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我的总体目标是确定在波士顿数据集上执行的 Superlearner 的变量重要性。但是，当我尝试使用 R 中的 VIP 包确定变量重要性时，我收到以下错误。我怀疑包含 SuperLeaner 对象的预测包装器是导致错误的原因，但我不确定。

# Call:  
# SuperLearner(Y = y_train, X = x_train, family = binomial(), SL.library =  # c("SL.mean",  
#    "SL.glmnet", "SL.ranger"), method = "method.AUC") 


#                    Risk      Coef
# SL.mean_All   0.55622189 0.3333333
# SL.glmnet_All 0.06240630 0.3333333
# SL.ranger_All 0.02745502 0.3333333
# Error in mean(actual == predicted, na.rm = FALSE): (list) object cannot be # coerced to type 'double'
# Traceback:

# 1. vi_permute(object = sl, method = "permute", feature_names = colnames, 
#  .     train = x_train, target = y_holdout, metric = "accuracy", 
#  .     type = "difference", nsim = 1, pred_wrapper = pred_wrapper)
# 2. vi_permute.default(object = sl, method = "permute", feature_names =    
#       colnames, 
#  .     train = x_train, target = y_holdout, metric = "accuracy", 
#  .     type = "difference", nsim = 1, pred_wrapper = pred_wrapper)
# 3. mfun(actual = train_y, predicted = pred_wrapper(object, newdata =  
#     train_x))
# 4. mean(actual == predicted, na.rm = FALSE)

我执行了以下操作：

library(MASS)
data(Boston, package = "MASS")

# Extract our outcome variable from the dataframe.
outcome = Boston$medv

# Create a dataframe to contain our explanatory variables.
data = subset(Boston, select = -medv)

set.seed(1)
# Reduce to a dataset of 150 observations to speed up model fitting.
train_obs = sample(nrow(data), 150)

# X is our training sample.
x_train = data[train_obs, ]

# Create a holdout set for evaluating model performance.
x_holdout = data[-train_obs, ]

# Create a binary outcome variable: towns in which median home value is > 22,000.
outcome_bin = as.numeric(outcome > 22)

y_train = outcome_bin[train_obs]
y_holdout = outcome_bin[-train_obs]

library(SuperLearner)
set.seed(1)
sl = SuperLearner(Y = y_train, X = x_train, family = binomial(),
  SL.library = c("SL.mean", "SL.glmnet", "SL.ranger"), method = "method.AUC")
sl

colnames <- colnames(x_train)
pred_wrapper <- function(sl, newdata) {
  predict(sl, x = as.matrix(y_holdout)) %>%
    as.vector()
}

# Plot VI scores
library(vip)
p1 <- vi_permute(object = sl, method = "permute", feature_names = colnames, train = x_train, 
          target = y_holdout,
          metric = "accuracy",
          type = "difference", 
          nsim = 1,
          pred_wrapper = pred_wrapper)
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1 回答 1

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对于该SuperLearner对象，您可以看到它返回一个概率列表

predict(sl,x_train[1:2,])
$pred
          [,1]
[1,] 0.4049966
[2,] 0.1905551

$library.predict
     SL.mean_All SL.glmnet_All SL.ranger_All
[1,]   0.3866667     0.5718232        0.2565
[2,]   0.3866667     0.1082986        0.0767

如果您阅读小插图 ( ?predict.SuperLearner)，我猜您想要来自超级学习者的预测。所以更改函数以提取概率并将它们转换为标签：

pred_wrapper <- function(sl, newdata) {
  ifelse(predict(sl,newdata)$pred>0.5,1,0)
}

我们简单检查一下结果：

table(pred_wrapper(sl,x_holdout),y_holdout)
   y_holdout
      0   1
  0 183  39
  1   9 125

并x_holdout用作火车：

p1 <- vi_permute(object = sl, method = "permute", feature_names = colnames, train = x_holdout, 
          target = y_holdout,
          metric = "accuracy",
          type = "difference", 
          nsim = 5,
          pred_wrapper = pred_wrapper) 

# A tibble: 13 x 3
   Variable Importance   StDev
   <chr>         <dbl>   <dbl>
 1 crim       0.00337  0.00126
 2 zn        -0.000562 0.00235
 3 indus      0.00337  0.00235
 4 chas       0.00674  0.00377
 5 nox        0.00225  0.00235
 6 rm         0.0315   0.0165 
 7 age        0.0213   0.0108 
 8 dis        0.0129   0.00944
 9 rad       -0.00169  0.00377
10 tax        0.00506  0.00126
11 ptratio    0.0174   0.0145 
12 black     -0.00281  0      
13 lstat      0.241    0.0204
于 2020-09-04T19:51:27.997 回答