R 在搭接中使用seq_

折叠交叉验证法，用于比较使用正向选择拟合的19个模型的残差。我在最后一步被卡住了。 代码如下所示：

library(ISLR)
summary(Hitters)  # Use the dataset of Hitters
Hitters = na.omit(Hitters)

library(leaps)

set.seed(11)
folds = sample(rep(1:10,length=nrow(Hitters)))  # used for cross validation later
table(folds)
cv.errors = matrix(NA,10,19)   
# store the errors from 10 validations, each contains an error for a model

# write a prediction function
predict.regsubsets = function(object,newdata,id,...){
  form = as.formula(object$call[[2]])   # extract the formula
  mat = model.matrix(form,newdata)      # extract the exploratory data
  coefi = coef(object,id=id)            # coefficients for the ith model
  return(mat[,names(coefi)]%*%coefi)    # manually get the predicted value
}
# write a function to extract the Mean of squared root of residuals
error = function(object,newdata,origin,num,...){
  pred = lapply(seq_along(1:num),function(x){predict.regsubsets(object,newdata,id=x)})
  sapply(pred,function(x){mean((x-origin)^2)})
}

# this gives error: $ operator is invalid for atomic vectors 
lapply(seq_along(1:10),function(X){
  best.fit = regsubsets(Salary~.,data=Hitters[folds!=X,],nvmax=19,method="forward")
  cv.errors[X,]=error(best.fit,newdata=Hitters[folds==X,],origin=Hitters$Salary[folds==X],num=19)
})

# this works well, except for being slow...
for(X in 1:10){
  best.fit = regsubsets(Salary~.,data=Hitters[folds!=X,],nvmax=19,method="forward")
  cv.errors[X,]=error(best.fit,newdata=Hitters[folds==X,],origin=Hitters$Salary[folds==X],num=19)
}

---

谢谢

这可能是一个范围界定问题。未尝试，但请尝试以下操作：

lapply(1:10,function(X){
  best.fit = regsubsets(Salary~.,data=Hitters[folds!=X,],nvmax=19,method="forward")
  cv.errors[X,] <- error(best.fit,newdata=Hitters[folds==X,],origin=Hitters$Salary[folds==X],num=19)
})

lappy（1:10，函数（X）{
best.fit=regsubsets（工资~，数据=Hitters[folds！=X，]，nvmax=19，method=“forward”）
简历错误[X]，不确定这是否能解决你的问题，但是：你真的需要在这里使用
seq\u吗？只要
1:10
就足够了。
seq\u-along（1:10）的值是1:10，所以SEQYALAN是多余的。考虑使用包<代码> CVoToS/<代码>而不是滚动自己。这里有什么问题？还不清楚。