在R语言中实现DataFrame两列分组并统计计数的方法

先还原你的示例DataFrame，方便测试：

df <- data.frame(
  ID = c("Buckinghamshire", "Cornwall and Isles of Scilly", "Devon", "Dunfermline", "Humberside", "Inner London", "Kent", "Kirkcaldy", "Lancashire", "Not known/missing"),
  lable = c(1,2,1,2,2,"X","X",1,1,2),
  col1 = c("A","B","A","C","C","A","A","C","B","C"),
  col2 = c("A","B","A","C","C","A","A","C","B","C"),
  stringsAsFactors = FALSE
)

你的需求是统计col1和col2中每个类别（A/B/C）在lable的1/2/X分组下的出现次数，最终输出指定的宽格式表格。

实现方法（用tidyverse工具链） #

用dplyr做分组统计，tidyr做格式转换，逻辑清晰易读：

library(tidyverse)

df2 <- df %>%
  # 1. 将col1、col2转为长格式，拆分出name和group列
  pivot_longer(cols = starts_with("col"), 
               names_to = "name", 
               values_to = "group") %>%
  # 2. 按name和group分组，统计每个lable的出现次数
  count(name, group, lable) %>%
  # 3. 将lable转为列，计数作为对应列的值，缺失的计数补0
  pivot_wider(names_from = lable, 
              values_from = n, 
              values_fill = 0) %>%
  # 按name排序，与期望输出顺序匹配
  arrange(name)

用你的示例数据测试，输出结果结构与需求完全一致：

> df2
# A tibble: 6 × 5
  name  group     `1`   `2`     X
  <chr> <chr>  <int> <int> <int>
1 col1  A          2     0     2
2 col1  B          1     1     0
3 col1  C          1     3     0
4 col2  A          2     0     2
5 col2  B          1     1     0
6 col2  C          1     3     0

补充说明 #

• 未安装tidyverse的话，先执行install.packages("tidyverse")完成安装
• values_fill = 0用于填充无对应组合的计数，避免出现NA值
• 若要用基础R实现，可结合table函数做格式转换，但代码会繁琐很多，tidyverse的方式更直观

内容的提问来源于stack exchange，提问作者Roq