User supplied arguments for ordering a data.frame using arrange

Question

Suppose I have this data.frame:

df = data.frame(strain=c("I","I","R","R"),sex=c("M","F","M","F"),age=c("8d","8d","64d","64d"))

and a user supplied data.frame that defines how to order df. For example:

order.df = data.frame(column = c("age","sex","strain"),order = c("+","-","+"))

I would like to use arrange of plyr package to order df according to the order defined by order.df. If it wasn't user supplied I would just do:

arrange(df, age, desc(sex), strain)

So my question is how do I achieve that with order.df? or what ever data structure is appropriate to store the user supplied order definition so it can work with arrange.

Answer 1

Here are two approach that will probably make @hadley kill some kittens...

# make sure that order.df contain character not factors
r <- lapply(order.df, as.character)
# create a list of names refering the columns
rl <- lapply(r[['column']],function(x)list(column = as.name(x)))
# create the appropriate list of arguments
rexp <- unname(Map (f=function(order,column){
  cm <- force(column)
  if (order =='-'){rn <-substitute(desc(column),cm)} else {
  rn <- substitute(column,cm)}
  rn
}, order=r[['order']],column = rl))



do.call(arrange, c(alist(df=df),rexp))
#  strain sex age
#1      R   M 64d
#2      R   F 64d
#3      I   M  8d
#4      I   F  8d



#alternatively, use as.quoted...
fmts <- ifelse(r[['order']]=='-', 'desc(%s)','%s')
rexp2 <- lapply(unname(Map(fmt = fmts, f= sprintf,r[['column']])), 
                function(x) as.quoted(x)[[1]])

do.call(arrange, c(alist(df=df),rexp2))
#  strain sex age
#1      R   M 64d
#2      R   F 64d
#3      I   M  8d
#4      I   F  8d