What is the difference between cross_val_score wit

问题:

I am confused about the difference between the cross_val_score scoring metric 'roc_auc' and the roc_auc_score that I can just import and call directly.

The documentation (http://scikit-learn.org/stable/modules/model_evaluation.html#scoring-parameter) indicates that specifying scoring='roc_auc' will use the sklearn.metrics.roc_auc_score. However, when I implement GridSearchCV or cross_val_score with scoring='roc_auc' I receive very different numbers that when I call roc_auc_score directly.

Here is my code to help demonstrate what I see:

# score the model using cross_val_score

rf = RandomForestClassifier(n_estimators=150,
                            min_samples_leaf=4,
                            min_samples_split=3,
                            n_jobs=-1)

scores = cross_val_score(rf, X, y, cv=3, scoring='roc_auc')

print scores
array([ 0.9649023 ,  0.96242235,  0.9503313 ])

# do a train_test_split, fit the model, and score with roc_auc_score

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.33)
rf.fit(X_train, y_train)

print roc_auc_score(y_test, rf.predict(X_test))
0.84634039111363313 # quite a bit different than the scores above!

I feel like I am missing something very simple here -- most likely a mistake in how I am implementing/interpreting one of the scoring metrics.

Can anyone shed any light on the reason for the discrepancy between the two scoring metrics?

回答1:

This is because you supplied predicted y's instead of the probability in roc_auc_score. This function takes a score, not the classified label. Try instead to do this:

print roc_auc_score(y_test, rf.predict_proba(X_test)[:,1])

It should give a similar result to previous result from cross_val_score. Refer to this post for more info.

回答2:

I just ran into a similar issue here. The key takeaway there was that cross_val_score uses the KFold strategy with default parameters for making the train-test splits, which means splits into consecutive chunks rather than shuffling. train_test_split on the other hand does a shuffled split.

The solution is to make the split strategy explicit and specify shuffling, like this:

shuffle = cross_validation.KFold(len(X), n_folds=3, shuffle=True)
scores = cross_val_score(rf, X, y, cv=shuffle, scoring='roc_auc')

回答3:

Ran into this problem myself and after digging a bit found the answer. Sharing for the love.

There is actually two and a half problems.

1. you need to use the same Kfold to compare scores (the same split of the train/test);
2. you need to feed the probabilities into the roc_auc_score (using the predict_proba() method). BUT, some estimators (like SVC) does not have a predict_proba() method, you then use the decision_function() method.

Here's a full example:

# Let's use the Digit dataset
digits = load_digits(n_class=4)
X,y = digits.data, digits.target
y[y==2] = 0 # Increase problem dificulty
y[y==3] = 1 # even more

Using two estimators

LR = LogisticRegression()
SVM = LinearSVC()

Split the train/test set. But keep it into a variable we can reuse.

fourfold = StratifiedKFold(n_splits=4, random_state=4)

Feed it to GridSearchCV and save scores. Note we are passing fourfold.

gs = GridSearchCV(LR, param_grid={}, cv=fourfold, scoring='roc_auc', return_train_score=True)
gs.fit(X,y)
gs_scores = np.array([gs.cv_results_[k][0] for k in gskeys])

Feed it to cross_val_score and save scores.

 cv_scores = cross_val_score(LR, X, y, cv=fourfold, scoring='roc_auc')

Sometimes, you want to loop and compute several different scores, so this is what you use.

loop_scores = list()
for idx_train, idx_test in fourfold.split(X, y):
  X_train, y_train, X_test, y_test = X[idx_train], y[idx_train], X[idx_test], y[idx_test]
  LR.fit(X_train, y_train)
  y_prob = LR.predict_proba(X_test)
  auc = roc_auc_score(y_test, y_prob[:,1])
  loop_scores.append(auc)

Do we have the same scores across the board?

print [((a==b) and (b==c)) for a,b,c in zip(gs_scores,cv_scores,loop_scores)]
>>> [True, True, True, True]

---

BUT, sometimes our estimator does not have a predict_proba() method. So, according to this example, we do this:

for idx_train, idx_test in fourfold.split(X, y):
  X_train, y_train, X_test, y_test = X[idx_train], y[idx_train], X[idx_test], y[idx_test]
  SVM.fit(X_train, y_train)
  y_prob = SVM.decision_function(X_test)
  prob_pos = (y_prob - y_prob.min()) / (y_prob.max() - y_prob.min())
  auc = roc_auc_score(y_test, prob_pos)