问题描述
我已经搜索了这个答案几个小时了。很多人都问过类似的问题，但是我没有找到一个简单的问题或一个简单的答案。这是我的方法：
I have been googling this answer for a few hours. A lot of people have asked similar questions, but I did not find either a simple enough question or a straightforward answer. Here is my approach:
假设我想在 data.table 中进行一个简单的分组： p>
Assume that I want to do a simple group by in data.table:
library(data.table)
mtcars = data.table(mtcars)
mtcars[,sum(mpg), gear]
# Here are the results
# gear V1
#1: 4 294.4
#2: 3 241.6
#3: 5 106.9
但是，如果我使用自定义函数执行此操作：
However, if I use a self-defined function to do this:
zz = function(data, var, group){
return(data[,sum(var), group])
}
zz(mtcars, mpg, gear)
我收到一条错误消息：
我尝试过替代， eval ， quote 和其他解决方案，但它们都不起作用。我想知道是否有人可以对此提供更直接的解决方案和解释。
I've tried substitute, eval, quote, and other solutions, but none of them works. I wonder if anyone could give a more straightforward solution and explanation to this.
谢谢你，万圣节快乐！
推荐答案
如果我们使用不带引号的参数，请使用替换和 eval uate
If we are using unquoted arguments, substitute and evaluate
zz <- function(data, var, group){
var <- substitute(var)
group <- substitute(group)
setnames(data[, sum(eval(var)), by = group],
c(deparse(group), deparse(var)))[]
# or use
# setnames(data[, sum(eval(var)), by = c(deparse(group))], 2, deparse(var))[]
}
zz(mtcars, mpg, gear)
# gear mpg
#1: 4 294.4
#2: 3 241.6
#3: 5 106.9
这篇关于一个简单的可重现示例，它在R中的自定义函数中将参数传递给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！