r - 用户提供的参数用于使用排列对 data.frame 进行排序

Question

假设我有这个 data.frame :

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

和一个用户提供的 data.frame定义如何订购 df .例如:

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

我想用 arrange的 plyr打包订购 df按照 order.df定义的顺序.如果不是用户提供的，我会这样做:

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

所以我的问题是如何使用 order.df 实现这一目标？或者什么数据结构适合存储用户提供的订单定义，以便它可以与安排一起工作。

Answer 1

这里有两种方法可能会让@hadley 杀死一些小猫......

# 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