Question

我想改进随机森林回归者的 GridSearchCV 的参数。

def Grid_Search_CV_RFR(X_train, y_train):
    from sklearn.model_selection import GridSearchCV
    from sklearn.model_selection import ShuffleSplit
    from sklearn.ensemble import RandomForestRegressor

    estimator = RandomForestRegressor()
    param_grid = { 
            "n_estimators"      : [10,20,30],
            "max_features"      : ["auto", "sqrt", "log2"],
            "min_samples_split" : [2,4,8],
            "bootstrap": [True, False],
            }

    grid = GridSearchCV(estimator, param_grid, n_jobs=-1, cv=5)

    grid.fit(X_train, y_train)

    return grid.best_score_ , grid.best_params_

def RFR(X_train, X_test, y_train, y_test, best_params):
    from sklearn.ensemble import RandomForestRegressor
    estimator = RandomForestRegressor(n_jobs=-1).set_params(**best_params)
    estimator.fit(X_train,y_train)
    y_predict = estimator.predict(X_test)
    print "R2 score:",r2(y_test,y_predict)
    return y_test,y_predict

def splitter_v2(tab,y_indicator):
    from sklearn.model_selection import train_test_split
    # Asignamos X e y, eliminando la columna y en X
    X = correlacion(tab,y_indicator)
    y = tab[:,y_indicator]
    # Separamos Train y Test respectivamente para X e y
    X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
    return X_train, X_test, y_train, y_test

我使用此代码函数 5次：

for i in range(5):
    print "Loop: " , i
    print "--------------"
    X_train, X_test, y_train, y_test = splitter_v2(tabla,1)
    best_score, best_params = Grid_Search_CV_RFR(X_train, y_train)
    y_test , y_predict = RFR(X_train, X_test, y_train, y_test, best_params)
    print "Best Score:" ,best_score
    print "Best params:",best_params

这是结果：

Loop:  0
--------------
R2 score: 0.900071279487
Best Score: 0.61802821072
Best params: {'max_features': 'log2', 'min_samples_split': 2, 'bootstrap': False, 'n_estimators': 10}
Loop:  1
--------------
R2 score: 0.993462885564
Best Score: 0.671309726329
Best params: {'max_features': 'log2', 'min_samples_split': 4, 'bootstrap': False, 'n_estimators': 10}
Loop:  2
--------------
R2 score: -0.181378339338
Best Score: -30.9012120698
Best params: {'max_features': 'log2', 'min_samples_split': 4, 'bootstrap': True, 'n_estimators': 20}
Loop:  3
--------------
R2 score: 0.750116663033
Best Score: 0.71472985391
Best params: {'max_features': 'log2', 'min_samples_split': 4, 'bootstrap': False, 'n_estimators': 30}
Loop:  4
--------------
R2 score: 0.692075744759
Best Score: 0.715012972471
Best params: {'max_features': 'sqrt', 'min_samples_split': 2, 'bootstrap': True, 'n_estimators': 30}

¿为什么我在 R2得分中得到不同结果？，¿这是因为我选择 CV = 5 ？，¿这是因为我没有在 RandomForestRegressor（）上确定 random_state = 0 ？

Answer 1

for model in models:
    m = str(model)
    print(m)
    # Наш Pipeline
    text_clf = Pipeline([('vect', CountVectorizer()),
                      ('tfidf', TfidfTransformer()),
                      ('clf', model),
    ])
    # Обучение    
    text_clf = text_clf.fit(X_train.to_numpy(), y_train)
    # Предсказание
    pred = text_clf.predict(X_test)
    # Метрики
    print('accuracy_score', accuracy_score(pred, y_test))
    print('recall_score', recall_score(pred, y_test, average="macro"))
    print('f1_score', f1_score(pred, y_test, average="macro"))

#lr
C = [1,10,25,50,100,150]
solver = ['newton-cg', 'sag', 'saga', 'lbfgs']

# rfc 
n_estimators = [50,100,200,300,500]
max_features = ["auto", "sqrt", "log2"]
max_depth = [3,6]

# Knc 
n_neighbors=[5,10,15,20]
p=[1,2]

GridSearchCV随机森林回归调整最佳参数

1 个答案: