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我打算用来care::sbf做单变量特征选择，我的输入是具有多个变量（又名，它的列）、候选特征列表和标签（又名，分类变量）的数据框。阅读caret包文档后，我尝试使用sbf,sbfController进行功能选择，但在下面遇到错误：

contrasts<-( , value = contr.funs *tmp*[1 + isOF[nn]]) 中的错误：
对比只能应用于具有 2 个或更多级别的因子

谁能指出我如何解决这个错误？caret::sbf使用什么来做特征选择是正确的？任何想法？

可重现的例子：

这是关于公共要点的可重现示例，我将其用作输入。

我目前的尝试：

library(caret)
library(e1071)
library(randomForest)

df=read.csv("df.csv", header=True)

sbfCtrl <- sbfControl(method = 'cv', number = 10, returnResamp = 'final', functions = caretFuncs, saveDetails = TRUE)

model <- sbf(form= ventil_status~ .,
                 data= df,
                 methods='knn',
                 trControl=trainControl(method = 'cv', classProbs = TRUE),
                 tuneGrid=data.frame(k=1:10),
                 sbfControl=sbfControl(functions = sbfCtrl,
                                       methods='repeatedcv', number = 10, repeats = 10))

print(model)
print(model$fit$results)

> model <- sbf(ventil_status~ ., data=df, sizes=c(1,5,10,20),
+              method= 'knn', trControl=trainControl(method = 'cv', classProbs = TRUE),
+              tuneGrid = data.frame(k=1:10),
+              sbfControl=sbfCtrl)
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : 
  contrasts can be applied only to factors with 2 or more levels

我用谷歌搜索了这个错误，但仍然无法克服它。任何想法使上述代码工作？使用 进行过滤器选择的正确方法是什么caret::sbf？

我想要的是输出数据框必须具有附加的 p 值的选定功能。所以这是我的尝试：

newdf <- df[ , -which(names(df) %in% c("subject"))]
p_value_vector <- sapply(names(newdf), function(i) 
    tryCatch(
        wilcox.test(newdf[newdf$ventil_status %in% "0", i], 
                        newdf[newdf$ventil_status %in% "1", i], 
                    na.action(na.omit))$p.value),
    warning = function(w) return(NA),
    error = function (e) return(NA)
)

预期输出：

我期待具有选定特征的输出数据框，其中返回的 p 值wilcox.test应附加到相应的特征。有什么想法可以在 r 中实现吗？如何caret::sbf正确操作特征选择？任何想法？

这是我的 R 会话信息：

> sessionInfo()
R version 3.6.3 (2020-02-29)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 18362)

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
 [1] ggpubr_0.2.5        magrittr_1.5        reshape2_1.4.3     
 [4] forcats_0.5.0       purrr_0.3.3         readr_1.3.1        
 [7] tibble_2.1.3        tidyverse_1.3.0     stringr_1.4.0      
[10] dplyr_0.8.5         scales_1.1.0        tidyr_1.0.2        
[13] aws.s3_0.3.20       randomForest_4.6-14 e1071_1.7-3        
[16] mlbench_2.1-1       caret_6.0-86        ggplot2_3.3.0      
[19] lattice_0.20-38
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1 回答 1

4

对于使用 sbf，您可以使用 caretSBF，然后添加您喜欢定义的分数和过滤器：

library(mlbench)
library(caret)

knnSBF = caretSBF
knnSBF$summary <- twoClassSummary
knnSBF$score <- function(x, y) {
    wilcox.test(x ~ y)$p.value
}
knnSBF$filter <- function(score, x, y) {
     score <= 0.05
}

然后定义训练参数和 sbf 参数：

sbfCtrl <- sbfControl(method = "cv",number = 3,
functions = knnSBF,saveDetails = TRUE)

trn_grid <- expand.grid(k=c(2,6,10))

trCtrl <-  trainControl(method = "cv",number = 3,
                        classProbs = TRUE,verboseIter = TRUE)

然后运行火车：

data(Sonar)
y = Sonar$Class
x = Sonar[,-ncol(Sonar)]
set.seed(111)
model1 <- sbf(x,y,trControl = trCtrl,
                sbfControl = sbfCtrl,
                method = "knn",
                tuneGrid = trn_grid)

model1$variables
$selectedVars
 [1] "V1"  "V2"  "V3"  "V4"  "V5"  "V6"  "V8"  "V9"  "V10" "V11" "V12" "V13"
[13] "V14" "V20" "V21" "V22" "V36" "V37" "V42" "V43" "V44" "V45" "V46" "V47"
[25] "V48" "V49" "V50" "V51" "V52" "V54" "V58"

$selectedVars
 [1] "V4"  "V5"  "V6"  "V9"  "V10" "V11" "V12" "V13" "V14" "V20" "V21" "V22"
[13] "V28" "V31" "V34" "V35" "V36" "V37" "V43" "V44" "V45" "V46" "V47" "V48"
[25] "V49" "V51" "V52"

$selectedVars
 [1] "V1"  "V2"  "V3"  "V4"  "V5"  "V6"  "V7"  "V8"  "V9"  "V10" "V11" "V12"
[13] "V13" "V14" "V21" "V22" "V23" "V34" "V35" "V36" "V37" "V43" "V44" "V45"
[25] "V46" "V47" "V48" "V49" "V50" "V51" "V52" "V53" "V56" "V58"

我不认为他们会返回 p 值，尽管我可能是错的。使用上面的示例，您可以计算 p 值

p_value_vector <- apply(x,2,function(i)wilcox.test(i~y)$p.value)
于 2020-03-31T21:36:33.790 回答