从命名列表和函数名称创建可解析的字符串

aL3*_*3xa 2 r

我显然是一个简单的问题,我的灰质目前拒绝掌握 - 说我有一个清单:

list(a = "foo", b = c("bar", "biz", "booze"))
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和一个功能fn.我怎样才能得到这样的字符串:

"fn(a = \"foo\", b = c(\"bar\", \"biz\", \"booze\"))"
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PS

我知道我会很遗憾在早上问这个......

jor*_*ran 5

这应该让你开始,对吧......?

deparse(list(a = "foo", b = c("bar", "biz", "booze")),control = NULL)
[1] "list(a = \"foo\", b = c(\"bar\", \"biz\", \"booze\"))"
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一个更完整的版本,我完成了@ aL3xa评论...

gsub("^list","fn",
    deparse(list(a = "foo", b = c("bar", "biz", "booze")),control = NULL))
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