CODEWITHSUNDEEP
rdata.table
colinfang
colinfang

Reputation: 21737

How to make a function to pass into j in the data.table in this simple case?

dt <- data.table(x=1:4, y=c(1,1,2,2), z=c(1,2,1,2))

I would likt to achieve this:

dt[,list(z, p=cumsum(x)), by=y]
   y z p
1: 1 1 1
2: 1 2 3
3: 2 1 3
4: 2 2 7

But via a function call like test(dt, z, x, y)

None of the following 2 ways works. data.table 1.8.10

test1 <- function(dt, a, b, c){
    dt[,list(eval(substitute(a), parent.frame()),
             p=cumsum(eval(substitute(b), parent.frame()))),
        by=eval(substitute(c)), verbose=TRUE]
}
test1(dt, z, x, y)
# Error in eval(expr, envir, enclos) : object 'a' not found
test2 <- function(dt, a, b, c){
    dt[,list(eval(substitute(a)),
             p=cumsum(eval(substitute(b)))),
        by=eval(substitute(c)), verbose=TRUE]
}
test2(dt, z, x, y)
# Error in eval(expr, envir, enclos) : object 'z' not found

What is a correct way to make it work?

Upvotes: 2

Views: 160

Answers (2)

Clayton Stanley
Clayton Stanley

Reputation: 7784

Here's a (IMO) cleaner version of @Chinmay Patil's second answer. It builds up the expression in a lispy way using R's backquote operator, and then evals the quoted expression.

test = function(dt, a, b, c) {
    z = substitute(a)
    x = substitute(b)
    y = substitute(c)
    expr = bquote(dt[, list(.(z), p=cumsum(.(x))), by=.(y)])
    eval(expr)
}

> test(dt, z, x, y)
   y z p
1: 1 1 1
2: 1 2 3
3: 2 1 3
4: 2 2 7
>

Upvotes: 1

CHP
CHP

Reputation: 17189

You can use deparse, substitute, eval and parse in following way. There maybe simpler solution, but following seem to work.

test1 <- function(dt, a, b, c){
  jvar <- paste0('list(',deparse(substitute(a)),', p=cumsum(',deparse(substitute(b)),'))')
  byvar <- paste0('list(', deparse(substitute(c)),')')
  dt[, eval(parse(text=jvar)), by=eval(parse(text=byvar))]

}

test1(dt, z, x, y)

##    y z p
## 1: 1 1 1
## 2: 1 2 3
## 3: 2 1 3
## 4: 2 2 7

or as @eddi sugguested 

test2 <- function(dt, a, b, c){
  eval(parse(text = paste0('dt[,', 'list(',deparse(substitute(a)),', p=cumsum(',deparse(substitute(b)),'))', ',by=', 'list(', deparse(substitute(c)),')', ']') ))  
}


test2(dt, z, x, y)
##    y z p
## 1: 1 1 1
## 2: 1 2 3
## 3: 2 1 3
## 4: 2 2 7

Upvotes: 3

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