r - 如何将可选参数传递给 do.call 中的函数

Question

我希望以编程方式创建一个变量名数组，例如：

desired_output <- c("first_purchase_date","last_purchase_date","largest_purchase_date",
                    "first_purchase_amount","last_purchase_amount","largest_purchase_amount") 

我相信我可以通过以下方式做到这do.call一点：

> do.call(paste, expand.grid(c("first","last","largest"),c("date","amount")))
[1] "first date"     "last date"      "largest date"   "first amount"   "last amount"    "largest amount"

但是我不太清楚如何将sep="_purchase_"参数传递给pastewithin do.call。在?do.call我读到

args是函数调用的参数列表。的names 属性args给出参数名称。

试图合并这一点，我试过：

df <- expand.grid(c("first","last","largest"), 
                  c("date","amount"), 
                  stringsAsFactors = FALSE)

do.call(paste, args = list(...=df, sep="_purchase_"))
# does not return desired output, but instead:
# [1] "c(\"first\", \"last\", \"largest\", \"first\", \"last\", \"largest\")"
# [2] "c(\"date\", \"date\", \"date\", \"amount\", \"amount\", \"amount\")"  

desired_output生产via的正确方法是什么do.call？

Answer 1

你可以做do.call(paste, c(df, sep="_purchase_"))，但也许apply(df, 1, paste, collapse="_purchase_")更直接。