我正在运行以下代码：
v.ctrl <- trainControl(method = "repeatedcv", repeats = 1,number = 3,
summaryFunction = twoClassSummary,
classProbs = TRUE,
allowParallel=T)
xgb.grid <- expand.grid(nrounds = 10000,
eta = c(0.01,0.05,0.1),
max_depth = c(2,4,6,8,10,14))
set.seed(45)
xgb_tune <-train(target~.,
data = train,
method = "xgbTree",
trControl = cv.ctrl,
tuneGrid = xgb.grid,
verbose = TRUE,
metric = "LogLoss",
nthread = 3)
错误很简单：
火车出错（目标〜。，数据=火车，方法=＆＃34; xgbTree＆＃34;，trControl = cv.ctrl，： 未使用的参数（data = train，method =＆＃34; xgbTree＆＃34;，trControl = cv.ctrl，tuneGrid = xgb.grid，verbose = T，metric =＆＃34; LogLoss＆＃34;，nthread = 3）< / p>
我的数据集
structure(list(feature19 = c(0.58776, 0.40764, 0.4708, 0.67577, 0.41681, 0.5291, 0.33197, 0.24138, 0.49776, 0.58293), feature6 = c(0.48424, 0.48828, 0.58975, 0.33185, 0.6917, 0.53813, 0.76235, 0.7036, 0.33871, 0.51928), feature10 = c(0.61347, 0.65801, 0.69926, 0.23311, 0.8134, 0.55321, 0.72926, 0.663, 0.49206, 0.55531), feature20 = c(0.39615, 0.49085, 0.50274, 0.6038, 0.37487, 0.53582, 0.62004, 0.63819, 0.37858, 0.40478), feature7 = c(0.55901, 0.38715, 0.50705, 0.76004, 0.3207, 0.54697, 0.31014, 0.21932, 0.4831, 0.52253), feature4 = c(0.5379, 0.52526, 0.44264, 0.28974, 0.65142, 0.41382, 0.44205, 0.47272, 0.6303, 0.56405), feature16 = c(0.41849, 0.45628, 0.37617, 0.39334, 0.46727, 0.36297, 0.3054, 0.41256, 0.6302, 0.41892), feature2 = c(0.62194, 0.5555, 0.61301, 0.27452, 0.74148, 0.49785, 0.5215, 0.46492, 0.54834, 0.58106), feature21 = c(0.32122, 0.37679, 0.35889, 0.74368, 0.18306, 0.47027, 0.40567, 0.47801, 0.41617, 0.35244), feature12 = c(0.56532, 0.55707, 0.49138, 0.24911, 0.69341, 0.42176, 0.41445, 0.45535, 0.62379, 0.5523), target = c(1L, 0L, 0L, 1L, 0L, 0L, 0L, 1L, 1L, 1L)), .Names = c("feature19", "feature6", "feature10", "feature20", "feature7", "feature4", "feature16", "feature2", "feature21", "feature12", "target"), row.names = c(NA, 10L), class = "data.frame")
有谁知道我是否必须重新处理xgbtree的数据？ 谢谢你！
答案 0 :(得分：0)
我意识到我在R / Caret /机器学习方面有点像菜鸟，但是在经常检查我的问题的答案后我看到了你的帖子，我设法让你的代码正常工作。我希望知识渊博的人能够完全回答你的问题，但与此同时，我也是这样做的。
首先，我将您的数据集输入到R并尝试运行您的代码。我相信你的控制功能可能会输入错误，你在“cv”中缺少一个“c”，这可能会导致你对未使用的参数产生问题。
然而，在我解决了这个问题之后，出现了多个错误和警告;例如，你正在使用twoClassSummary但是指定了logLoss（注意这里的语法，它不是LogLoss以防万一）...而是我把这个summaryFunction切换到mnlog来正确调用logLoss函数，就像我从中读到的那样2ClassSummary使用AUC作为其指标。另外，我用训练集中的“目标”变量替换了一个简单的字符变量，在本例中为“Y”或“N”。您可以下载csv文件here。
之后，我一直收到有关您的调优网格的错误，说明基本上您缺少xgBoost方法的调整参数，可以在插入符号文档（可用模型）中找到。我只是添加了其余参数的默认值（其中大多数是1）。我使用的调谐网格可以找到here。
我用来实际训练xgb模型的最终代码如下：
control = trainControl(method = "repeatedcv", repeats = 1, number = 3,
summaryFunction = mnLogLoss,
classProbs = TRUE,
allowParallel=T)
tune = train(x=set[,1:10], y=set[,11], method="xgbTree", trControl=control,
tuneGrid = xgb.grid, verbose=TRUE, metric="logLoss", nthread=3)
输出显示在这里：
tune
eXtreme Gradient Boosting
10 samples
10 predictors
2 classes: 'N', 'Y'
No pre-processing
Resampling: Cross-Validated (3 fold, repeated 1 times)
Summary of sample sizes: 6, 8, 6
Resampling results across tuning parameters:
eta max_depth logLoss
0.01 2 0.6914816
0.01 4 0.6914816
0.01 6 0.6914816
0.01 8 0.6914816
0.01 10 0.6914816
0.01 14 0.6914816
0.05 2 0.6848399
0.05 4 0.6848399
0.05 6 0.6848399
0.05 8 0.6848399
0.05 10 0.6848399
0.05 14 0.6848399
0.10 2 0.6765847
0.10 4 0.6765847
0.10 6 0.6765847
0.10 8 0.6765847
0.10 10 0.6765847
0.10 14 0.6765847
Tuning parameter 'nrounds' was held constant at a value of 10000
Tuning parameter 'gamma' was held constant at a
value of 0
Tuning parameter 'colsample_bytree' was held constant at a value of 1
Tuning parameter
'min_child_weight' was held constant at a value of 1
Tuning parameter 'subsample' was held constant at a value of 1
logLoss was used to select the optimal model using the smallest value.
The final values used for the model were nrounds = 10000, max_depth = 2, eta
= 0.1, gamma = 0, colsample_bytree =
1, min_child_weight = 1 and subsample = 1.
我希望这会有所帮助，而且正是你所寻求的。如果我正确地执行了日志丢失命令，我有点怀疑，因为它会显示最大深度字面上对日志丢失没有影响。我使用不同的度量标准AUC重新调整模型，结果显示无论改变什么都没有效果，与Cohen的Kappa相同。我猜这只是因为十个样本，但希望有人能够解释我做了什么，所以这不仅仅是代码转储。