如何返回每个分类实例的概率？

Question

Let's say that I already fitted scikit's SGDC , from the documentation I read that predict_proba() function return a vector of probability estimates, Thus I did the follwing:

In:
proba = clf.predict_proba(X_test)

print('proba:',proba.shape)
print(type(prediction))

Out:
proba: (292683, 39)
<class 'numpy.ndarray'>

However, I do not understand why proba has that dimention ( 292683, 39 ), insted of (292683,) . So, my question is how should I return the probability for each classified instance?. For example a vector full of the probabilities for each classified insance:

.9098
.6789
.2346
.4545
...
.9076

Update

This is my actual output:

0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38
1.6032895251736538e-09,0.0027001605689774967,1.3127275209812045e-05,0.0004133169272159469,6.421335538574734e-05,0.01244940641130727,4.971270475822253e-05,0.06927362982555345,0.05447770875726582,0.0002585581503775057,1.30512865257421e-05,0.00015347845576367026,0.004231831363568738,0.003134713706992086,0.00017618959500039568,0.004525087952898131,0.07230938415776024,0.004255936398577753,0.0006231217282368267,0.07381737590135892,1.7062740932146373e-05,0.04873946029933614,2.2579270275470988e-05,0.04738213671381574,0.011041250070307537,0.06786077438113797,0.008012001696580576,0.0009697583063038865,0.002640793732663328,0.00041955324710243576,0.005333452308762462,0.0023973060671898918,0.24386456744298726,1.2930500605063882e-05,0.010271860113445061,0.10478318644646997,0.1096803752152842,0.029709960729470408,0.0039009845913073
...
2.70775531177066e-05,0.056826721550724914,0.00021452452508401623,0.005773421211249144,0.03601322253697087,0.03387846954273534,0.0002233544773721261,0.0009621520077239175,0.005573279378280768,0.0011059321386392307,0.00014906386779747047,0.0007207742574711379,0.018149812871977058,0.017479374046348212,0.0004917497325634417,0.009446560753589354,0.37652447022205116,0.008895752894288417,0.00136242543496297,0.1961349850670937,0.011158949542858676,0.0010422870520728268,4.0487954942671204e-05,0.013908461124574075,0.005521009748034979,0.019087261334748272,0.00355886145992077,0.0054657023293853595,0.004395464092632666,0.00018729724505224616,0.0015209690844465442,0.003930224604070839,0.03922346296961368,2.1100171629256666e-05,0.001026959174556334,0.09177893762051553,0.021131552685297615,0.0007056741594152797,0.006342213576191516

Answer 1

predict_proba returns a vector of form P(y=y_i|x) for each y_i (class). Consequently, you can extract many measures from it. For example, if you are asking "how probable is my model's current classification" (thus your model's certainty in its own prediction) all you have to do is to index this array row-wise with your predictions so you get P(y=pred(x)|x), which is more or less:

for probs, pred in zip(clf.predict_proba(x), clf.predict(x)):
  print probs[pred]

you might also ask for probability of the correct class (meaning "according to my model, what is the probability of belonging to a valid class") analogously by (I am assuming y holds indexes of valid classes)

for probs, truth in zip(clf.predict_proba(x), y):
  print probs[truth]

Answer 2

I guess 39 is the number of different classes a sample could belong to.As you have done predict_proba. Its going to give you a probability of belonging to each particular class.

There is never going to be a single probability associated with each sample.

So, the error metric generally used for such situations is multi class log loss.