如何从R表达式提取变量以在data.frame上下文中求值

Question

I have expressions in character that are supposed to be evaluated in a data.table (not important just context). To make sure all the required columns are present I would like to extract the said columns within the R expression.

What I want:

library(data.table)
DT <- data.table(p001=rnorm(10),p002=rnorm(10),p003=rnorm(10))
expr <- 'p001+mean(p001,na.rm=TRUE)-weighted.mean(p002,w=p003)+someRandomOtherColumn'

# DT[,test:=p001+mean(p001,na.rm=TRUE)-weighted.mean(p002,w=p003)+someRandomOtherColumn]
# would fail as p004 is not in the columns

Basically I am looking for a way (probably a regex) that would extract from expr p001,p002,p003,someRandomOtherColumn .

My view on it: The way I see it I should be able to capture p001,p001,TRUE,p002,p003,someRandomOtherColumn with some regex that would capture things within f(,) and then filter for 'allowed' column names ( TRUE is not in that case).

Nested f(,,) are not an issue as I can call the same function recursively and nested f(,(),) are also fine.

What I have: From now this is what I have, this can be made to work but this feels bad

expr <- 'p001+mean(p001,na.rm=TRUE)-weighted.mean(p002,w=p003)+someRandomOtherColumn'
clean <- function(string) gsub(string, pattern='[_|\\.|a-zA-z]+\\(([^)]*)\\)', replacement='\\1', perl=TRUE)
clean(expr)
[1] "p001+p001,na.rm=TRUE-p002,w=p003+someRandomOtherColumn"
# Then I can remove =* than split on ,|+|-|*

Answer 1

When you add a ~ to your expression, you can create a valid R formula expression:

expr <- '~ p001+mean(p001,na.rm=TRUE)-weighted.mean(p002,w=p003)+someRandomOtherColumn'

This string can be converted to a formula with as.formula . Afterwards, the variable names can be extracted with all.vars :

all.vars(as.formula(expr))
# [1] "p001"             "p002"             "p003"             "someRandomOtherColumn"