r - 如何在 R 中使用 SVM 进行递归特征消除

Question

我有一个如下所示的数据集

    ID  885038  885039  885040  885041  885042  885043  885044  Class
1267359 2       0       0       0       0       1       0      0
1295720 0       0       0       0       0       1       0      0
1295721 0       0       0       0       0       1       0      0
1295723 0       0       0       0       0       1       0      0
1295724 0       0       0       1       0       1       0      0
1295725 0       0       0       1       0       1       0      0
1295726 2       0       0       0       0       1       0      1
1295727 2       0       0       0       0       1       0      1
1295740 0       0       0       0       0       1       0      1
1295742 0       0       0       0       0       1       0      1
1295744 0       0       0       0       0       1       0      1
1295745 0       0       0       0       0       1       0      1
1295746 0       0       0       0       0       1       0      1

为了进行递归特征消除，我遵循了以下步骤

1. 训练 SVM 分类器
2. 计算所有功能的排名标准
3. 删除排名值最小的特征
4. 转到 1。

以下是我为执行相同操作而编写的 R 代码，但是，它没有显示任何错误，并且循环以训练集的长度继续。

data <- read.csv("dummy - Copy.csv", header = TRUE)
rownames(data) <- data[,1]
data<-data[,-1]

for (k in 1:length(data)){

  inTraining <- createDataPartition(data$Class, p = .70, list = FALSE)
  training <- data[ inTraining,]
  testing  <- data[-inTraining,]

  ## Building the model ####
  svm.model <- svm(Class ~ ., data = training, cross=10,metric="ROC",type="eps-regression",kernel="linear",na.action=na.omit,probability = TRUE)

  ###### auc  measure #######

  #prediction and ROC
  svm.model$index
  svm.pred <- predict(svm.model, testing, probability = TRUE)

  #calculating auc
  c <- as.numeric(svm.pred)
  c = c - 1
  pred <- prediction(c, testing$Class)
  perf <- performance(pred,"tpr","fpr")
  plot(perf,fpr.stop=0.1)
  auc <- performance(pred, measure = "auc")
  auc <- auc@y.values[[1]]

  #compute the weight vector
  w = t(svm.model$coefs)%*%svm.model$SV

  #compute ranking criteria
  weight_matrix = w * w

  #rank the features
  w_transpose <- t(weight_matrix)
  w2 <- as.matrix(w_transpose[order(w_transpose[,1], decreasing = FALSE),])
  a <- as.matrix(w2[which(w2 == min(w2)),]) #to get the rows with minimum values
  row.names(a) -> remove
  data<- data[,setdiff(colnames(data),remove)]
  print(length(data))
  length <- (length(data))
  cols_names <- colnames(data)
  print(auc)
  output <- paste(length,auc,sep=";")
  write(output, file = "output.txt",append = TRUE)
  write(cols_names, file = paste(length,"cols_selected", ".txt", sep=""))
}

打印输出如下

[1] 3
[1] 0.5
[1] 2
[1] 0.5
[1] 2
[1] 0.5
[1] 2
[1] 0.75
[1] 2
[1] 1
[1] 2
[1] 0.75
[1] 2
[1] 0.5
[1] 2
[1] 0.75

但是当我选择任何特征子集时，例如特征 3 并使用上面的代码(没有循环)构建 SVM 模型，我没有得到相同的 AUC 值 0.75。

data <- read.csv("3.csv", header = TRUE)
rownames(data) <- data[,1]
data<-data[,-1]

  inTraining <- createDataPartition(data$Class, p = .70, list = FALSE)
  training <- data[ inTraining,]
  testing  <- data[-inTraining,]

  ## Building the model ####
  svm.model <- svm(Class ~ ., data = training, cross=10,metric="ROC",type="eps-regression",kernel="linear",na.action=na.omit,probability = TRUE)

  ###### auc  measure #######

  #prediction and ROC
  svm.model$index
  svm.pred <- predict(svm.model, testing, probability = TRUE)

  #calculating auc
  c <- as.numeric(svm.pred)
  c = c - 1
  pred <- prediction(c, testing$Class)
  perf <- performance(pred,"tpr","fpr")
  plot(perf,fpr.stop=0.1)
  auc <- performance(pred, measure = "auc")
  auc <- auc@y.values[[1]]

  print(auc)

prints output 
    [1] 3
    [1] 0.75 (instead of 0.5)

