Setting several values at once in a list r - r

Question

This should be really simple. I am currently trying to make a list I am building slightly more efficient. Instead of having to write out:
list('1'= value1, '2' =value1, '3' = value1)
how would I condense this to be able to simply list the numbers I want to be equal to value1. e.g. '1:4' =value1 or '1,2,3,4' =value1
EDIT:
So, for background, I am currently trying to create custom formatting for an excel file using the xlsx package.
wb = createWorkbook()
sheet =createSheet(wb,sheetName = "TestFormatting")
dfcurrency = DataFormat("[$$-409]#,##0_ ;[Red]-[$$-409]#,##0 ")
dfdate = DataFormat("m/d/yyyy")
currency = CellStyle(wb, dataFormat = dfcurrency)
date = CellStyle(wb, dataFormat = dfdate)
datastyle = setNames(as.list(c(currency,date)),rep(c(3,4),c(1)))
data = addDataFrame(table,sheet, colStyle = datastyle)
Is what I am currently running, thanks to akrun's help. This gives the error:
Error in thisColStyle$ref : no field, method or inner class called 'ref'
And just in case it's useful, here is the data structure of table:
structure(list(workingdate = structure(c(1458518400, 1458604800,
1458691200, 1458777600, 1458864000, 1459119600), class = c("POSIXct",
"POSIXt"), tzone = ""), trader = structure(c(1L, 1L, 1L, 1L,
1L, 1L), .Label = c("a", "b", "c",
"d", "e"), class = "factor"), pnl.1d = c(3,
-573.7978, -107.1941, 1128.3061, -0.709699999999998, 3.55990000000003
), rt.1d.Util = c(0, -3.82531866666667e-05, -7.14627333333333e-06,
7.52204066666667e-05, -4.73133333333332e-08, 2.37326666666669e-07
)), .Names = c("workingdate", "trader", "pnl.1d", "rt.1d.Util"
), row.names = c(NA, 6L), class = "data.frame")

Answer 1

Here's a very general way to do similar things. This solution is likely more convoluted than the best solution, but it will work and can be extended to similar problems. It is based on eval and parse. parse turns a string into an unevaluated expression, eval evaluates it.
So, eval(parse(text="5+5")) will return 10.
If we can create the string "list('1'=value1, '2'=value1, '3'=value1)", we can then use eval(parse(text= to turn it into the list you want.
The following code will create the above string:
value1 <- 'asdf'
paste(
'list(', paste(sapply(seq_len(4),
function(n) { paste("'", n,"'", "=", "value1", sep="")}),
collapse = ","),
')')
So, combining everything, call
eval(parse(text=
paste(
'list(', paste(sapply(seq_len(4),
function(n) { paste("'", n,"'", "=", "value1", sep="")}),
collapse = ","),
')')))
And you get the list you want.

Answer 2

Thanks to Julian's comment I was able to create a solution to this. I will accept Julian's comment as the answer but will give my own (less general) solution as an example. It basically applies his solution so as to create more customisability in an albeit very roundabout way:
#if no columns need a type of format enter 0
a =paste(sapply(list(c(
#enter column numbers formatted as currency eg. 1:5, 8, 10
3
)),
function(n) { paste("'", n,"'", "=", "currency", sep="")}))
b =paste(sapply(list(c(
#columns formatted as date
1
)),
function(n) { paste("'", n,"'", "=", "date", sep="")}))
You can continue in this fashion with this general formula for as many variables as you like. You can then combine them into one text file ready to be parsed:
text = paste( 'list(',paste(c(a,b,c,d), collapse = ","),')')
datastyle = eval(parse(text = text))
where you simply enter all your formats or styles in a,b,c,d,...
Hopefully this will help someone who finds a similar problem.