r - 将相同的因子水平应用于 R 中具有不同水平数量的多个变量

Question

我有一个 data.table与 168 variables和 8,278 observations .变量 69:135最初存储为字符串。他们应该成为区域假人，我想以 2 级(=是，公司在这里运营)和 1 级(=否，公司不在这里运营)结束。问题是原始变量中有三种不同的输入组合:1) "TRUE", "1", "0", "FALSE", 2) "TRUE", "FALSE", 和 3) "1"，“0”。此外，约。 5 个变量只有一个值，“0”或“1”。这里给出了一个例子:

#generating replicable data
structure(list(
  region1 = structure(c("TRUE", "FALSE", "0", "1", NA), class = "character"), 
  region2 = structure(c("1", "1", "0", NA, NA), class = "character"), 
  region3 = structure(c(NA, "FALSE", "TRUE", NA, "FALSE"), class = "character"),
  region4 = structure(c(NA, "0", "0", NA, "0"), class = "character")),
  .Names = c("region1", "region2", "region3", "region4"), row.names = c(NA, 5), class = "data.table")

#this gives:
#   region1 region2 region3 region4
#1    TRUE       1    <NA>    <NA>
#2   FALSE       1   FALSE       0
#3       0       0    TRUE       0
#4       1    <NA>    <NA>    <NA>
#5    <NA>    <NA>   FALSE       0                                                                                      

我正在寻找一种方法来一次性将所有变量的“TRUE”和“1”替换为 2，将“FALSE”和“0”替换为 1。所以想要的结果是:

#   region1 region2 region3 region4
#1:       2       2      NA      NA
#2:       1       2       1       1
#3:       1       1       2       1
#4:       2      NA      NA      NA
#5:      NA      NA       1       1

我已经看过了
Apply factor levels to multiple columns with missing factor levels
和
Change level of multiple factor variables .
但是，这对我没有帮助。
我使用嵌套的 ifelse() 尝试了以下操作命令:

library(data.table)
library(forcats)

check <- cbind(dt[1:68], as.data.table(apply(dt[69:135], 2, function(x) {
  ifelse("1" %in% x & "TRUE" %in% x,
         fct_collapse(x,
                      "2" = c("TRUE",
                              "1"),
                      "1" = c("FALSE",
                              "0")
         ),
         ifelse("1" %in% x & !("TRUE" %in% x),
                fct_collapse(x,
                             "2" = "1",
                             "1" = "0"),
                fct_collapse(x,
                             "2" = "TRUE",
                             "1" = "FALSE"
                )))
}
)), dt[136:168]) 

但是之前的代码没有给我想要的结果。它运行了，但我收到一条警告消息，并且在检查相应的变量时，它们仍然作为带有原始输入的字符串存储。

# examples of warnings
1: Unknown levels in `f`: TRUE, FALSE
2: Unknown levels in `f`: TRUE, FALSE
3: Unknown levels in `f`: TRUE, FALSE
4: Unknown levels in `f`: 0
5: Unknown levels in `f`: TRUE, FALSE
6: Unknown levels in `f`: TRUE, FALSE
7: Unknown levels in `f`: 0

单独使用和不与 fct_collapse 结合使用时也是如此嵌套 ifelse()命令完成这项工作:

#the ifelse statement works
ifelse("TRUE" %in% dt$region1, 2, "FALSE")
ifelse(5 %in% dt$region1, 2, "FALSE")

#also the nested ifelse statement works
ifelse("1" %in% dt$region1 & "TRUE" %in% dt$region1,
              0,
              ifelse("1" %in% dt$region1 & !("TRUE" %in% dt$region1),
                     1,
                     2
              ))


ifelse("1" %in% dt$region2 & "TRUE" %in% dt$region2,
               0,
               ifelse("1" %in% dt$region2 & !("TRUE" %in% dt$region2),
                      1,
                      2
               ))

有谁知道如何解决这个问题？
非常感谢您提前提供任何建议!

Answer 1

这是 set() 的方法在 for 中调用环形。

library(data.table)

f <- function(x){
  x <- as.character(x)
  i1 <- x %in% c("TRUE", "1")
  i0 <- x %in% c("FALSE", "0")
  x[which(i1)] <- "2"
  x[which(i0)] <- "1"
  as.integer(x)
}

for (j in seq_along(dt)) set(dt, j = j, value = f(dt[[j]]))

dt
#   region1 region2 region3 region4
#1:       2       2      NA      NA
#2:       1       2       1       1
#3:       1       1       2       1
#4:       2      NA      NA      NA
#5:      NA      NA       1       1

感谢 jangorecki's comment一个更简单的方法是

dt[, names(dt) := lapply(dt, f)]