是否可以在 Jupyter Notebook 中看到 GridSearchCV 的进度？我正在 python 中运行这个脚本:
param_grid = {'learning_rate': [0.05, 0.10, 0.15, 0.20, 0.25, 0.30] ,
'max_depth' : [3, 4, 5, 6, 8, 10, 12, 15],
'min_child_weight' : [1, 3, 5, 7],
'gamma' : [0.0, 0.1, 0.2 , 0.3, 0.4],
'colsample_bytree' : [0.3, 0.4, 0.5 , 0.7],
'verbose' : [100] }
xgboost_reg = XGBRegressor()
grid_search = GridSearchCV(xgboost_reg, param_grid, cv=5, scoring='neg_mean_squared_error', return_train_score=True)
grid_search.fit(my_data, my_labels, verbose=False)
我只能在单元格的输出中看到一些警告。
最佳答案
您需要verbose参数:
grid_search = GridSearchCV(xgboost_reg, param_grid, cv=5, scoring='neg_mean_squared_error', return_train_score=True, verbose=2)
grid_search.fit(my_data, my_labels, verbose=False)
我从玩具数据中获得的示例:
Fitting 3 folds for each of 5 candidates, totalling 15 fits
[CV] C=0.1 ...........................................................
[CV] ............................................ C=0.1, total= 0.0s
[CV] C=0.1 ...........................................................
[CV] ............................................ C=0.1, total= 0.0s
[CV] C=0.1 ...........................................................
[CV] ............................................ C=0.1, total= 0.0s
[CV] C=0.5 ...........................................................
[CV] ............................................ C=0.5, total= 0.0s
[CV] C=0.5 ...........................................................
[CV] ............................................ C=0.5, total= 0.0s
[CV] C=0.5 ...........................................................
[CV] ............................................ C=0.5, total= 0.0s
关于python - Jupyter Notebook 中的 GridSearchCV 进展，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56495946/