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)

我收到此警告：

警告消息：在nominalTrainWorkflow（x = x，y = y，wts =重量， info = trainInfo ，：重新采样中缺少值绩效指标。

我正在尝试使用正面情绪和负面情绪的数量来区分电影是击打还是失败。这是我的数据

  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'

因此，您的翻牌结果为5，而预测变量之一是具有40个不同值的变量。考虑到您的每个案例都是唯一的，并且结果严重不平衡，这似乎并不令人惊讶。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，那就是钳工中缺少代码，这会说出“真的吗？您认为统计软件包应该能够解决严重的数据不足吗？”

顺便说一句：应该（但不足为奇的是不会清除警告）：

(split = "information")

如果将交叉验证箱的数量更改为允许将触发器分配到各个箱中的数量，那么您将得到非警告结果。鉴于样本量较小，它是否会有效仍然值得怀疑：

> 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

我的训练数据未形成R中的决策树

1 个答案: