用 NA data.table 替换一些字符

Question

我正在构建一个函数来将一些字符（如“-”）替换为 R 中 data.table 内的正确 NA。

我的功能如下：

na_replacer <- function(data_set, characters_to_replace) {
  text_features <- names(data_set)[sapply(data_set, class) %in% c("character","factor")]
  for (x in text_features) {
    data_set[, lapply(.SD, function(x) replace(x, which(x==any(characters_to_replace)), NA))]
  }
  return (data_set)
}

当我运行这个函数时，我得到了以下异常：

charToDate(x) 中的错误：
字符串不在标准中 明确的格式

你能帮我让这个功能按预期工作吗？或者也许有一个更短的版本来做我尝试执行的操作？

这是一个调用函数的示例数据集

DT = data.table(ID = c("foo","bar","-","foo","[]","bah"), a = 1:6, b = 7:12, c = 13:18, d = c("aaa", "bbb", "ccc", "_", "eeee", "ffff"))
DT <- na_replacer(data_set = DT, characters_to_replace = c('-', '_', '[]'))

之前的数据集：

    ID a  b  c    d
1: foo 1  7 13  aaa
2: bar 2  8 14  bbb
3:   - 3  9 15  ccc
4: foo 4 10 16    _
5:  [] 5 11 17 eeee
6: bah 6 12 18 ffff

之后的预期数据集：

    ID a  b  c    d
1: foo 1  7 13  aaa
2: bar 2  8 14  bbb
3:  NA 3  9 15  ccc
4: foo 4 10 16   NA
5:  NA 5 11 17 eeee
6: bah 6 12 18 ffff

Answer 1

请测试在data.table 上运行的修改后的函数。

na_replacer <- function(data_set, characters_to_replace = c('-', '_')) {
    library(data.table)
    setDT(data_set)
    text_features <- names(data_set)[sapply(data_set, class) %in% c("character", "factor")]
    for (x in text_features) {
        foo <- data_set[, get(x)]
        data_set[, eval(x) := ifelse(foo %in% characters_to_replace, NA, foo)]
    }
    return(data_set)
}

Answer 2

OP 已请求将 data.table 的 character 或 factor 类型的所有列中的某些字符串替换为 NA。

previously accepted answer 在因子列中失败。

以下两种方法也适用于因子列：

加入更新

library(data.table)
options(datatable.print.class = TRUE)

for (col in DT[, names(.SD)[lapply(.SD, class) %in% c("character", "factor")]]) {
  DT[.(chr = c("-", "_", "[]")), on = paste0(col, "==chr"), (col) := NA_character_][]
}
DT

       ID     a     b     c      d
   <char> <int> <int> <int> <fctr>
1:    foo     1     7    13    aaa
2:    bar     2     8    14    bbb
3:     NA     3     9    15    ccc
4:    foo     4    10    16     NA
5:     NA     5    11    17   eeee
6:    bah     6    12    18   ffff

使用set()

for (col in DT[, names(.SD)[lapply(.SD, class) %in% c("character", "factor")]]) {
  set(DT, DT[get(col) %in% c("-", "_", "[]"), which = TRUE], col, NA_character_)
}
DT

       ID     a     b     c      d
   <char> <int> <int> <int> <fctr>
1:    foo     1     7    13    aaa
2:    bar     2     8    14    bbb
3:     NA     3     9    15    ccc
4:    foo     4    10    16     NA
5:     NA     5    11    17   eeee
6:    bah     6    12    18   ffff

数据

OP 在最新更新中提供的示例数据集正在使用，但有一处修改：列 d 被强制转换为 factor：

DT <- data.table(ID = c("foo", "bar", "-", "foo", "[]", "bah"), 
                 a = 1:6, b = 7:12, c = 13:18, 
                 d = factor(c("aaa", "bbb", "ccc", "_", "eeee", "ffff")))

Answer 3

检查一下：

solution <- function(dt, replacer) {
  result <- do.call(cbind, lapply(dt, function(x) lapply(x, function(x) {  ifelse(is.na(x), replacer, x) } )))
  as.data.frame(result)
}

# example:
dt <- data.frame(x = c(1, 4, NA, NA, 54), y = c(5, NA, -1, 0, 5))
cat("before:")
dt
cat("after:")
solution(dt, "-")

它将所有NA 值替换为data.frame 中的给定符号。

Answer 4

这样的东西能用吗

na_replacer <- function(data_set, characters_to_replace) {
  text_features <- names(data_set)[sapply(data_set, class) %in% c("character","factor")]
  for (x in text_features) {
    data_set[[x]][grep(paste0('[',characters_to_replace,']',collapse  =""),data_set[[x]])] <- NA
  }
  return (data_set)
}