Shortcut for evaluating strings in data.table assignments

Question

Wrangling data with data.table's, I find myself writing lines like this third line a lot:

DT = data.table(a = 1:10)
name = 'a'
DT[,eval(parse(text=sprintf('%s_plus_one := %s + 1',name,name)))]

which I had hoped to reduce to

DT[,s('%s_plus_one := %s + 1',name,name)]

using a function like:

# s is *very* short for substitute and evalute
s <- function(...)
    eval(parse(text=sprintf(...)))

but then I get this error:

> DT[,s('%s_plus_one := %s + 1',name,name)]
    Check that is.data.table(DT) == TRUE. Otherwise, := and `:=`(...) are defined for use in j, once only and in particular ways. See help(":=").

I know I can do this:

# sp is short for substitute and parse
sp <- function(...)
    parse(text=sprintf(...))

DT[,eval(sp('%s_plus_one := %s + 1',name,name))]
DT
#>      a a_plus_one
#>  1:  1          2
#>  2:  2          3
#>  3:  3          4
#>  4:  4          5

but building strings to be evaluated in a data.table assignment is so common, that I'd hoped minimize typing as much as possible.

why not build call from language objects instead of parsing string? it gives kind of validation before it is being evaluated. — jangorecki
– jangorecki, Commented Apr 1, 2016 at 2:41

thelatemail · Accepted Answer · 2016-04-01 00:59:42Z

7

Much of the time there shouldn't even be a need to go into the world of eval(parse()). I.e., for this example, put the name in parentheses on the left-hand side and use get on the right-hand side:

DT[, (paste0(name,"_plus_one")) := get(name) + 1]

DT
#     a a_plus_one
# 1:  1          2
# 2:  2          3
# 3:  3          4
# 4:  4          5
# 5:  5          6
# 6:  6          7
# 7:  7          8
# 8:  8          9
# 9:  9         10
#10: 10         11

edited Apr 1, 2016 at 0:59

answered Apr 1, 2016 at 0:53

thelatemail

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2 Comments

eddi Over a year ago

I didn't try, but I'm pretty sure the parentheses are not needed on lhs

thelatemail Over a year ago

@eddi - in this case, yep, but I tend to do it out of reflex as DT[, (name) := get(name) + 1] and DT[, name := get(name) + 1] give different results.

akrun · Accepted Answer · 2016-04-01 03:29:30Z

2

We could use .SDcols

DT[,(paste0(name,"_plus_one")) := .SD+1, .SDcols=name]
DT
#     a a_plus_one
# 1:  1          2
# 2:  2          3
# 3:  3          4
# 4:  4          5
# 5:  5          6
# 6:  6          7
# 7:  7          8
# 8:  8          9
# 9:  9         10
#10: 10         11

As @thelatemail mentioned this would also work with multiple columns

DT = data.table(a = 1:10, b=2:11)
name <- c("a","b")
DT[,(paste0(name,"_plus_one")) := .SD + 1, .SDcols=name]

edited Apr 1, 2016 at 3:29

answered Apr 1, 2016 at 3:14

akrun

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2 Comments

Jthorpe Over a year ago

Bonus question: is there a way to use this solution when merging two data.tables which contain variables of the same name, as in this question?

akrun Over a year ago

@Jthorpe Perhaps you may check .EACHI . Without an example, it is not clear how you want to do it.

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