如何通过可选参数do.call

Question

我希望以编程方式创建的变量名，如数组内的功能：

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

我相信我可以用do.call做到这一点，建立在这样的：

> 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" 

但是我不能完全弄清楚如何在do.call范围内将sep="_purchase_"参数传递给paste。在?do.call我读了

args是参数的函数调用列表。 args的names 属性给出参数名称。

试图将这一，我曾尝试：

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\")" 

什么是产生通过do.calldesired_output正确的方法是什么？

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。