问题描述
我想在函数调用中反复执行data.table任务:我的问题类似于或,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 character
s 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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