这两个代码是相同的(一个有递归循环，另一个没有任何递归循环)，但同一特征子集的 AUC 值仍然存在差异。

两个代码的 3 个特征(885041、885043 和 Class)是相同的，但它给出了不同的 AUC 值。

Answer 1

我认为仅使用交叉验证就可以了。在您的代码中，您已经使用 10 倍 CV 来测试错误。拆分数据集似乎没有必要。

由于您没有提及调整参数，因此 cost 或 gamma 将设置为默认值。

library(tidyverse)
library(e1071)
library(caret)
library(ROCR)
library(foreach)

<小时/>

特征名称是数字，似乎svm()在拟合过程后更改了其中的名称。为了之后匹配，我会首先更改列名称。

其次，可以使用 caret::creadeFolds() 而不是 createDataPartition() 来指定折叠。

set.seed(1)
k <- 5 # 5-fold CV
mydf3 <-
  mydf %>% 
  rename_at(.vars = vars(-ID, -Class), .funs = function(x) str_c("X.", x, ".")) %>% 
  mutate(fold = createFolds(1:n(), k = k, list = FALSE)) # fold id column

# the number of features-------------------------------
x_num <-
  mydf3 %>% 
  select(-ID, -Class, -fold) %>% 
  ncol()

要进行迭代，foreach() 可以是另一种选择。

cl <- parallel::makeCluster(2)
doParallel::registerDoParallel(cl, cores = 2)
parallel::clusterExport(cl, c("mydf3", "x_num"))
parallel::clusterEvalQ(cl, c(library(tidyverse), library(ROCR)))
#---------------------------------------------------------------
svm_rank <-
  foreach(j = seq_len(x_num), .combine = rbind) %do% {
    mod <-
      foreach(cv = 1:k, .combine = bind_rows, .inorder = FALSE) %dopar% { # parallization
        tr <-
          mydf3 %>% 
          filter(fold != cv) %>% # train
          select(-fold, -ID) %>% 
          e1071::svm( # fitting svm
           Class ~ .,
           data = .,
           kernel = "linear",
           type = "eps-regression",
           probability = TRUE,
           na.action = na.omit
          )
        # auc
        te <-
          mydf3 %>% 
          filter(fold == cv) %>% 
          predict(tr, newdata = ., probability = TRUE)
        predob <- prediction(te, mydf3 %>% filter(fold == cv) %>% select(Class))
        auc <- performance(predob, measure = "auc")@y.values[[1]]
        # ranking - your formula
        w <- t(tr$coefs) %*% tr$SV
        if (is.null(names(w))) colnames(w) <- attr(tr$terms, "term.labels") # when only one feature left
        (w * w) %>%
          tbl_df() %>%
          mutate(auc = auc)
      }
    auc <- mean(mod %>% select(auc) %>% pull()) # aggregate cv auc
    w_mat <- colMeans(mod %>% select(-auc)) # aggregate cv ranking
    remove <- names(which.min(w_mat)) # minimum rank
    used <-
      mydf3 %>% 
      select(-ID, -Class, -fold) %>% 
      names() %>% 
      str_c(collapse = " & ")
    mydf3 <-
      mydf3 %>%
      select(-remove) # remove feature for next step
    tibble(used = used, delete = remove, auc = auc)
  }
#---------------------------------------------------
parallel::stopCluster(cl)

每一步，你都可以获得

svm_rank
#> # A tibble: 7 x 3
#>   used                                                      delete     auc
#>   <chr>                                                     <chr>    <dbl>
#> 1 X.885038. & X.885039. & X.885040. & X.885041. & X.885042… X.88503…   0.7
#> 2 X.885038. & X.885040. & X.885041. & X.885042. & X.885043… X.88504…   0.7
#> 3 X.885038. & X.885041. & X.885042. & X.885043. & X.885044. X.88504…   0.7
#> 4 X.885038. & X.885041. & X.885043. & X.885044.             X.88504…   0.7
#> 5 X.885038. & X.885041. & X.885043.                         X.88504…   0.7
#> 6 X.885038. & X.885041.                                     X.88503…   0.7
#> 7 X.885041.                                                 X.88504…   0.7