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所以我有一个我无法弄清楚的特定错误。通过搜索，我发现模型和交叉验证集没有相同级别的数据来拟合模型。我试图完全理解我的用例。基本上我正在建立一个 QDA 模型来根据数值预测车辆国家。此代码适用于任何人，因为它是一个公开的 google 表格文档。对于那些在 YouTube 上关注 Doug Demuro 的人，您可能会觉得这有点有趣。

#load dataset into r
library(gsheet)
url = 'https://docs.google.com/spreadsheets/d/1KTArYwDWrn52fnc7B12KvjRb6nmcEaU6gXYehWfsZSo/edit'
doug_df = read.csv(text=gsheet2text(url, format='csv'), stringsAsFactors=FALSE,header=FALSE)

#begin cleanup. remove first blank rows of data
doug_df = doug_df[-c(1,2,3), ]
attach(doug_df)

#name columns appropriately
names(doug_df) = c("year","make","model","styling","acceleration","handling","fun factor","cool factor","total weekend score","features","comfort","quality","practicality","value","total daily score","dougscore","video duration","filming city","filming state","vehicle country")

#removing categorical columns and columns not being used for discriminate analysis to include totals columns
library(dplyr)
doug_df = doug_df %>% dplyr::select (-c(make,model,`total weekend score`,`total daily score`,dougscore,`video duration`,`filming city`,`filming state`))

#convert from character to numeric
num.cols <- c("year","styling","acceleration","handling","fun factor","cool factor","features","comfort","quality","practicality","value")
doug_df[num.cols] <- sapply(doug_df[num.cols], as.numeric)
`vehicle country` = as.factor(`vehicle country`)

#create a new column to reflect groupings for response variable
doug_df$country.group=ifelse(`vehicle country`=='Germany','Germany',
                          ifelse(`vehicle country`=='Italy','Italy',
                                 ifelse(`vehicle country`=='Japan','Japan',
                                        ifelse(`vehicle country`=='UK','UK',
                                               ifelse(`vehicle country`=='USA','USA','Other')))))

#remove the initial country column
doug_df = doug_df %>% dplyr::select (-c(`vehicle country`))

#QDA with multiple predictors
library(MASS)
qdafit1 = qda(country.group~styling+acceleration+handling+`fun factor`+`cool factor`+features+comfort+quality+value,data=doug_df)

#predict using model and compute error
n=dim(doug_df)[1]
fittedclass = predict(qdafit1,data=doug_df)$class
table(doug_df$country.group,fittedclass)
Error = sum(doug_df$country.group != fittedclass)/n; Error

#conduct k 10 fold cross validation
allpredictedCV1 = rep("NA",n)

cvk = 10
groups = c(rep(1:cvk,floor(n/cvk)))
set.seed(4)
cvgroups = sample(groups,n,replace=TRUE)

for (i in 1:cvk)  {
  qdafit1 = qda(country.group~styling+acceleration+handling+`fun factor`+`cool factor`+features+comfort+quality+value,data=doug_df,subset=(cvgroups!=i))
  newdata1i = data.frame(doug_df[cvgroups==i,])
  allpredictedCV1[cvgroups==i] = as.character(predict(qdafit1,newdata1i)$class)
}
table(doug_df$country.group,allpredictedCV1)
CVmodel1 = sum(allpredictedCV1!=doug_df$country.group)/n; CVmodel1

这是使用交叉验证的代码的最后一部分引发错误：

model.frame.default(as.formula(delete.response(Terms))、newdata 中的错误：可变长度不同（为“有趣的因素”找到）

有人可以帮我更深入地解释一下发生了什么吗？我认为变量乐趣因子在交叉验证的每一折中都没有与模型相同的水平。现在我需要知道我的选项来修复它。提前致谢！

编辑

除了上述之外，当我尝试预测虚拟汽车评论时，我遇到了一个非常相似的错误。

#build a dummy review and predict it using multiple models
dummy_review = data.frame(year=2014,styling=8,acceleration=6,handling=6,`fun factor`=8,`cool factor`=8,features=4,comfort=4,quality=6,practicality=3,value=5)

#predict vehicle country for dummy data using model 1
predict(qdafit1,dummy_review)$class

这将返回以下错误：

model.frame.default(as.formula(delete.response(Terms))、newdata 中的错误：可变长度不同（为“有趣的因素”找到）

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