将一个因子转换为二进制虚拟变量，但不是所有存在的因子

Question

我有许多数据帧，其中包含一个因子，我希望将其扩展为多个二进制等价物（一个热编码）。然而，在每个数据框中并不是所有可能的因素都存在，但我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对象添加到每个数据帧中。

从下面的代码中，我可以在每个数据框中创建虚拟对象，但不是所有可能的虚拟对象。例如，set1.df 在类别 "E" 或 "F" 中没有任何人，而 set2.df 在类别 "D" 中没有任何人。需要的是 set1.df 中全为 0 的 set1.dfE set1.dfF 列和全为零的 set2.df 列 set2.dfD。在创建虚拟对象之前我不能 rbind set1.df 和 set2.df 因为我需要在 rbinding 之前使用二进制变量对每个数据帧进行一些处理。只是重申一下，我事先知道我的数据可能有哪些级别，例如 "A" 到 "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)
  }
}

您应该创建一个空数据框，其行数与 ID 数相同，列数与您在 set1.df 中的级别数相同。然后，使用循环计算每一列中的 person_cat 。只有当 person_cat 等于列名 (category_level) 时，单元格的值才会为 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进制虚拟变量，但不是所有存在的因子

Convert a factor into binary dummies but not all factors present

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