如何将用户定义函数的参数传递为data.table中的列名？

Question

How can I pass an argument to be a column name of data.table in side the function? For example, I have a data called data1 with columns called 'hours' and 'location'. In the output, I want to find the outliers by location and named by 'hours'. I tried use substitute(y) and so forth, The output always uses 'y' as the column name. Could anyone help me? Thank you.

mf<-function(data, y){
newy<-as.name(deparse(substitute(y)))
output<-data[,.(y=boxplot.stats(eval(newy))$out),by=.(location)]
return(output)
}
mf(data=data1,y=hours)

Answer 1

It's better to write functions which take character values for choosing columns. In this case, your function can be rewritten as:

mf <- function(data, y){
  output <- data[, boxplot.stats(get(y))['out'], by = .(location)]
  setnames(output, 'out', y)
  return(output)
}

By using [ to subset the output of boxplot.stats , a named list with one element ( 'out' ) is returned. So output will have two columns: location and out . Then you just need to change out to be whatever was given for y .

Example:

set.seed(100)
data1 <- data.table(
  location = state.name,
  hours    = rpois(1000, 12)
)
mf(data = data1, y = 'hours')
#           location hours
#  1:       Delaware    25
#  2:        Georgia    21
#  3:          Idaho     4
#  4:  Massachusetts     5
#  5:       Missouri     7
#  6: South Carolina     5
#  7: South Carolina     6
#  8:   South Dakota    20
#  9:          Texas     5
# 10:           Utah    22

Non-standard evaluation is tricky and only worth the effort if you can get something out of it. data.table uses it for optimization behind the scenes. tidyverse packages use it to allow in-database processing. If there's no benefit (besides not having to type a few quotation marks), there's only a cost.