2

我正在尝试使用 NSE 构建一个公式，以便我可以轻松地在列中进行管道传输。以下是我想要的用例：

df %>% make_formula(col1, col2, col3)

[1] "col1 ~ col2 + col3"

我首先做了这个功能：

varstring <- function(...) {
 as.character(match.call()[-1])
}

这适用于单个对象或多个对象：

varstring(col)

[1] "col"

varstring(col1, col2, col3)

[1] "col1" "col2" "col3"

我创建了我的函数来创建接下来的公式：

formula <- function(df, col, ...) {
 group <- varstring(col)
 vars <- varstring(...)

 paste(group,"~", paste(vars, collapse = " + "), sep = " ")
}

但是，函数调用formula(df, col, col1, col2, col3)会产生[1] "group ~ ..1 + ..2 + ..3".

我知道该公式实际上是在评估varstring(group)，varstring(...)而不是像我想要的那样实际替换用户提供的对象进行评估。但我无法弄清楚如何按预期进行这项工作。

4

4 回答 4

3

您可以通过使用将任意数量的参数与二进制函数连接起来reduce()

make_formula <- function(lhs, ..., op = "+") {
  lhs <- ensym(lhs)
  args <- ensyms(...)

  n <- length(args)

  if (n == 0) {
    rhs <- 1
  } else if (n == 1) {
    rhs <- args[[1]]
  } else {
    rhs <- purrr::reduce(args, function(out, new) call(op, out, new))
  }

  # Don't forget to forward the caller environment
  new_formula(lhs, rhs, env = caller_env())
}

make_formula(disp)
#> disp ~ 1

make_formula(disp, cyl)
#> disp ~ cyl

make_formula(disp, cyl, am, drat)
#> disp ~ cyl + am + drat

make_formula(disp, cyl, am, drat, op = "*")
#> disp ~ cyl * am * drat

使用表达式的一大优势是它对小鲍比表 ( https://xkcd.com/327/ ) 很健壮：

# User inputs are always interpreted as symbols (variable name)
make_formula(disp, `I(file.remove('~'))`)
#> disp ~ `I(file.remove('~'))`

# With `paste()` + `parse()` user inputs are interpreted as arbitrary code
reformulate(c("foo", "I(file.remove('~'))"))
#> ~foo + I(file.remove("~"))
于 2020-08-28T08:23:24.493 回答
2

我建议使用rlang::enquo(s) 并rlang::as_name实现这一目标：

library(rlang)

formula <- function(df, col, ...) {
  group <- enquo(col)
  vars <- enquos(...)

  group_str <- rlang::as_name(group)
  vars_str <- lapply(vars, rlang::as_name)
  
  paste(group_str,"~", paste(vars_str, collapse = " + "), sep = " ")
}

formula(mtcars, col, col1, col2, col3)
#> [1] "col ~ col1 + col2 + col3"
于 2020-08-27T20:06:45.990 回答
0

我们可以使用reformulate

formula_fn <- function(dat, col, ...) {
           deparse(reformulate(purrr::map_chr(ensyms(...), rlang::as_string), 
                 response = rlang::as_string(ensym(col) )))
      
 }
formula_fn(mtcars, col, col1, col2, col3)
#[1] "col ~ col1 + col2 + col3"
于 2020-08-27T20:45:25.393 回答
0

我已经接受了上面@LionelHenry 的建议，并创建了以下功能，其中包含我最初的问题中未要求的一些附加功能。

#' Create a formula
#'
#' Creates a new formula object to be used anywhere formulas are used (i.e, `glm`).
#'
#' @param ... any number of arguments to compose the formula
#' @param lhs a boolean indicating if the formula has a left hand side of the argument
#' @param op the operand acting upon the arguments of the right side of the formula.
#' @param group an argument to use as a grouping variable to facet by
#'
#' @return a formula
#'
#' @details If `lhs` is `TRUE`, the first argument provided is used as the left hand side of the formula.
#' The `group` paramenter will add `| group` to the end of the formula. This is useful for packages that support faceting by grouping variables for the purposes of tables or graphs.
#'
#' @export
#'
#' @examples
#' make_formula(var1, var2, var3)
#' make_formula(var1, var2, var3, lhs = FALSE)
#' make_formula(var1, var2, var3, lhs = FALSE, group = var4)
#'
make_formula <- function(..., lhs = TRUE, op = "+", group = NULL) {
  args <- rlang::ensyms(...)
  n <- length(args)
  group <- rlang::enexpr(group)

  if(lhs) {
    left <- args[[1]]
    if (n == 1) {
      right <- 1
    } else if (n == 2) {
      right <- args[[2]]
    } else {
      right <- purrr::reduce(args[-1], function(out, new) call(op, out, new))
    }
  } else {
    left <- NULL
    if (n == 1) {
      right <- args[[1]]
    } else {
      right <- purrr::reduce(args, function(out, new) call(op, out, new))
    }
  }

  if(!is.null(group)) {
    group <- rlang::ensym(group)
    right <- purrr::reduce(c(right, group), function(out, new) call("|", out, new))
  }

  rlang::new_formula(left, right, env = rlang::caller_env()) # Forward to the caller environment
}
于 2020-08-30T06:08:07.983 回答