how to create a quoted expression from strings

Question

Given a vector of strings, I would like to create an expression without the quotation marks.

# eg, I would like to go from 
c("string1", "string2")

# to...  (notice the lack of '"' marks)
quote(list(string1, string2))

I am encountering some difficulty dropping the quotation marks

input <- c("string1", "string2")
output <- paste0("quote(list(", paste(input, collapse=","), "))")

# not quite what I am looking for.     
as.expression(output)
expression("quote(list(string1,string2))")

---

This is for use in data.table column selection, in case relevant.
What I am looking for should be able to fit into data.table as follows:

library(data.table)
mydt <- data.table(id=1:3, string1=LETTERS[1:3], string2=letters[1:3])

result <- ????? # some.function.of(input)
> mydt[ , eval( result )]
   string1 string2
1:       A       a
2:       B       b
3:       C       c

Answer 1

I tend to use as.quoted from the plyr package

 outputString <- sprintf('list(%s)', paste(input, collapse = ', ')) 


 library(plyr)
  output <- as.quoted(outputString)[[1]]

  mydt[, eval(output)]
   string1 string2
1:       A       a
2:       B       b
3:       C       c

However if it is simply column selection, you can pass the string and use with = FALSE

mydt[, input, with = FALSE]
   string1 string2
1:       A       a
2:       B       b
3:       C       c

Answer 2

Here is what I'd do:

## Create an example of a data.table "dt" whose columns you want to index 
## using a character vector "xx"
library(data.table)
dt <- data.table(mtcars)
xx <- c("wt", "mpg")

## Construct a call object identical to that produced by quote(list("wt", "mpg"))
jj <- as.call(lapply(c("list", xx), as.symbol))

## Try it out
dt[1:5,eval(jj)]
#       wt  mpg
# 1: 2.620 21.0
# 2: 2.875 21.0
# 3: 2.320 22.8
# 4: 3.215 21.4
# 5: 3.440 18.7

---

When "computing on the language" like this, it's often helpful to have a look at the structure of the object you're trying to construct. Based on the following (and once you know about as.call() and as.symbol()), creating the desired language object becomes a piece of cake:

x <- quote(list(wt, mpg))

str(x)
#  language list(wt, mpg)

class(x)
# [1] "call"

str(as.list(x))
# List of 3
#  $ : symbol list
#  $ : symbol wt
#  $ : symbol mpg

Answer 3

I found these answers helpful but incomplete for using variables and multiple lines within the expression. To create a quoted expression from strings, with variables and multiple lines make use of quote(), atop() and subsititute():

  # Prepare variables
  samp_skewness = round(skewness(dv),2)
  samp_kurtosis = round(kurtosis(dv),2)
  samp_var = round(var(dv))
  samp_summ <- summary(dv)
  num_samples = length(dv)

  # Prepare quotes containing math annotations
  q1 = quote(paste(mu,"="))
  q2 = quote(paste(sigma,"="))
  q3 = quote(paste(gamma[1],"="))
  q4 = quote(paste(gamma[2],"="))

# Use subsitition to construct the expression, passing in the variables and quoted expressions
  title = substitute(atop(paste("Top Title, # samples: ", ns), 
            paste(e1,v1,", ",e2,v2,", ",e3,v3,", ",e4,v4)),
            list(ns=num_samples,v1=round(samp_summ['Mean']),v2=samp_var,
            v3=samp_skewness,v4=samp_kurtosis,e1=q1,e2=q2,e3=q3,e4=q4))

In ggplot: ...

labs(title = title) +

...