嵌套函数中的Quosure

Question

我正在努力编写一个函数fun2，它使用fun1.继续犯错误。下面我写了一个简化的例子。这是我第一次处理“整洁的评价”，不知道它的来龙去脉。

示例数据格式：

d1 = data.frame(
  ID = c("A", "A", "A", "B", "B", "C", "C", "C", "C"),
  EXPR = c(2, 8, 3, 5, 7, 20, 1, 5, 4)
)

d2 = data.frame(
  ID = c("A", "B", "C"),
  NUM = c(22, 50, 31)
)

第一函数

fun1 <- function(
  df1 = "df 1", 
  df2 = "df 2",
  t1 = "threshold 1",
  expr_col = "expr column", 
  id_col = "sample column - must be present in df1 and df2") {

  # dataframes
  df <- df1
  db <- df2

  # quosure
  enquo_id <- enquo(id_col)
  enquo_expr <- enquo(expr_col)

  # classify
  df <- df %>% 
    mutate(threshold = t1) %>% 
    mutate(class = ifelse(!!enquo_expr > t1, "positive", "negative")) %>% 
    mutate(class = factor(class, levels = c("positive", "negative")))

  # calculate sample data
  df.sum <- df %>% 
    group_by(!!enquo_id, class) %>% 
    summarise(count = n()) %>% 
    complete(class, fill = list(count = 0)) %>% 
    mutate(total = sum(count), freq = count/total)

  # merge dataframes
  df.sum <- left_join(df.sum, db, by = quo_name(enquo_id))

  # return
  return(df.sum)
}

如果我运行这方面的测试，我将得到一个数据帧作为回报，就像预期的那样。

test <- fun1(df1 = d1, df2 = d2, t1 = 3, expr_col = EXPR, id_col = ID)

第二函数现在使用fun2，我尝试在for循环中使用fun1来迭代seq向量的ti到tf：

fun2 <- function(
  df1 = "df 1", 
  df2 = "df 2",
  expr_col = "expr column", 
  id_col = "sample column - must be present in df1 and df2",
  ti = "initial value",
  tf = "final value",
  res = "resolution") {

  # define variables for fun1
  var1 <- enquo(d1)
  var2 <- enquo(d2)
  var3 <- enquo(t1)
  var4 <- enquo(EXPR)
  var5 <- enquo(ID)

  # get sequence of values
  seq <- seq(from = ti, to = tf, by = res)

  # open list
  t.list <- list()

  # Loop ----
  for (i in seq_along(seq)){
    t1 <- seq[i]
    t.list[[i]] <- fun1(df1 = var1,
                        df2 = var2,
                        t1 = var3,
                        expr_col = var4,
                        id_col = var5)
  }
  df.out <- plyr::ldply(t.list, rbind)

  ### Return ---
  return(df.out)
}

但如果我做这个

test <- fun2(df1 = d1, df2 = d2, expr_col = EXPR, id_col = ID, ti = 1, tf = 10, res = 1)

我收到一条错误消息

Error in (function (x)  : object 'EXPR' not found

我尝试了很多东西..。我被困在这里了。我想我并没有正确地使用enquo()。我可以通过不使用varX并在fun1参数中直接放置每个元素的实际适当名称来使其工作，但对我来说，这样做的全部意义是使其“泛化”，因此只在fun2中指定参数，然后将其传递给fun1。

任何帮助都将不胜感激。

Answer 1

非常感谢你的回答，史密斯。现在使用以下代码对我进行排序：

fun2 <- function(
  df1 = "df 1", 
  df2 = "df 2",
  expr_col = "expr column", 
  id_col = "sample column - must be present in df1 and df2",
  ti = "initial value",
  tf = "final value",
  res = "resolution") {

  # define variables for fun1
  var4 <- enquo(expr_col)
  var5 <- enquo(id_col)

  # get sequence of values
  seq <- seq(from = ti, to = tf, by = res)

  # open list
  t.list <- list()

  ### Loop --------------------------------------------------------------
  for (i in seq_along(seq)){
    t1 <- seq[i]
    t.list[[i]] <- fun1(df1 = df1,
                        df2 = df2,
                        t1 = t1,
                        expr_col = !!var4,
                        id_col = !!var5)
  }
  df.out <- plyr::ldply(t.list, rbind)

  ### Return ---
  return(df.out)
}

# TEST FUN2
test <- fun2(df1 = d1, df2 = d2, expr_col = EXPR, id_col = ID, ti = 1, tf = 10, res = 1)