我在学习R语言，并且通过使用caret包来尝试学习机器学习。
问题 – 在创建dummies并移除NZV变量后，当我将Y即预测变量作为因子添加回数据框时，它会在同一列中生成NA（问题步骤5-6）。那么，如何在最终数据框中保持Y变量为因子呢？
1. 数据（来自uci/kaggle的银行营销响应数据）
str(data)'data.frame': 4119 obs. of 21 variables: $ age : int 30 39 25 38 47 32 32 41 31 35 ... $ job : Factor w/ 12 levels "admin.","blue-collar",..: 2 8 8 8 1 8 1 3 8 2 ... $ marital : Factor w/ 4 levels "divorced","married",..: 2 3 2 2 2 3 3 2 1 2 ... $ education : Factor w/ 8 levels "basic.4y","basic.6y",..: 3 4 4 3 7 7 7 7 6 3 ... $ default : Factor w/ 3 levels "no","unknown",..: 1 1 1 1 1 1 1 2 1 2 ... $ housing : Factor w/ 3 levels "no","unknown",..: 3 1 3 2 3 1 3 3 1 1 ... $ loan : Factor w/ 3 levels "no","unknown",..: 1 1 1 2 1 1 1 1 1 1 ... $ contact : Factor w/ 2 levels "cellular","telephone": 1 2 2 2 1 1 1 1 1 2 ... $ month : Factor w/ 10 levels "apr","aug","dec",..: 7 7 5 5 8 10 10 8 8 7 ... $ day_of_week : Factor w/ 5 levels "fri","mon","thu",..: 1 1 5 1 2 3 2 2 4 3 ... $ duration : int 487 346 227 17 58 128 290 44 68 170 ... $ campaign : int 2 4 1 3 1 3 4 2 1 1 ... $ pdays : int 999 999 999 999 999 999 999 999 999 999 ... $ previous : int 0 0 0 0 0 2 0 0 1 0 ... $ poutcome : Factor w/ 3 levels "failure","nonexistent",..: 2 2 2 2 2 1 2 2 1 2 ... $ emp.var.rate : num -1.8 1.1 1.4 1.4 -0.1 -1.1 -1.1 -0.1 -0.1 1.1 ... $ cons.price.idx: num 92.9 94 94.5 94.5 93.2 ... $ cons.conf.idx : num -46.2 -36.4 -41.8 -41.8 -42 -37.5 -37.5 -42 -42 -36.4 ... $ euribor3m : num 1.31 4.86 4.96 4.96 4.19 ... $ nr.employed : num 5099 5191 5228 5228 5196 ... $ y : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...2. 保存X和Y变量
Y = subset(data, select = y)X = subset(data, select = -y)dim(X)dim(Y)[1] 4119 20[1] 4119 13. 创建dummies
pp_dummy <- dummyVars(y ~ ., data = data)data <- predict(pp_dummy, newdata = data)data <- data.frame(data)4. 使用接近零方差移除变量
nzv_list <- nearZeroVar(data) %>% as.vector()data <- data[, -nzv_list ]str(data)'data.frame': 4119 obs. of 44 variables: $ age : num 30 39 25 38 47 32 32 41 31 35 ... $ job.admin. : num 0 0 0 0 1 0 1 0 0 0 ... $ job.blue.collar : num 1 0 0 0 0 0 0 0 0 1 ... $ job.management : num 0 0 0 0 0 0 0 0 0 0 ... $ job.services : num 0 1 1 1 0 1 0 0 1 0 ... $ job.technician : num 0 0 0 0 0 0 0 0 0 0 ... $ marital.divorced : num 0 0 0 0 0 0 0 0 1 0 ... $ marital.married : num 1 0 1 1 1 0 0 1 0 1 ... $ marital.single : num 0 1 0 0 0 1 1 0 0 0 ... $ education.basic.4y : num 0 0 0 0 0 0 0 0 0 0 ... $ education.basic.6y : num 0 0 0 0 0 0 0 0 0 0 ... $ education.basic.9y : num 1 0 0 1 0 0 0 0 0 1 ... $ education.high.school : num 0 1 1 0 0 0 0 0 0 0 ... $ education.professional.course: num 0 0 0 0 0 0 0 0 1 0 ... $ education.university.degree : num 0 0 0 0 1 1 1 1 0 0 ... $ default.no : num 1 1 1 1 1 1 1 0 1 0 ... $ default.unknown : num 0 0 0 0 0 0 0 1 0 1 ... $ housing.no : num 0 1 0 0 0 1 0 0 1 1 ... $ housing.yes : num 1 0 1 0 1 0 1 1 0 0 ... $ loan.no : num 1 1 1 0 1 1 1 1 1 1 ... $ loan.yes : num 0 0 0 0 0 0 0 0 0 0 ... $ contact.cellular : num 1 0 0 0 1 1 1 1 1 0 ... $ contact.telephone : num 0 1 1 1 0 0 0 0 0 1 ... $ month.apr : num 0 0 0 0 0 0 0 0 0 0 ... $ month.aug : num 0 0 0 0 0 0 0 0 0 0 ... $ month.jul : num 0 0 0 0 0 0 0 0 0 0 ... $ month.jun : num 0 0 1 1 0 0 0 0 0 0 ... $ month.may : num 1 1 0 0 0 0 0 0 0 1 ... $ month.nov : num 0 0 0 0 1 0 0 1 1 0 ... $ day_of_week.fri : num 1 1 0 1 0 0 0 0 0 0 ... $ day_of_week.mon : num 0 0 0 0 1 0 1 1 0 0 ... $ day_of_week.thu : num 0 0 0 0 0 1 0 0 0 1 ... $ day_of_week.tue : num 0 0 0 0 0 0 0 0 1 0 ... $ day_of_week.wed : num 0 0 1 0 0 0 0 0 0 0 ... $ duration : num 487 346 227 17 58 128 290 44 68 170 ... $ campaign : num 2 4 1 3 1 3 4 2 1 1 ... $ previous : num 0 0 0 0 0 2 0 0 1 0 ... $ poutcome.failure : num 0 0 0 0 0 1 0 0 1 0 ... $ poutcome.nonexistent : num 1 1 1 1 1 0 1 1 0 1 ... $ emp.var.rate : num -1.8 1.1 1.4 1.4 -0.1 -1.1 -1.1 -0.1 -0.1 1.1 ... $ cons.price.idx : num 92.9 94 94.5 94.5 93.2 ... $ cons.conf.idx : num -46.2 -36.4 -41.8 -41.8 -42 -37.5 -37.5 -42 -42 -36.4 ... $ euribor3m : num 1.31 4.86 4.96 4.96 4.19 ... $ nr.employed : num 5099 5191 5228 5228 5196 ...5. 问题：在将y作为因子添加到数据时，会在列中产生NA。
data$y <- as.factor(Y)str(data)'data.frame': 4119 obs. of 45 variables: $ age : num 30 39 25 38 47 32 32 41 31 35 ... $ job.admin. : num 0 0 0 0 1 0 1 0 0 0 ... $ job.blue.collar : num 1 0 0 0 0 0 0 0 0 1 ... $ job.management : num 0 0 0 0 0 0 0 0 0 0 ... $ job.services : num 0 1 1 1 0 1 0 0 1 0 ... $ job.technician : num 0 0 0 0 0 0 0 0 0 0 ... $ marital.divorced : num 0 0 0 0 0 0 0 0 1 0 ... $ marital.married : num 1 0 1 1 1 0 0 1 0 1 ... $ marital.single : num 0 1 0 0 0 1 1 0 0 0 ... $ education.basic.4y : num 0 0 0 0 0 0 0 0 0 0 ... $ education.basic.6y : num 0 0 0 0 0 0 0 0 0 0 ... $ education.basic.9y : num 1 0 0 1 0 0 0 0 0 1 ... $ education.high.school : num 0 1 1 0 0 0 0 0 0 0 ... $ education.professional.course: