在一个函数中改变多个具有不同名称的变量...？

Question

我正在尝试使用按日期索引的变量的最近可用版本来替换 x、q、z 的更新值。在 STATA 中，我们可以轻松地将其作为一个 for 循环来完成（参见下面的示例代码）。 STATA

• y指的是日期值20191125

local y 20191125

foreach v in attend child sibling{
replace `v'=`v'`y' if !missing(`v'`y')
}

数据

+----+----+---+-----------+-----------+-----------+
|attend  | child | sibling | attend20191125 | child20191125 | sibling20191125 |
+----+----+---+-----------+-----------+-----------+
| 1  | 2  | 3 |         6 |         8 |         0 |
| 1  | NA | 0 |         1 |         1 |         1 |
| NA | 0  | 1 |         5 |         4 |         2 |
+----+----+---+-----------+-----------+-----------+

潜在输出：

+----+----+---+-----------+-----------+-----------+
|attend  | child | sibling | attend20191125 | child20191125 | sibling20191125 |
+----+----+---+-----------+-----------+-----------+
| 1  | 2  | 3 |         6 |         8 |         0 |
| 1  |1   | 0 |         1 |         1 |         1 |
| 5  | 0  | 1 |         5 |         4 |         2 |
+----+----+---+-----------+-----------+-----------+

我知道如何从另一列替换一列的 NA 值。如何动态使用 purrr 做同样的事情？如何告诉 R 替换具有相同名称 + 日期前缀的变量中的值？我的数据集中有大约 25 个变量需要这个。

#Method 1: I can do it one variable at a time
df%<>%
mutate(attend=ifelse(is.na(attend)==T, attend20191125, attend),
       child=ifelse(is.na(child)==T, child20191125, child),
       sibling=ifelse(is.na(sibling)==T, sibling20191125, sibling))

#Method 2: using mutate_at (but not sure how I can dynamically refer to the date indexed variable?).

df%<>%
mutate_at(c("attend", "child", "sibling"), .=ifelse(is.na(.)==T, var20191125, .))

Answer 1

这是一种方式：

get_replacement <- function(.) {
  name <- names(.)
  vals <- .[[1]]
  out <- ifelse(is.na(vals), df[[str_c(name, "20191125")]], vals)
  rlang::list2(!!name := out)  # dynamically set column name
}

df %>%
  lmap_at(vars(attend, child, sibling), get_replacement)