将变量名作为参数传递给data.table

本文介绍了将变量名作为参数传递给data.table的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在尝试创建一个修改data.table的函数，并希望使用一些非标准的评估，但是我意识到我并不真正知道如何在data.tables中使用它。
我的功能基本上是这样的：

I'm trying to create a function that modifies a data.table and wanted to use some non-standard evaluation but I realised that I don't really know how to work with it inside data.tables.My function is basically something like this:

do_stuff <- function(dt, col) {
  copy(dt)[, new_col := some_fun(col)][]
}

，我这样称呼它：

and I want to call it thus:

do_stuff(data, column)

其中列为数据内部存在的列的名称。如果我运行该函数，则会出现错误：

Where "column" is the name of the column that exists inside "data". If I run that function I get an error:

#> Error in some_fun(col) : object 'column' not found

哪个对我说data.table显然是将正确的名称传递给函数（列），但是由于某种原因，找不到它。这是一个最小的可复制示例

Which says to me that data.table is apparently passing the correct name to the function ("column") but for some reason it's not finding it. Here's a minimal reproducible example

library(data.table)

data <- data.table(x = 1:10, y = rnorm(10))

plus <- function(x, y) {
   x + y
}

add_one <- function(data, col) {
   copy(data)[, z := plus(col, 1)][]
}

add_one(data, y)
#> Error in plus(col, 1): object 'y' not found

使用 deparse（substitute（col））似乎不起作用，不幸的是：（

Using deparse(substitute(col)) doesn't seem to work, unfortunately :(

add_one <- function(data, col) {
   copy(data)[, z := plus(deparse(substitute(col)), 1)][]
}

add_one(data, y)
#> Error in x + y: non-numeric argument to binary operator

推荐答案

通常，引用和评估将起作用：

Generally, quote and eval will work:

library(data.table)
plus <- function(x, y) {
   x + y
}

add_one <- function(data, col) {
   expr0 = quote(copy(data)[, z := plus(col, 1)][])

   expr  = do.call(substitute, list(expr0, list(col = substitute(col))))
   cat("Evaluated expression:\n"); print(expr); cat("\n")

   eval(expr)
}

set.seed(1)
library(magrittr)
data.table(x = 1:10, y = rnorm(10)) %>%
   add_one(y)

which gives

Evaluated expression:
copy(data)[, `:=`(z, plus(y, 1))][]

     x          y         z
 1:  1 -0.6264538 0.3735462
 2:  2  0.1836433 1.1836433
 3:  3 -0.8356286 0.1643714
 4:  4  1.5952808 2.5952808
 5:  5  0.3295078 1.3295078
 6:  6 -0.8204684 0.1795316
 7:  7  0.4874291 1.4874291
 8:  8  0.7383247 1.7383247
 9:  9  0.5757814 1.5757814
10: 10 -0.3053884 0.6946116

这篇关于将变量名作为参数传递给data.table的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

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