r中的决策树与我的训练数据不一致

Question

library(caret)
library(rpart.plot)
car_df <- read.csv("TrainingDataSet.csv", sep = ',', header = TRUE)
str(car_df)
set.seed(3033)
intrain <- createDataPartition(y = car_df$Result, p= 0.7, list = FALSE)
training <- car_df[intrain,]
testing <- car_df[-intrain,]
dim(training)
dim(testing)
anyNA(car_df)
trctrl <- trainControl(method = "repeatedcv", number = 10, repeats = 3)
set.seed(3333)
dtree_fit <- train(Result ~., data = training, method = "rpart",
parms = list(split = "infromation"),
trControl=trctrl,
tuneLength = 10)

我收到这个警告：

警告消息：在nomialTrainWorkflow(x=x，y=y，wts=权重，info=trainInfo，：重采样中缺少值绩效衡量标准。

我试图用积极和消极情绪的数量来分类一部电影是成功还是失败。这里是我的数据

dput(car_df) 
structure(list(MovieName = structure(c(20L, 5L, 31L, 26L, 27L, 
12L, 36L, 29L, 38L, 4L, 6L, 8L, 10L, 15L, 18L, 21L, 24L, 34L, 
35L, 7L, 37L, 25L, 23L, 2L, 11L, 40L, 33L, 28L, 14L, 3L, 17L, 
16L, 32L, 22L, 30L, 1L, 19L, 39L, 9L, 13L), .Label = c("#96Movie", 
"#alphamovie", "#APrivateWar", "#AStarIsBorn", "#BlackPanther", 
"#BohemianRhapsody", "#CCV", "#Creed2", "#CrimesOfGrindelwald", 
"#Deadpool2", "#firstman", "#GameNight", "#GreenBookMovie", "#grinchmovie", 
"#Incredibles2", "#indivisiblemovie", "#InstantFamily", "#JurassicWorld", 
"#KolamaavuKokila", "#Oceans8", "#Overlord", "#PariyerumPerumal", 
"#RalphBreaksTheInternet", "#Rampage", "#Ratchasan", "#ReadyPlayerOne", 
"#RedSparrow", "#RobinHoodMovie", "#Sarkar", "#Seemaraja", "#Skyscraper", 
"#Suspiria", "#TheLastKey", "#TheNun", "#ThugsOfHindostan", "#TombRaider", 
"#VadaChennai", "#Venom", "#Vishwaroopam2", "#WidowsMovie"), class = "factor"), 
PositivePercent = c(40.10554, 67.65609, 80.46796, 71.34831, 
45.36082, 68.82591, 46.78068, 63.85787, 47.20497, 32.11753, 
63.7, 39.2, 82.76553, 88.78613, 72.18274, 72.43187, 31.0089, 
38.50932, 38.9, 19.9, 84.26854, 29.4382, 58.13953, 86.9281, 
64.54965, 56, 0, 56.61914, 58.82353, 54.98891, 78.21682, 
90, 64.3002, 85.8, 51.625, 67.71894, 92.21557, 53.84615, 
40.12158, 68.08081), NegativePercent = c(11.34565, 21.28966, 
6.408952, 13.10861, 26.80412, 17.10526, 18.61167, 10.55838, 
46.48033, 56.231, 9.9, 12.1, 9.018036, 6.473988, 13.90863, 
16.77149, 63.20475, 42.54658, 40.9, 5.4, 3.907816, 2.022472, 
10.51567, 3.267974, 15.12702, 15.3, 100, 18.12627, 11.76471, 
13.41463, 5.775076, 10, 20.08114, 2.1, 5.5, 7.739308, 0, 
34.61538, 12.86727, 10.70707), Result = structure(c(2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("Flop", "Hit"
), class = "factor")), class = "data.frame", row.names = c(NA, 
-40L))

Answer 1

> str(car_df)
'data.frame':   40 obs. of  4 variables:
$ MovieName      : Factor w/ 40 levels "#96Movie","#alphamovie",..: 20 5 31 26 27 12 36 29 38 4 ...
$ PositivePercent: num  40.1 67.7 80.5 71.3 45.4 ...
$ NegativePercent: num  11.35 21.29 6.41 13.11 26.8 ...
$ Result         : Factor w/ 2 levels "Flop","Hit": 2 2 2 2 2 2 2 2 2 1 ...
> with(car_df, table( Result))
Result
Flop  Hit 
5   35 
> dtree_fit
CART 
29 samples
3 predictor
2 classes: 'Flop', 'Hit' 

所以你有一个5次失败的结果，其中一个预测因子是一个有40个不同值的变量。这似乎并不奇怪，因为你的每一个案例都是独一无二的，而且你的结果严重不平衡。数据的存在并不能保证得出实质性结论的可能性。如果这里有任何错误，那就是fitter中缺乏代码，会说"真的吗？你认为统计包应该能够解决严重缺乏数据的问题吗？">

BTW：应该是(但毫不奇怪没有清除警告(：

(split = "information")

如果将交叉验证bin的数量更改为允许flop在各个bin之间分布的数量，则可以获得非警告结果。考虑到样本量较小，它是否会有很大的有效性仍然值得怀疑：

> trctrl <- trainControl(method = "repeatedcv", number = 3, repeats = 3)
set.seed(3333)
dtree_fit <- train(Result ~., data = training, method = "rpart",
parms = list(split = "infromation"),
trControl=trctrl,
tuneLength = 10)
# no warning on one of my runs