光 GBM 回归 CV 解释结果答案

【问题标题】：Light GBM Regression CV Interpreting Results光 GBM 回归 CV 解释结果
【发布时间】：2021-07-03 18:43:01
【问题描述】：

我查看了文档，但找不到我的问题的答案，希望这里有人知道。下面是一些示例代码：

N_FOLDS= 5

model = lgb.LGBMClassifier()
default_params = model.get_params()

#overwriting a param
default_params['objective'] = 'regression'

cv_results = lgb.cv(default_params, train_set, num_boost_round = 100000, nfold = N_FOLDS, 
                    early_stopping_rounds = 100, metrics = 'rmse', seed = 50, stratified=False)

我得到一个这样的字典，其中包含 6 个不同的值：

{'rmse-mean': [635.2078190031074,
  632.0847253839236,
  629.6661071275558,
  627.9721515847672,
  626.6712284533291,
  625.293530527769],
 'rmse-stdv': [197.5088741303537,
  198.66960690389863,
  199.56134068525006,
  200.25929541235243,
  200.8251430042537,
  201.50213772830526]}

起初，我认为该字典中的值对应于每个折叠的 RMSE（在本例中为 5），但似乎并非如此。字典看起来像是 RMSE 值的递减。

有谁知道每个值对应什么？

【问题讨论】：

标签： python machine-learning regression cross-validation lightgbm

【解决方案1】：

它不对应于折叠，而是对应于每个提升轮的 cv 结果（所有测试折叠的 RMSE 的平均值），如果我们只说 5 轮并打印每轮的结果，您可以非常清楚地看到这一点：

import lightgbm as lgb
from sklearn.datasets import load_boston
X, y = load_boston(return_X_y=True)
train_set = lgb.Dataset(X,label = y)

params = {'learning_rate': 0.05,'num_leaves': 4,'subsample': 0.5}

cv_results = lgb.cv(params, train_set, num_boost_round = 5, nfold = N_FOLDS, verbose_eval  = True,
                    early_stopping_rounds = None, metrics = 'rmse', seed = 50, stratified=False)

[LightGBM] [Info] Total Bins 1251
[LightGBM] [Info] Number of data points in the train set: 404, number of used features: 13
[LightGBM] [Info] Start training from score 22.585149
[LightGBM] [Info] Start training from score 22.109406
[LightGBM] [Info] Start training from score 22.579703
[LightGBM] [Info] Start training from score 22.784158
[LightGBM] [Info] Start training from score 22.599010
[1] cv_agg's rmse: 8.86903 + 0.88135
[2] cv_agg's rmse: 8.58355 + 0.860252
[3] cv_agg's rmse: 8.31477 + 0.842578
[4] cv_agg's rmse: 8.06201 + 0.82627
[5] cv_agg's rmse: 7.8268 + 0.800053

import pandas as pd
pd.DataFrame(cv_results)

    rmse-mean   rmse-stdv
0   8.869030    0.881350
1   8.583552    0.860252
2   8.314774    0.842578
3   8.062014    0.826270
4   7.826800    0.800053

在您的帖子中，您设置了 early_stopping_rounds = 100 并使用了默认值 learning rate = 0.1，这取决于您的数据可能有点高，所以它很可能在 6 轮后停止。

使用上面的相同示例，如果我们设置early_stopping_rounds = 100，它会每 100 轮评估一次指标的改进，并在停止前 100 轮返回结果：

cv_results = lgb.cv(params, train_set, num_boost_round = 2000, nfold = N_FOLDS, 
verbose_eval  = True,early_stopping_rounds = 100, metrics = 'rmse',
seed = 50, stratified=False)

[...]
[1475]  cv_agg's rmse: 3.20605 + 0.50213
[1476]  cv_agg's rmse: 3.20616 + 0.501997
[1477]  cv_agg's rmse: 3.20607 + 0.501998
[1478]  cv_agg's rmse: 3.20636 + 0.501865
[1479]  cv_agg's rmse: 3.20631 + 0.501905
[1480]  cv_agg's rmse: 3.20633 + 0.501731
[1481]  cv_agg's rmse: 3.20659 + 0.501494
[1482]  cv_agg's rmse: 3.2068 + 0.502046
[1483]  cv_agg's rmse: 3.20687 + 0.50213
[1484]  cv_agg's rmse: 3.20701 + 0.502265
[1485]  cv_agg's rmse: 3.20717 + 0.502096
[1486]  cv_agg's rmse: 3.2072 + 0.501779
[1487]  cv_agg's rmse: 3.20722 + 0.501613
[1488]  cv_agg's rmse: 3.20718 + 0.501308
[1489]  cv_agg's rmse: 3.20701 + 0.501232

pd.DataFrame(cv_results).shape
(1389, 2)

如果您想从模型中估计 rmse，请取最后一个值。

【讨论】：