我重新创建了一个新的Ensemble方法，可以在我的三个分类器之间手动进行投票 . （丹尼尔的Courtesty帮助我从这里开始实现这个功能：Improving the prediction score by use of confidence level of classifiers on instances） .

此手动投票的目的是为最有信心的分类器接受每个实例的答案 . 以下是具有准确度分数的代码：

# parameters for random forest
rfclf_params = {
    'n_estimators': 500, 
    'bootstrap': True, 
    'class_weight':None, 
    'criterion':'gini',
    'max_depth':None, 
    'max_features':'auto',
    'warm_start': True,
    'random_state': 41
    # ... fill in the rest you want here
}

# Fill in svm params here
svm_params = {
    'C': 100,
    'probability':True,
    'random_state':42
}

# KNeighbors params go here
kneighbors_params= {
    'n_neighbors': 5,
    'weights':'distance'
}

y_test_classes = (y_test_sl, y_test_lim, y_test_shale, y_test_sandlim, y_test_ss, y_test_dol, y_test_sand)
classifiers = [RandomForestClassifier, SVC, KNeighborsClassifier]
params = [rfclf_params, svm_params, kneighbors_params]
y_trains_classes= (y_train_sl, y_train_lim, y_train_shale, y_train_sandlim, 
                   y_train_ss, y_train_dol, y_train_sand)
y_classes_names = ("shaly limestone", "limestone", "shale", "sandy lime", 
                   "shaly sandstone", "dolomite", "sandstone")

#Just get predictions
for y_trains, y_test, y_strings in zip(y_trains_classes, y_test_classes, y_classes_names):
    y_preds_test = ensemble_test(classifiers, params, X_train, y_trains, X_test_prepared)
    print("\n","Accuracy score for", y_strings, "=", accuracy_score(y_test, y_preds_test))
    print("f1_score for", y_strings, "=", f1_score(y_test, y_preds_test,
                                                        average = 'weighted', labels=np.unique(y_preds_test)))
    print("roc auc score for", y_strings, "=", roc_auc_score(y_test, y_preds_test,
                                                                  average = 'weighted'))

Accuracy score for shaly limestone = 0.949514563107
f1_score for shaly limestone = 0.949653574035
roc auc score for shaly limestone = 0.933362369338

 Accuracy score for limestone = 0.957281553398
f1_score for limestone = 0.957272532095
roc auc score for limestone = 0.957311555515

 Accuracy score for shale = 0.95145631068
f1_score for shale = 0.948556595316
roc auc score for shale = 0.845505617978

 Accuracy score for sandy lime = 0.998058252427
f1_score for sandy lime = 0.998008114117
roc auc score for sandy lime = 0.95

 Accuracy score for shaly sandstone = 0.996116504854
f1_score for shaly sandstone = 0.998054474708
roc auc score for shaly sandstone = 0.5

 Accuracy score for dolomite = 1.0
f1_score for dolomite = 1.0
roc auc score for dolomite = 1.0

 Accuracy score for sandstone = 0.996116504854
f1_score for sandstone = 0.996226826208
roc auc score for sandstone = 0.997995991984

当我想绘制一个ROC曲线时，我知道我需要从这个函数再次获得 predict_probas ，参考之前链接的建议，我将函数返回概率：

def ensemble_proba(classifiers, params, X_train, y_train, X_test):
    best_preds_test = np.zeros((len(X_test), 2))
    classes_test = np.unique(y_train)

    for i in range(len(classifiers)):
        # Construct the classifier by unpacking params 
        # store classifier instance
        clf_test = classifiers[i](**params[i])
        # Fit the classifier as usual and call predict_proba
        clf_test.fit(X_train, y_train)
        y_preds_test = clf_test.predict_proba(X_test)
        # Take maximum probability for each class on each classifier 
        # This is done for every instance in X_test
        # see the docs of np.maximum here: 
        # https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.maximum.html
        best_preds_test = np.maximum(best_preds_test, y_preds_test)

    # map the maximum probability for each instance back to its corresponding class
    preds_test = np.array([classes_test[np.argmax(pred)] for pred in best_preds_test])
    return np.array([np.amax(pred_probs) for pred_probs in best_preds_test])

现在，由于我想为测试集中的所有类绘制ROC曲线，我做了以下操作，得到的ROC曲线看起来与我预期的非常不同，因为我的ROC-AUC分数非常好，除了“shaly”砂岩“类 .

for y_trains, y_test, y_strings in zip(y_trains_classes, y_test_classes, y_classes_names):
    y_scores_ensemble_all = ensemble_proba(classifiers, params, X_train, y_trains, X_test_prepared)
    fpr_ensemble_all, tpr_ensemble_all, thresholds_ensemble_all = roc_curve(y_test_all,
                                                                              y_scores_ensemble_all)

    plt.figure(figsize=(8, 6))
    plot_roc_curve(fpr_ensemble_all, tpr_ensemble_all, "Ensemble manual voting")
    plt.legend(loc="lower right", fontsize=16)
    plt.title('ROC curve of Ensemble manual voting of  %s'%(y_strings))
    plt.axis([-0.01, 1.01, -0.01, 1.01])
    plt.show()

First two classes

Next two classes

Next two classes

Last class

为什么曲线看起来像这样，当他们的F1分数和ROC-AUC分数对几乎所有类别都相当好但是他们在ROC曲线上表现不佳？当我从函数中返回概率时，我做错了什么，或者由于某些原因曲线应该是这样的？

python machine-learning scikit-learn data-visualization multiclass-classification