SMOTE平衡数据集后LightGBM训练仅单类问题修复请求

问题背景 #

原数据集为二分类不平衡数据集，Target变量中类别"0"有61471条记录，类别"1"有4456条记录。为缓解不平衡，使用SMOTE对训练集进行过采样处理。

SMOTE处理情况 #

执行SMOTE代码 #

trainSMOTE <- SMOTE(target ~ ., train, perc.over = 400, k = 5)
dim(trainSMOTE)

输出结果：57928 67

平衡验证 #

table(trainSMOTE$target)

输出结果：

0     1 
35648 22280

可见SMOTE后数据集已达到平衡状态。

LightGBM训练异常情况 #

训练代码片段 #

创建LightGBM数据集

train_data <- lgb.Dataset(data.matrix(trainSMOTE[, -9]), label = trainSMOTE[, trainSMOTE$target])
test_data <- lgb.Dataset(data.matrix(test[, -9]), label = test[, test$target])

模型参数定义

params <- list(
  objective = "binary",
  metric = 'auc', 
  boosting_type = "gbdt",
  num_leaves = 100, 
  learning_rate = 0.05,
  feature_fraction = 0.9,
  bagging_fraction = 0.8,
  bagging_freq = 5,
  min_data_in_leaf = 50,
  max_depth = -1,
  verbose = -1
)

启动训练

model <- lgb.train(params = params,
                   data = train_data,
                   valids = list(test = test_data),
                   early_stopping_rounds = 50)

模型报错信息 #

[LightGBM] [Warning] Contains only one class[LightGBM].
[Info] Number of positive: 57928, number of negative: 0.
[LightGBM] [Warning] Auto-choosing col-wise multi-threading, the overhead of testing was 0.050367 seconds.
You can set force_col_wise=true to remove the overhead.
[LightGBM] [Info] Total Bins 15250.
[LightGBM] [Info] Number of data points in the train set: 57928, number of used features: 65.
[LightGBM] [Info] [binary:BoostFromScore]: pavg=1.000000 -> initscore=34.539576.
[LightGBM] [Info] Start training from score 34.539576.
[LightGBM] [Info] [binary:BoostFromScore]: pavg=1.000000 -> initscore=34.539576.
[LightGBM] [Warning] Stopped training because there are no more leaves that meet the split requirements.
[1] "[1]: test's auc:0.5"...........

问题原因 #

核心错误出在标签列的提取逻辑：
代码中label = trainSMOTE[, trainSMOTE$target]的写法完全错误，trainSMOTE$target是一个长度为57928的向量，用它作为列索引时，会对每一行提取对应索引的列值，最终得到的不是标签列，而是一个混乱的矩阵，导致LightGBM识别到的所有样本标签被错误统一为同一类。

解决方案 #

修正标签提取逻辑 #

将标签列的提取方式改为直接取目标列，以下两种方式均可：

方式1：按列名提取（更健壮，避免列索引变化）

# 训练集
train_data <- lgb.Dataset(
  data = data.matrix(trainSMOTE[, -which(names(trainSMOTE) == "target")]),
  label = trainSMOTE$target
)
# 测试集
test_data <- lgb.Dataset(
  data = data.matrix(test[, -which(names(test) == "target")]),
  label = test$target
)

方式2：按列索引提取（已知target是第9列时使用）

# 训练集
train_data <- lgb.Dataset(
  data = data.matrix(trainSMOTE[, -9]),
  label = trainSMOTE[, 9]
)
# 测试集
test_data <- lgb.Dataset(
  data = data.matrix(test[, -9]),
  label = test[, 9]
)

额外验证步骤 #

在创建数据集前，先确认标签分布是否正确，避免再次出错：

# 确认训练集标签分布
table(trainSMOTE$target)
# 确认提取的label是否与原标签一致
table(trainSMOTE[, 9]) # 对应方式2的列索引

内容的提问来源于stack exchange，提问作者Guillermo Mansilla