分组内重新定义因子水平与顺序的R语言方案问询

需求说明 #

需要按预设顺序呈现数据汇总结果，核心要求：

• 针对col1的不同分组，单独指定col2的排序规则
• 保留各col1分组内未出现在原始数据中的col2因子水平（类似group_by(..., .drop=FALSE)的补全效果）
• col2的值可跨分组重复，且无全局统一排序逻辑，属于双水平因子

示例输入数据 #

df <- read.table(
  header = TRUE,
  sep=",",
  text = "
col1,col2
Tunnels,Dick
Tunnels,Tom
Tunnels,Tom
Beatles,George
Beatles,Paul
Beatles,Ringo
Beatles,Ringo
UK Artists,Gilbert
"
)

期望输出 #

col1       col2        n
 Beatles    John        0
 Beatles    Paul        1
 Beatles    George      1
 Beatles    Ringo       2
 UK Artists Gilbert     1
 UK Artists George      0
 Tunnels    Tom         2
 Tunnels    Dick        1
 Tunnels    Harry       0

无效实现代码 #

以下代码无法满足需求：全局统一的col2因子水平无法适配不同分组的自定义排序，还会保留跨分组的多余水平，导致结果不符合预期。

col2_tunnels <- c("Tom", "Dick", "Harry")
col2_beatles <- c("John", "Paul", "George", "Ringo")
col2_artists <- c("Gilbert", "George")
col2_order <- unique(c(col2_tunnels, col2_beatles, col2_artists)) # 无法保留重复项
col1_order <- c("Beatles", "UK Artists", "Tunnels")

df %>%
  mutate(
    col1 = factor(col1, levels = col1_order),
    col2 = factor(col2, levels = col2_order)
  ) %>%
  group_by(col1, col2, .drop = FALSE) %>%
  summarise(n = n(), )

现有可行方案 #

通过按col1分组拆分数据，结合命名列表定义每个分组的col2因子顺序，可实现需求：

col2_fctlist <- list(
  Tunnels = c("Tom", "Dick", "Harry"),
  Beatles = c("John", "Paul", "George", "Ringo"),
  'UK Artists' = c("Gilbert", "George")
)

x <- lapply(col1_order, function(col1grp)
  df %>% filter(col1==col1grp) %>% 
    mutate(col2 = factor(col2, levels = col2_fctlist[[col1grp]])) %>% 
    group_by(col1, col2, .drop = FALSE) %>%
    summarise(n = n(), )
)

do.call(rbind, x)

更优实现方法 #

方法1：使用dplyr::group_modify（推荐） #

利用group_modify在分组内直接处理因子水平，无需手动拆分合并数据，保持管道操作的连贯性：

col2_fctlist <- list(
  Beatles = c("John", "Paul", "George", "Ringo"),
  'UK Artists' = c("Gilbert", "George"),
  Tunnels = c("Tom", "Dick", "Harry")
)
col1_order <- names(col2_fctlist) # 直接复用列表名作为col1顺序

df %>%
  mutate(col1 = factor(col1, levels = col1_order)) %>%
  group_by(col1, .drop = FALSE) %>%
  group_modify(function(.x, .y) {
    # 获取当前分组对应的col2因子水平
    col2_levels <- col2_fctlist[[.y$col1]]
    .x %>%
      mutate(col2 = factor(col2, levels = col2_levels)) %>%
      count(col2, .drop = FALSE)
  }) %>%
  ungroup()

优势：代码连贯易读，减少冗余步骤，逻辑清晰，无需手动处理数据拆分与合并。

方法2：结合tidyr::complete补全水平 #

适合习惯tidyr工具链的场景，直接补全分组内缺失的col2水平后统计数量：

library(tidyr)

col2_fctlist <- list(
  Beatles = c("John", "Paul", "George", "Ringo"),
  'UK Artists' = c("Gilbert", "George"),
  Tunnels = c("Tom", "Dick", "Harry")
)
col1_order <- names(col2_fctlist)

df %>%
  mutate(col1 = factor(col1, levels = col1_order)) %>%
  group_by(col1) %>%
  # 补全当前分组的所有col2水平
  complete(col2 = col2_fctlist[[cur_group()$col1]]) %>%
  # 统计有效数量，缺失值对应0
  summarise(n = sum(!is.na(col2)), .groups = "drop_last") %>%
  # 按分组内的col2顺序排序
  arrange(factor(col2, levels = col2_fctlist[[cur_group()$col1]])) %>%
  ungroup()

内容的提问来源于stack exchange，提问作者moreQthanA