Question

我有许多数据帧，其中包含一个因子，希望将其扩展为许多二进制等价物（一种热编码）。但是，在每个数据帧中，并非所有可能的因素都存在，但是我确实知道所有可能的因素是什么（有70个这样的因素）。我想将所有可能的二进制虚拟变量添加到每个数据帧。

从下面的代码中，我可以在每个数据帧中创建虚拟对象，但不能创建所有可能的虚拟对象。例如，set1.df没有任何人属于“ E”或“ F”类别，而set2.df没有任何人属于“ D”类别。需要的是set1.df中的其他列set1.dfE set1.dfF均为0，set2.df中的列set2.dfD均为零。在创建虚拟变量之前，我无法rbind set1.df和set2.df，因为在绑定之前我需要使用二进制变量对每个数据帧进行一些处理。再次重申一下，我知道事前数据中可能存在哪些级别，例如“ A”至“ F”。

library(dummies)

person_id <- c(1,2,3,4,5,6,7,8,9,10)
person_cat <- c("A","B","C","A","B","C","D","A","A","A")
set1.df <- data.frame(person_id,person_cat)

person_id <- c(11,12,13,14,15,16,17,18,19,20)
person_cat <- c("A","B","C","A","B","C","E","E","F","A")
set2.df <- data.frame(person_id,person_cat)

dummies1 <- dummy(set1.df[,2])
dummies2 <- dummy(set2.df[,2])

dummies1
dummies2

预期输出为：

> dummies1
      set1.dfA set1.dfB set1.dfC set1.dfD set1.dfE set1.dfF
 [1,]        1        0        0        0        0        0
 [2,]        0        1        0        0        0        0
 [3,]        0        0        1        0        0        0
 [4,]        1        0        0        0        0        0
 [5,]        0        1        0        0        0        0
 [6,]        0        0        1        0        0        0
 [7,]        0        0        0        1        0        0
 [8,]        1        0        0        0        0        0
 [9,]        1        0        0        0        0        0
[10,]        1        0        0        0        0        0
> dummies2
      set2.dfA set2.dfB set2.dfC set2.df$D set2.dfE set2.dfF
 [1,]        1        0        0        0        0        0
 [2,]        0        1        0        0        0        0
 [3,]        0        0        1        0        0        0
 [4,]        1        0        0        0        0        0
 [5,]        0        1        0        0        0        0
 [6,]        0        0        1        0        0        0
 [7,]        0        0        0        0        1        0
 [8,]        0        0        0        0        1        0
 [9,]        0        0        0        0        0        1
[10,]        1        0        0        0        0        0

Answer 1

这是一种解决方案：

levels <- c('A', 'B', 'C', 'D', 'E', 'F')

data <- data.frame(matrix(NA, nrow = length(person_id), ncol = length(levels)))
names(data) <- levels 
for (i in 1:nrow(data)) {
  for (j in 1:length(data)){
    data[i, j] <- ifelse(set1.df[i, 2] == names(data)[j], 1, 0)
  }
}

您应该创建一个空的数据框，其行数与ID相同，列数与set1.df中的级别相同。然后，使用循环来评估每一列中的person_cat。仅当person_cat等于列名（category_level）时，单元格的值才为1。

Answer 2

 library(dummies)

person_id <- c(1,2,3,4,5,6,7,8,9,10)
person_cat <- c("A","B","C","A","B","C","D","A","A","A")
person_cat < -factor(person_cat,levels=c("A","B","C","D","E","F"))
set1.df <- data.frame(person_id,person_cat)

person_id <- c(11,12,13,14,15,16,17,18,19,20)
person_cat <- c("A","B","C","A","B","C","E","E","F","A")
person_cat <- factor(person_cat,levels=c("A","B","C","D","E","F"))
set2.df <- data.frame(person_id,person_cat)

dummies1 <- dummy(set1.df[,2],drop=FALSE)
dummies2 <- dummy(set2.df[,2],drop=FALSE)

dummies1
dummies2

将一个因子转换为二元虚拟变量，但并非所有因子都存在

2 个答案: