R语言实现按ID、Location、Customer分组拼接Issue列并去重

刚好碰到过类似的需求，给你分享两种实用的R实现方法，都能完美达成你的目标——按ID、Location、Customer分组，对Issue列去重后拼接成逗号分隔的字符串：

方法一：使用dplyr包（tidyverse生态，推荐） #

dplyr是数据处理的常用工具，代码可读性高，操作简洁：

# 先安装并加载dplyr包（首次使用需安装）
# install.packages("dplyr")
library(dplyr)

# 构造你的示例数据框
df <- data.frame(
  ID = c(1, 2, 3, 1, 2, 2),
  Location = c("x", "y", "z", "x", "y", "y"),
  Customer = c("a", "b", "c", "a", "b", "b"),
  Issue = c("Issue1", "Issue2", "Issue3", "Issue4", "Issue5", "Issue6")
)

# 核心操作：分组 → 去重拼接Issue
result <- df %>%
  group_by(ID, Location, Customer) %>%  # 指定分组列
  summarize(
    Issues = paste(unique(Issue), collapse = ", "),  # 去重后拼接
    .groups = "drop"  # 取消分组状态，避免后续警告
  )

# 查看结果
print(result)

关键说明： #

• unique(Issue)：自动过滤分组内重复的Issue值，确保每个Issue只保留一次
• paste(..., collapse = ", ")：把去重后的字符串用「逗号+空格」连接成一个字符串
• .groups = "drop"：分组完成后取消分组标记，让结果回归普通数据框

方法二：使用base R（无需额外安装包） #

如果不想加载第三方包，用base R的aggregate函数也能实现：

# 构造示例数据框
df <- data.frame(
  ID = c(1, 2, 3, 1, 2, 2),
  Location = c("x", "y", "z", "x", "y", "y"),
  Customer = c("a", "b", "c", "a", "b", "b"),
  Issue = c("Issue1", "Issue2", "Issue3", "Issue4", "Issue5", "Issue6")
)

# 分组聚合操作
result_base <- aggregate(
  Issue ~ ID + Location + Customer, 
  data = df,
  FUN = function(x) paste(unique(x), collapse = ", ")
)

# 将结果列名改为需求的"Issues"
colnames(result_base)[4] <- "Issues"

# 查看结果
print(result_base)

两种方法运行后都会得到你期望的结果：

ID Location Customer               Issues
1  1        x        a        Issue1, Issue4
2  2        y        b Issue2, Issue5, Issue6
3  3        z        c                Issue3

内容的提问来源于stack exchange，提问作者Partha sarathi