Question

我想知道如何使用sklearn运行多类，多标签，序数分类。我想预测目标群体的排名，范围从某个位置上最普遍的一个（1）到最不普遍的一个（7）。我似乎无法正确处理。你能帮我吗？


# Random Forest Classification

# Import
import numpy as np
import pandas as pd
from sklearn.model_selection import GridSearchCV, cross_val_score, train_test_split
from sklearn.metrics import make_scorer, accuracy_score, confusion_matrix, f1_score
from sklearn.preprocessing import StandardScaler
from sklearn.ensemble import RandomForestClassifier

# Import dataset
dataset = pd.read_excel('alle_probs_edit.v2.xlsx')
X = dataset.iloc[:,4:-1].values
Y = dataset.iloc[:,-1].values

# Split in Train and Test
X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state = 42 )

# Scaling the features (alle Variablen auf eine gleiche Ebene), necessary depend on the choosen method
sc_X = StandardScaler()
X_train = sc_X.fit_transform(X_train)
X_test = sc_X.transform(X_test)

# Creat classifier
classifier =  RandomForestClassifier(criterion = 'entropy')

# Choose some parameter combinations to try
parameters = {'bootstrap': [True, False],
 'max_depth': [50],
 'max_features': ['auto', 'sqrt'],
 'min_samples_leaf': [1, 2, 3, 4],
 'min_samples_split': [9, 10, 11, 12, 13],
 'n_estimators': [500,1000,1500]}

# Type of scoring used to compare parameter combinations
acc_scorer = make_scorer(accuracy_score)

# Run the grid search
grid_obj = GridSearchCV(classifier, parameters, scoring=acc_scorer, cv = 3, n_jobs = -1)
grid_obj = grid_obj.fit(X_train, Y_train)

# Set the classifier to the best combination of parameters
classifier = grid_obj.best_estimator_

# Fit the best algorithm to the data
classifier.fit(X_train, Y_train)

#Prediction the Test data
Y_pred = classifier.predict(X_test)

#Confusion Matrix
cm = pd.DataFrame(confusion_matrix(Y_test, Y_pred))

#Accuracy
accuracy1 = accuracy_score(Y_test, Y_pred)
print("Accuracy1: %.2f%%" % (accuracy1 * 100.0))

# k-Fold Class Validation
accuracy1 = cross_val_score(estimator = classifier, X = X_train, y = Y_train, cv = 10)
kfold = accuracy1.mean()
accuracy1.std()

Answer 1

这可能不是您要找的准确答案，this article概述了一种技术，如下所示：

我们可以通过将k类有序回归问题转换为k-1个二元分类问题来利用有序类的值，将有序值V1，V2，V3，…，Vk的有序属性A *转换为k -1个二进制属性，每个与原始属性的前k-1个值相对应。第ith个二进制属性表示测试A *> Vi

基本上，聚合多个二进制分类器（预测目标> 1，目标> 2，目标> 3，目标> 4）可以预测目标是1、2、3、4还是5。作者创建了一个在Python字典中存储多个二进制分类器的OrdinalClassifier类。

class OrdinalClassifier():

    def __init__(self, clf):
        self.clf = clf
        self.clfs = {}

    def fit(self, X, y):
        self.unique_class = np.sort(np.unique(y))
        if self.unique_class.shape[0] > 2:
            for i in range(self.unique_class.shape[0]-1):
                # for each k - 1 ordinal value we fit a binary classification problem
                binary_y = (y > self.unique_class[i]).astype(np.uint8)
                clf = clone(self.clf)
                clf.fit(X, binary_y)
                self.clfs[i] = clf

    def predict_proba(self, X):
        clfs_predict = {k:self.clfs[k].predict_proba(X) for k in self.clfs}
        predicted = []
        for i,y in enumerate(self.unique_class):
            if i == 0:
                # V1 = 1 - Pr(y > V1)
                predicted.append(1 - clfs_predict[y][:,1])
            elif y in clfs_predict:
                # Vi = Pr(y > Vi-1) - Pr(y > Vi)
                 predicted.append(clfs_predict[y-1][:,1] - clfs_predict[y][:,1])
            else:
                # Vk = Pr(y > Vk-1)
                predicted.append(clfs_predict[y-1][:,1])
        return np.vstack(predicted).T

    def predict(self, X):
        return np.argmax(self.predict_proba(X), axis=1)

该技术起源于A Simple Approach to Ordinal Classification

Answer 2

这是一个使用 KNN 的示例，它应该可以在 sklearn 管道或网格搜索中进行调整。

from sklearn.neighbors import KNeighborsClassifier
from sklearn.base import clone, BaseEstimator, ClassifierMixin
from sklearn.utils.validation import check_X_y, check_is_fitted, check_array
from sklearn.utils.multiclass import check_classification_targets

class KNeighborsOrdinalClassifier(BaseEstimator, ClassifierMixin):
    def __init__(self, n_neighbors=5, *, weights='uniform', 
                 algorithm='auto', leaf_size=30, p=2, 
                 metric='minkowski', metric_params=None, n_jobs=None):
        
        self.n_neighbors = n_neighbors
        self.weights = weights
        self.algorithm = algorithm
        self.leaf_size = leaf_size
        self.p = p
        self.metric = metric
        self.metric_params = metric_params
        self.n_jobs = n_jobs
        
    def fit(self, X, y):
        X, y = check_X_y(X, y)
        check_classification_targets(y)
        
        self.clf_ = KNeighborsClassifier(**self.get_params())
        self.clfs_ = {}
        self.classes_ = np.sort(np.unique(y))
        if self.classes_.shape[0] > 2:
            for i in range(self.classes_.shape[0]-1):
                # for each k - 1 ordinal value we fit a binary classification problem
                binary_y = (y > self.classes_[i]).astype(np.uint8)
                clf = clone(self.clf_)
                clf.fit(X, binary_y)
                self.clfs_[i] = clf
        return self
    
    def predict_proba(self, X):
        X = check_array(X)
        check_is_fitted(self, ['classes_', 'clf_', 'clfs_'])
        
        clfs_predict = {k:self.clfs_[k].predict_proba(X) for k in self.clfs_}
        predicted = []
        for i,y in enumerate(self.classes_):
            if i == 0:
                # V1 = 1 - Pr(y > V1)
                predicted.append(1 - clfs_predict[y][:,1])
            elif y in clfs_predict:
                # Vi = Pr(y > Vi-1) - Pr(y > Vi)
                 predicted.append(clfs_predict[y-1][:,1] - clfs_predict[y][:,1])
            else:
                # Vk = Pr(y > Vk-1)
                predicted.append(clfs_predict[y-1][:,1])
        return np.vstack(predicted).T
    
    def predict(self, X):
        X = check_array(X)
        check_is_fitted(self, ['classes_', 'clf_', 'clfs_'])
        
        return np.argmax(self.predict_proba(X), axis=1)

sklearn的多类，多标签，序数分类

2 个答案: