将函数从参数传递到嵌套函数

I a nutshell I would like to run the following minimal piece of code:

简单地说，我想运行以下最小的代码段:

library(plyr)

surround = function(my.df, my.var, my.val, my.method) {
    ddply(my.df, my.var, summarize, value = my.method(as.name(my.val)))
}

my.df = data.frame(group = rep(letters[1:4], times = 25),
                   x = rnorm(100))

surround(my.df, "group", "x", mean)

However, this leads to Error: could not find function "my.method". I do realize that this is a scoping issue and I should be using eval or substitute but I can't figure it out.

但是，这会导致错误:无法找到函数“my.method”。我确实意识到这是一个范围问题，我应该使用eval或代换，但我搞不清楚。

1 个解决方案

#1

If you use a customised function instead of summarise, it would work.

如果您使用定制的函数而不是摘要，它将会工作。

surround <- function(my.df, my.var, my.val, my.method) {
  ddply(my.df, my.var, function(x) c(value = my.method(x[[my.val]])))
}

my.df <- data.frame(group=rep(letters[1:4], times=25),
                   x=rnorm(100))

surround(my.df, "group", "x", mean)