我想改进随机森林回归变量 * a的GridSearchCV**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
我在以下代码中使用了函数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
以下是结果:
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}
为什么我得到不同的结果在R2评分?,这是因为我选择CV=5?,这是因为我没有确定一个随机_状态=0对我的随机森林回归()?
2条答案
按热度按时间bd1hkmkf1#
qxsslcnc2#
定义调整_r2(r2,n,p):返回1-((1-r2)*(n-1)/(n-p-1))
i在范围(45,120,1)内时:对于范围(2,16,1)中的j:对于范围(10,30,1)内的k:rf =随机森林回归变量(n_估计量= k,随机_状态=i,最大_深度=j)