如何用purrr或across结合mutate按条件生成新数据列

方法1：使用dplyr::across()（推荐） #

这是最简洁的原生dplyr方案，无需额外依赖purrr即可批量生成目标列：

library(dplyr)

df <- tibble(year = c("2018", "2018", "2019", "2019"), 
             observed = c("YES", "NO", "NO", "YES"))

# 提取数据中所有唯一的年份
target_years <- unique(df$year)

# 批量生成列
df_processed <- df %>%
  mutate(
    across(
      all_of(target_years),
      ~if_else(observed == "YES" & year == cur_column(), 1, 0),
      .names = "observed_{.col}"
    )
  )

df_processed

输出结果：

# A tibble: 4 × 4
  year  observed observed_2018 observed_2019
  <chr> <chr>            <dbl>         <dbl>
1 2018  YES                  1             0
2 2018  NO                   0             0
3 2019  NO                   0             0
4 2019  YES                  0             1

• all_of(target_years)指定要基于哪些年份生成新列
• cur_column()获取当前迭代的年份值，用于条件判断
• .names参数定义新列的命名规则，自动生成observed_年份格式的列名

方法2：使用purrr包实现 #

如果需要用purrr工具，可以通过map_dfc批量生成列后与原数据绑定：

library(dplyr)
library(purrr)

df <- tibble(year = c("2018", "2018", "2019", "2019"), 
             observed = c("YES", "NO", "NO", "YES"))

target_years <- unique(df$year)

# 用map_dfc生成列并合并，再与原数据绑定
df_processed <- df %>%
  bind_cols(
    map_dfc(target_years, ~{
      col_name <- paste0("observed_", .x)
      tibble(!!col_name := if_else(observed == "YES" & year == .x, 1, 0))
    })
  )

df_processed

或者用reduce逐个迭代添加列：

df_processed <- reduce(
  target_years,
  function(current_df, y) {
    current_df %>%
      mutate(!!paste0("observed_", y) := if_else(observed == "YES" & year == y, 1, 0))
  },
  .init = df
)

内容的提问来源于stack exchange，提问作者nicholas