r data.table在函数调用中的用法 [英] r data.table usage in function call

问题描述

我想在函数调用中反复执行data.table任务：减少大分类变量的级别数量我的问题类似于 Data.table和get（）命令或使用R中的变量传递data.table中的列名称，但我无法使用它

I want to perform a data.table task over and over in a function call: Reduce number of levels for large categorical variables My problem is similar to Data.table and get() command (R) or pass column name in data.table using variable in R but I can't get it to work

没有函数调用，这很好：

Without a function call this works just fine:

# 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"

但包装方式类似于

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

我只得到如下错误：

 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

推荐答案

一种可能性是定义自己使用 data.table :: setattr 重新调平函数，这将修改 dt 。像

One possibility is to define your own re-leveling function using data.table::setattr that will modify dt in place. Something like

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

Then use it within another predefined function

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"

---

其他 data.table 替代品

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))]
}

或使用 / .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]]))
}

或与基准R组合（不修改原始数据集）

Or combining with base R (doesn't modify original data set)

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，我们可以坚持我们字符 > type 是一个字符），你可以简单地做

Alternatively, we could stick we characters instead (if type is a character), you could simply do

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

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