could it be a numerical error?
>
> I don't know why the test passes for the default data.
>
>
>>
>> Alex
>>
>> On Thu, May 24, 2012 at 4:50 PM, Ian Goodfellow
>> <[email protected]> wrote:
>>> On Thu, May 24, 2012 at 10:47 AM, Olivier Grisel
>>> <[email protected]> wrote:
>>>> 2012/5/24 Ian Goodfellow <[email protected]>:
>>>>> Well that's the thing, coef_ and intercept_ seem wrong, given the
>>>>> results of my script below. My implementation of predict based on
>>>>> coef_ only agrees with predict 50% of the time.
>>>>> Does anyone know if coef_ and intercept_ are just getting set wrong?
>>>>> Or does predict implement a different decision function than
>>>>> (X*coef_+intercept_)>0?
>>>>
>>>> If X is dense then this should be, for binary classification with a
>>>> linear kernel:
>>>>
>>>> np.dot(X, clf.coef_) - clf.intercept > 0 (if I remember correctly the
>>>> sign of the intercept)
>>>
>>> I've tried both signs of the intercept and neither one agrees with the
>>> predict function.
>>>
>>>>
>>>> Off course this formula is specific to the dense linear case. It won't
>>>> work for the SVC models with kernels.
>>>
>>> Yes, of course. You'll notice that my debugging script uses kernel = 
>>> 'linear'...
>>>
>>>>
>>>> --
>>>> Olivier
>>>> http://twitter.com/ogrisel - http://github.com/ogrisel
>>>>
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