在 data.table 的“by”中使用符号列表

Question

我想写一个函数outer_fun()，它做一些事情，还调用另一个函数inner_fun()。来自outer_fun() 的所有参数都传递给inner_fun()。

inner_fun() 对 data.table 进行一些计算（这是两个函数的参数）。另一个要通过函数传递的参数是by。

这是我所拥有的草图：

library(data.table)

data("CO2")
setDT(CO2)

outer_fun <- function(DT, by) {
    # some other stuff
    by <- substitute(by)
    inner_fun(DT, by)
}

inner_fun <- function(DT, by) {
    DT[, .(mean = mean(uptake)),
    by = list(Plant, by)]
}

outer_fun(CO2, by = Type)

这会引发错误：

Error in `[.data.table`(DT, , .(mean = mean(uptake)), by = list(Plant,  : 
  column or expression 2 of 'by' or 'keyby' is type language. Do not quote column names. Usage: DT[,sum(colC),by=list(colA,month(colB))] 

据我了解这个问题，我必须将by 中的两个列表正确组合到inner_fun() 中。另一种尝试是这样的：

outer_fun <- function(DT, by) {
    # some other stuff
    by <- substitute(by)
    inner_fun(DT, by)
}
inner_fun <- function(DT, by) {
    .by <- append(by, substitute(Plant), after = 0L)
    DT[, .(mean = mean(uptake)),
       by = eval(as.expression(list(.by)))]
}

outer_fun(CO2, by = Type)

这会引发类似的错误：

 Error in `[.data.table`(DT, , .(mean = mean(uptake)), by = eval(as.expression(list(.by)))) : 
  column or expression 1 of 'by' or 'keyby' is type symbol. Do not quote column names. Usage: DT[,sum(colC),by=list(colA,month(colB))] 

我在这里挣扎了两天，非常感谢您的帮助！

编辑： 看来，我还不清楚所需的解决方案应该具备什么能力。结果应该适用于data.table中允许的各种by，例如：

outer_fun(CO2, by = Type)
outer_fun(CO2, by = "Type")
outer_fun(CO2, by = .(Type, Treatment))
outer_fun(CO2, by = c("Type", "Treatment"))
outer_fun(CO2, by = "Type,Treatment")
outer_fun(CO2, by = Treatment == "chilled")
outer_fun(CO2, by = cut(conc, breaks = quantile(conc)))
...

Answer 1

我认为让[.data.table 函数自己处理by 参数会更好。这将使inner_fun 和outer_fun 更容易。缺点是Plant 等其他固定分组变量应在outer_fun 的by 参数中提供。

outer_fun <- function(DT, ...) {
    inner_fun(DT, ...)
}
inner_fun <- function(DT, ...) {
    DT[, .(mean = mean(uptake)), ...]
}

那么，下面所有的例子都是可能的：

outer_fun(CO2, by = .(Plant, Type, Treatment))
outer_fun(CO2, by = c("Plant", "Type", "Treatment"))
outer_fun(CO2, by = "Plant,Type,Treatment")
outer_fun(CO2, by = .(Plant, Treatment == "chilled"))
outer_fun(CO2, by = .(Plant, cut(conc, breaks = quantile(conc), include.lowest = T)))

Answer 2

我会通过最终将 by 信息作为字符提供来解决这个问题

library(data.table)

data("CO2")
setDT(CO2)

outer_fun <- function(DT, by) {
  # some other stuff
  by <- substitute(by)
  inner_fun(DT, by)
}

inner_fun <- function(DT, by) {

  # add Plant to the by Information
  byFun <- c("Plant", as.character(by))

  # remove list oder c()- function names
  byFun <- byFun[!byFun %in% c(".", "list", "c")] 

  DT[, .(mean = mean(uptake)),
     by = byFun]
 }

# single column name unquoted
outer_fun(CO2, by = Type)[1:3]
#>    Plant   Type     mean
#> 1:   Qn1 Quebec 33.22857
#> 2:   Qn2 Quebec 35.15714
#> 3:   Qn3 Quebec 37.61429


#list of column names unquoted
outer_fun(CO2, by = .(Type, Treatment))[1:3]
#>    Plant   Type  Treatment     mean
#> 1:   Qn1 Quebec nonchilled 33.22857
#> 2:   Qn2 Quebec nonchilled 35.15714
#> 3:   Qn3 Quebec nonchilled 37.61429

outer_fun(CO2, by = list(Type, Treatment))[1:3]
#>    Plant   Type  Treatment     mean
#> 1:   Qn1 Quebec nonchilled 33.22857
#> 2:   Qn2 Quebec nonchilled 35.15714
#> 3:   Qn3 Quebec nonchilled 37.61429


# single column name as string
outer_fun(CO2, by = "Type")[1:3]
#>    Plant   Type     mean
#> 1:   Qn1 Quebec 33.22857
#> 2:   Qn2 Quebec 35.15714
#> 3:   Qn3 Quebec 37.61429


# multiple column names as string
outer_fun(CO2, by = c("Type", "Treatment"))[1:3]
#>    Plant   Type  Treatment     mean
#> 1:   Qn1 Quebec nonchilled 33.22857
#> 2:   Qn2 Quebec nonchilled 35.15714
#> 3:   Qn3 Quebec nonchilled 37.61429

outer_fun(CO2, by = list("Type", "Treatment"))[1:3]
#>    Plant   Type  Treatment     mean
#> 1:   Qn1 Quebec nonchilled 33.22857
#> 2:   Qn2 Quebec nonchilled 35.15714
#> 3:   Qn3 Quebec nonchilled 37.61429

Answer 3

这是一个使用Chapter 6: Computing on the language的选项：

outer_fun <- function(DT, outby) {
    inner_fun(DT, substitute(outby))
}

inner_fun <- function(DT, inby) {
    l <- as.list(inby)

    if (l[[1L]] == as.symbol("list") || l[[1L]] == as.symbol(".")) 
        l <- l[-1L]

    if (is.character(l[[length(l)]])) {
        if (length(l) == 1L)
            l <- strsplit(l[[1L]],",")[[1L]]
        l <- lapply(l, as.name)

        if (l[[1L]] == as.symbol("c"))
            l <- l[-1L]
    }

    BY <- as.call(c(as.symbol("list"), as.name("Plant"), l))
    eval(bquote(DT[, mean(uptake), .(BY)]))
}


outer_fun(CO2, outby = list(Type, Treatment))
outer_fun(CO2, outby = list(Type))
outer_fun(CO2, outby = .(Type, Treatment))
outer_fun(CO2, outby = list(Treatment == "chilled"))
outer_fun(CO2, outby = list(cut(conc, breaks = quantile(conc))))
outer_fun(CO2, outby = Type)

outer_fun(CO2, outby = "Type")
outer_fun(CO2, outby = c("Type", "Treatment"))
outer_fun(CO2, outby = "Type,Treatment")