r data.table 在函数调用中的用法

Question

我想在函数调用中一遍又一遍地执行 data.table 任务:Reduce number of levels for large categorical variables我的问题类似于 Data.table and get() command (R)或 pass column name in data.table using variable in R但我无法让它工作

没有函数调用，这工作得很好:

# Load data.table
require(data.table)

# Some data
set.seed(1)
dt <- data.table(type = factor(sample(c("A", "B", "C"), 10e3, replace = T)),
                 weight = rnorm(n = 10e3, mean = 70, sd = 20))

# Decide the minimum frequency a level needs...
min.freq <- 3350

# Levels that don't meet minumum frequency (using data.table)
fail.min.f <- dt[, .N, type][N < min.freq, type]

# Call all these level "Other"
levels(dt$type)[fail.min.f] <- "Other"

但包裹得像

reduceCategorical <- function(variableName, min.freq){
  fail.min.f <- dt[, .N, variableName][N < min.freq, variableName]
  levels(dt[, variableName][fail.min.f]) <- "Other"
}

我只会收到如下错误:

 reduceCategorical(dt$x, 3350)
Fehler in levels(df[, variableName][fail.min.f]) <- "Other" : 
 trying to set attribute of NULL value

有时

Error is: number of levels differs

Answer 1

一种可能性是使用 data.table::setattr 定义您自己的重新调平函数，这将修改 dt。有点像

DTsetlvls <- function(x, newl)  
   setattr(x, "levels", c(setdiff(levels(x), newl), rep("other", length(newl))))

然后在另一个预定义函数中使用它

f <- function(variableName, min.freq){
  fail.min.f <- dt[, .N, by = variableName][N < min.freq, get(variableName)]
  dt[, DTsetlvls(get(variableName), fail.min.f)]
  invisible()
}

f("type", min.freq)
levels(dt$type)
# [1] "C"     "other"

---

一些其他的data.table替代方案

f <- function(var, min.freq) {
  fail.min.f <- dt[, .N, by = var][N < min.freq, get(var)]
  dt[get(var) %in% fail.min.f, (var) := "Other"]
  dt[, (var) := factor(get(var))]
}

或者使用set/.I

f <- function(var, min.freq) {
  fail.min.f <- dt[, .I[.N < min.freq], by = var]$V1
  set(dt, fail.min.f, var, "other")
  set(dt, NULL, var, factor(dt[[var]]))
}

或者结合base R(不修改原始数据集)

f <- function(df, variableName, min.freq){
  fail.min.f <- df[, .N, by = variableName][N < min.freq, get(variableName)]
  levels(df$type)[fail.min.f] <- "Other"
  df
} 

或者，我们可以用 character 代替(如果 type 是 character)，你可以简单地做

f <- function(var, min.freq) dt[, (var) := if(.N < min.freq) "other", by = var]