num 0 0 0 0 0 0 0 0 1 0 ... $ education.university.degree : num 0 0 0 0 1 1 1 1 0 0 ... $ default.no : num 1 1 1 1 1 1 1 0 1 0 ... $ default.unknown : num 0 0 0 0 0 0 0 1 0 1 ... $ housing.no : num 0 1 0 0 0 1 0 0 1 1 ... $ housing.yes : num 1 0 1 0 1 0 1 1 0 0 ... $ loan.no : num 1 1 1 0 1 1 1 1 1 1 ... $ loan.yes : num 0 0 0 0 0 0 0 0 0 0 ... $ contact.cellular : num 1 0 0 0 1 1 1 1 1 0 ... $ contact.telephone : num 0 1 1 1 0 0 0 0 0 1 ... $ month.apr : num 0 0 0 0 0 0 0 0 0 0 ... $ month.aug : num 0 0 0 0 0 0 0 0 0 0 ... $ month.jul : num 0 0 0 0 0 0 0 0 0 0 ... $ month.jun : num 0 0 1 1 0 0 0 0 0 0 ... $ month.may : num 1 1 0 0 0 0 0 0 0 1 ... $ month.nov : num 0 0 0 0 1 0 0 1 1 0 ... $ day_of_week.fri : num 1 1 0 1 0 0 0 0 0 0 ... $ day_of_week.mon : num 0 0 0 0 1 0 1 1 0 0 ... $ day_of_week.thu : num 0 0 0 0 0 1 0 0 0 1 ... $ day_of_week.tue : num 0 0 0 0 0 0 0 0 1 0 ... $ day_of_week.wed : num 0 0 1 0 0 0 0 0 0 0 ... $ duration : num 487 346 227 17 58 128 290 44 68 170 ... $ campaign : num 2 4 1 3 1 3 4 2 1 1 ... $ previous : num 0 0 0 0 0 2 0 0 1 0 ... $ poutcome.failure : num 0 0 0 0 0 1 0 0 1 0 ... $ poutcome.nonexistent : num 1 1 1 1 1 0 1 1 0 1 ... $ emp.var.rate : num -1.8 1.1 1.4 1.4 -0.1 -1.1 -1.1 -0.1 -0.1 1.1 ... $ cons.price.idx : num 92.9 94 94.5 94.5 93.2 ... $ cons.conf.idx : num -46.2 -36.4 -41.8 -41.8 -42 -37.5 -37.5 -42 -42 -36.4 ... $ euribor3m : num 1.31 4.86 4.96 4.96 4.19 ... $ nr.employed : num 5099 5191 5228 5228 5196 ... $ y : Factor w/ 1 level "1:2": NA NA NA NA NA NA NA NA NA NA ...6. 如果我直接添加Y，它不会立即生成NA，但当我将其转换为因子时，它会生成NA
data$y <- Y # as.factor(Y)data <- data %>% mutate(y = as.factor(y))str(data)(更新)
7. 如果我不将其转换为因子，那么我总是需要使用pull(data$y)而不是直接使用data$y。以下是示例：
subsets <- c(7, 10, 12, 15, 20)control <- rfeControl(functions = rfFuncs, method = "cv", verbose = FALSE)system.time( RFE_res <- rfe(x = data[, 1:44], # subset(train, select = -y) y = pull(data$y), sizes = subsets, rfeControl = control )) 如何避免使用pull(data$y)，而直接使用data$y呢？
回答：
这与pull()无关。
你不能将一个数据框转换为向量，即使只有1列：
X = subset(iris,select=-Species)Y = subset(iris,select=Species)as.factor(Y)Species <NA> Levels: 1:3.valid.factor(Y)[1] "factor levels must be \"character\""levels(Y)NULL你需要调用数据框的列：
X$y = as.factor(Y$Species)# or X %>% mutate(y = as.factor(Y$Species))> str(X)'data.frame': 150 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ y : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...