ranger和classProbs = TRUE时缺少所有精度值

Question

library(dplyr)
library(caret)
library(doParallel)

cl <- makeCluster(3, outfile = '')
registerDoParallel(cl)
set.seed(2019)
fit1 <- train(x = X_train %>% head(1000) %>% as.matrix(),
              y = y_train %>% head(1000),
              method = 'ranger', 
              verbose = TRUE,
              trControl = trainControl(method = 'oob',
                                       verboseIter = TRUE,
                                       allowParallel = TRUE,
                                       classProbs = TRUE),
              tuneGrid = expand.grid(mtry = 2:3,
                                     min.node.size = 1, 
                                     splitrule = 'gini'),
              num.tree = 100,
              metric = 'Accuracy',
              importance = 'permutation')
stopCluster(cl)

上面的代码导致错误：

汇总结果出了点问题;缺少所有准确度指标值：准确度Kappa 闵。 ：NA Min。 ：NA 1 Qu。：NA 1st Qu。：NA 中位数：NA中位数：NA 平均值：NaN平均值：NaN 第三曲：NA第三曲：NA 最大。 ：NA Max。 ：NA NA：2 NA：2 错误：停止

我已经搜索过这个错误了，发现背后有很多可能的原因。不幸的是，我没有找到适用于我案件的任何内容。在这里，问题似乎与classProbs = TRUE - 当我删除这个和FALSE的默认值使用模型成功训练。但是，我不明白为什么它可能是一个问题，因为根据文件：

合乎逻辑的;应该为每个重采样中的分类模型（以及预测值）计算类概率吗？

数据样本：

X_train <- structure(list(V5 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), V1 = c(41.5, 
5.3, 44.9, 58.7, 67.9, 56.9, 3.7, 43.4, 38.6, 34.2, 42.3, 29.1, 
27.6, 44.2, 55.6, 53.7, 48, 58.4, 54, 7.1, 35.9, 36, 61.2, 24.1, 
20.3, 10.8, 13, 69.4, 71.5, 45.6, 34.4, 17.1, 30.1, 68.9, 25.1, 
37.4, 55.5, 58.9, 49.8, 47.2, 29.5, 19.9, 24.1, 27, 33.3, 41.9, 
33.2, 27.9, 48.4, 41.2), V2 = c(33.1, 35.4, 66.2, 1.8, 5, -0.9, 
32.8, 35.8, 36, 4, 65.5, 64, 61, 68.9, 69.3, 59.7, 29.8, 24.4, 
62.7, 12.2, 6, -1.2, 63.5, 7.5, 22.9, 40.5, 47.3, 1.6, -1.5, 
33.3, 53.3, 23.7, 2.7, 61, 2.4, 13.5, 8.1, 55.1, 29.6, 36.8, 
26.8, 26, 30.8, 53.8, 10.6, 1.9, 10.2, 29.1, 51.4, 33.1), V3 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0), V4 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -50L))
y_train <- structure(c(2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 
2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 
1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L), .Label = c("plus", "minus"), class = "factor")

Answer 1

基于对https://stats.stackexchange.com/questions/23763/is-there-a-way-to-disable-the-parameter-tuning-grid-feature-in-caret的回应，我尝试按照建议将trainControl“方法”设置为“无法允许成功执行。第二个答案回答暗示随机森林方法不应该使用复杂的网格。”我还设置了'mtry'参数单个值，但我不确定是否有必要。）（我之前曾尝试删除使用并行集群而不会对错误产生任何影响。）现在您可以添加功能，因为您的代码没有抛出错误。

fit1 <- train(form=y~., x = X_train[,2:3] ,
              y = factor(y_train) ,
              method = 'ranger', 
              verbose = TRUE,
              trControl=trainControl(method="none"),
              tuneGrid = expand.grid(mtry = 2,
                                     min.node.size = 1, 
                                     splitrule = 'gini'
                                     ),
              num.tree = 100,
              metric = 'Accuracy',
              importance = 'permutation')