如何在data.table的.SD调用中获取列名并实现文本追加

需求：为表格中包含字符"a"的单元格追加后缀" by {列名}"，其中{列名}是该单元格所在列的原始变量名，需对所有列执行此操作。

输入表格 #

library(data.table)
# Input 
ip <- data.table(x = c("ab", "cd", "ac", "de"),
                 y = c("fr", "ad", "fa", "we"))

ip[]
#>     x  y
#> 1: ab fr
#> 2: cd ad
#> 3: ac fa
#> 4: de we

期望输出 #

# Desired Output table
op <- data.table(x = c("ab by x", "cd", "ac by x", "de"),
                 y = c("fr", "ad by y", "fa by y", "we"))
op[]
#>          x       y
#> 1: ab by x      fr
#> 2:      cd ad by y
#> 3: ac by x fa by y
#> 4:      de      we

尝试过的方法及问题 #

最初尝试用deparse(substitute(x))获取列名，单独调用函数时正常：

add_if_pattern <- function(x, pattern) {
  y <- deparse(substitute(x))
  fifelse(test = grepl(pattern, x),
          paste(x, "by",  y),
          x)
}

pattern <- "a"
z <- "blah"
q <- "bleh"
add_if_pattern(z, pattern) ## 符合预期
#> [1] "blah by z"
add_if_pattern(q, pattern) ## 符合预期
#> [1] "bleh"

但将该函数放入lapply(.SD)中执行时，得到的是X[[i]]而非原始列名：

tp <- copy(ip)
ip <- copy(tp)

vars <- names(ip)
ip[, (vars) := lapply(.SD,add_if_pattern, pattern)]
ip[]
#>               x            y
#> 1: ab by X[[i]]           fr
#> 2:           cd ad by X[[i]]
#> 3: ac by X[[i]] fa by X[[i]]
#> 4:           de           we

尝试用names(.SD)获取列名时，因作用域和参数含义错误报错：

ip <- copy(tp)
ip[, (vars) := lapply(.SD,
                      \(x){
                        fifelse(test = grepl("classified", x),
                                paste(x, "by",  names(.SD)[..x]),
                                x)
                      })]
#> Error in `[.data.table`(ip, , `:=`((vars), lapply(.SD, function(x) {: Variable 'x' is not found in calling scope. Looking in calling scope because this symbol was prefixed with .. in the j= parameter.

解决方案 #

方法1：使用Map同时遍历列值与列名 #

Map可以同时传入列值（.SD）和对应的列名（names(.SD)），这样就能在函数中直接拿到当前列的名称：

ip <- copy(tp)
pattern <- "a"
ip[, (vars) := Map(function(col_val, col_name) {
  fifelse(grepl(pattern, col_val), paste(col_val, "by", col_name), col_val)
}, .SD, names(.SD))]
ip[]

方法2：遍历列索引获取列名 #

通过seq_along(.SD)遍历列的索引，再通过索引从.SD中取列值、从names(.SD)中取列名：

ip <- copy(tp)
ip[, (vars) := lapply(seq_along(.SD), function(i) {
  col_val <- .SD[[i]]
  col_name <- names(.SD)[i]
  fifelse(grepl(pattern, col_val), paste(col_val, "by", col_name), col_val)
})]
ip[]

问题原因解释 #

• deparse(substitute(x))在lapply(.SD)中时，x代表的是.SD中的元素，内部表现为X[[i]]，因此解析后得到的是这个字符串而非原始列名。
• 第二种报错是因为x是列的单元格值，不是列的索引，用names(.SD)[..x]试图通过值取列名逻辑错误，且..前缀用于引用上层作用域的变量，这里使用场景不对。

内容的提问来源于stack exchange，提问作者R.Andres Castaneda