如何在使用folds.split(train.values, target.values)时,在交叉验证中使用tqdm?

4

我正在使用lightgbm模型进行交叉验证,具体步骤如下。
同时,我想在for循环中使用tqdm来查看进程。

folds = KFold(n_splits=num_folds, random_state=2319)
oof = np.zeros(len(train))
getVal = np.zeros(len(train))
predictions = np.zeros(len(target))
feature_importance_df = pd.DataFrame()

print('Light GBM Model')
for fold_, (trn_idx, val_idx) in enumerate(folds.split(train.values, target.values)):

    X_train, y_train = train.iloc[trn_idx][features], target.iloc[trn_idx]
    X_valid, y_valid = train.iloc[val_idx][features], target.iloc[val_idx]


    print("Fold idx:{}".format(fold_ + 1))
    trn_data = lgb.Dataset(X_train, label=y_train, categorical_feature=categorical_features)
    val_data = lgb.Dataset(X_valid, label=y_valid, categorical_feature=categorical_features)

    clf = lgb.train(param, trn_data, 1000000, valid_sets = [trn_data, val_data], verbose_eval=5000, early_stopping_rounds = 4000)
    oof[val_idx] = clf.predict(train.iloc[val_idx][features], num_iteration=clf.best_iteration)
    getVal[val_idx]+= clf.predict(train.iloc[val_idx][features], num_iteration=clf.best_iteration) / folds.n_splits

    fold_importance_df = pd.DataFrame()
    fold_importance_df["feature"] = features
    fold_importance_df["importance"] = clf.feature_importance()
    fold_importance_df["fold"] = fold_ + 1
    feature_importance_df = pd.concat([feature_importance_df, fold_importance_df], axis=0)

    predictions += clf.predict(test[features], num_iteration=clf.best_iteration) / folds.n_splits

print("CV score: {:<8.5f}".format(roc_auc_score(target, oof)))

我尝试过使用 tqdm(enumerate(folds.split(train.values, target.values))) 或者 enumerate(tqdm(folds.split(train.values, target.values))),但是都没有起作用。
我猜测这是因为 enumerate 没有长度。
但我想知道在这种情况下如何使用 tqdm。
能否有人帮助我?
先谢谢了。

1个回答

16

在k折迭代过程中进行进度条(desc参数是可选的):

from tqdm import tqdm
for train, test in tqdm(kfold.split(x, y), total=kfold.get_n_splits(), desc="k-fold"):
   # Your code here

输出结果将会是这个样子:

k-fold: 100%|██████████| 10/10 [02:26<00:00, 16.44s/it]

你给我提供了一个非常好的答案,真心感谢! - Bowen Peng

网页内容由stack overflow 提供, 点击上面的
可以查看英文原文,
原文链接