如何将3列年龄数据及2列活动数据合并为单列且无重复遗漏？

1. 合并年龄列（under30、age30to64、age65plus） #

先看这三列的取值逻辑：

• under30 标记是否为30岁以下
• age30to64 标记是否为30-64岁
• age65plus 标记是否为65岁以上

用case_when就能精准匹配出每个样本的年龄分组，不会遗漏或重复：

library(dplyr)

df <- df %>%
  mutate(
    combined_age = case_when(
      under30 == "under 30" ~ "under 30",
      age30to64 == "age 30 to 64" ~ "age 30 to 64",
      age65plus == "65 plus" ~ "65 plus",
      TRUE ~ NA_character_ # 兜底处理异常情况，确保没有漏网数据
    ) %>% factor(levels = c("under 30", "age 30 to 64", "65 plus"))
  )

验证下：原数据里第4行的ageCat是over64，对应age65plus是"65 plus"，合并后combined_age会是"65 plus"，其他行都是"age 30 to 64"，和原分类完全一致。

2. 合并活动列（active、active1） #

active列管的是“是否活跃”，active1列是“活跃程度”——看数据就能发现，active为"Not active"时，active1都是"Low"；active为"Active"时，active1会给出具体强度（Moderate/Vigorous）。

你可以选择两种合并方式：

方式一：保留完整状态描述 #

把“是否活跃”和“强度”结合，信息更直观：

df <- df %>%
  mutate(
    combined_active = case_when(
      active == "Not active" ~ "Not active",
      active == "Active" & active1 == "Low" ~ "Active (Low)",
      active == "Active" & active1 == "Moderate" ~ "Active (Moderate)",
      active == "Active" & active1 == "Vigorous" ~ "Active (Vigorous)",
      TRUE ~ NA_character_
    ) %>% factor()
  )

方式二：简化分类 #

因为Not active已经对应Low，直接简化成：不活跃就标"Not active"，活跃就用active1的强度值：

df <- df %>%
  mutate(
    combined_active = case_when(
      active == "Not active" ~ "Not active",
      TRUE ~ as.character(active1)
    ) %>% factor(levels = c("Not active", "Low", "Moderate", "Vigorous"))
  )

完整代码 #

把两步整合到一起，还可以删掉没用的原列：

library(dplyr)

# 加载你的数据
df <- structure(list(fruits = c(0, 0, 0, 0, 1), veggies = c(0, 1, 1, 1, 1), age = structure(c(7L, 8L, 9L, 10L, 6L), levels = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13"), class = "factor"), under30 = structure(c(1L, 1L, 1L, 1L, 1L), levels = c("30 plus", "under 30"), class = "factor"), age30to64 = structure(c(2L, 2L, 2L, 1L, 2L), levels = c("under 30 or 65 plus", "age 30 to 64"), class = "factor"), age65plus = structure(c(1L, 1L, 1L, 2L, 1L), levels = c("under 65", "65 plus"), class = "factor"), arthritis = structure(c(1L, 2L, 1L, 1L, 1L), levels = c("No arthritis", "Arthritis"), class = "factor"), gender = structure(c(2L, 2L, 2L, 1L, 2L), levels = c("male", "female"), class = "factor"), genhealth = structure(c(3L, 3L, 2L, 3L, 2L), levels = c("Excellent", "Very good", "Good", "Fair", "Poor"), class = "factor"), education = structure(c(5L, 6L, 4L, 6L, 6L), levels = c("1", "2", "3", "4", "5", "6"), class = "factor"), income = structure(c(8L, 8L, 7L, 6L, 8L), levels = c("1", "2", "3", "4", "5", "6", "7", "8"), class = "factor"), active = structure(c(2L, 1L, 2L, 1L, 2L), levels = c("Not active", "Active"), class = "factor"), active1 = structure(c(2L, 1L, 2L, 1L, 3L), levels = c("Low", "Moderate", "Vigorous"), class = "factor"), bmi = c(18.2199993133545, 27.4599990844727, 21.9699993133545, 35.939998626709, 39.8600006103516), bmicat = structure(c(1L, 3L, 2L, 4L, 4L), levels = c("Underweight", "Normal", "Overweight", "Obese"), class = "factor"), activetimes = c(20, 0, 5, 0, 8), ageCat = structure(c(2L, 2L, 2L, 3L, 2L), levels = c("under30", "age30to64", "over64"), class = "factor")), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"))

# 合并+清理数据
df_clean <- df %>%
  # 合并年龄列
  mutate(
    combined_age = case_when(
      under30 == "under 30" ~ "under 30",
      age30to64 == "age 30 to 64" ~ "age 30 to 64",
      age65plus == "65 plus" ~ "65 plus",
      TRUE ~ NA_character_
    ) %>% factor(levels = c("under 30", "age 30 to 64", "65 plus"))
  ) %>%
  # 合并活动列（用简化方式）
  mutate(
    combined_active = case_when(
      active == "Not active" ~ "Not active",
      TRUE ~ as.character(active1)
    ) %>% factor(levels = c("Not active", "Low", "Moderate", "Vigorous"))
  ) %>%
  # 删掉冗余的原列
  select(-under30, -age30to64, -age65plus, -active, -active1)

# 查看处理后的结果
print(df_clean)

结果说明 #

• combined_age完美对应三个年龄分组，和原数据的ageCat逻辑一致，没有重复或遗漏。
• combined_active整合了两列的关键信息：既保留了“不活跃”的状态，又区分了活跃时的不同强度，所有有效数据都没丢。

内容的提问来源于stack exchange，提问作者Costa Wesele