R语言：按指定公式计算数据框中分类因子的频率占比

R语言实现方案 #

你给出的公式存在括号缺失的笔误，正确计算逻辑应为每个ID分组下，(Y和O的Counts之和 ÷ 该ID所有Counts总和) × 100，以下是两种可直接运行的实现方法：

方法1：使用dplyr包（语法简洁易读）

# 未安装包先运行 install.packages("dplyr")
library(dplyr)

df <- data.frame( ID = c("A","A","A","B","B","B","C","C","C"), 
                  levels = c( "Y", "R", "O","Y", "R", "O","Y", "R", "O" ),
                  Counts=c(5,1,5,10,2,1,3,5,8))

result <- df %>%
  group_by(ID) %>%
  summarise(
    total = sum(Counts),
    sum_YO = sum(Counts[levels %in% c("Y", "O")]),
    freq = sum_YO / total * 100
  ) %>%
  select(ID, freq)

运行后得到的实际计算结果如下：

ID  freq
  <chr> <dbl>
1 A      90.9
2 B      84.6
3 C      68.8

你给出的预期输出示例数值为占位值，可根据实际业务需求调整公式细节。

方法2：基础R实现（无需安装额外依赖）

df <- data.frame( ID = c("A","A","A","B","B","B","C","C","C"), 
                  levels = c( "Y", "R", "O","Y", "R", "O","Y", "R", "O" ),
                  Counts=c(5,1,5,10,2,1,3,5,8))

result <- do.call(rbind, lapply(split(df, df$ID), function(group_df) {
  total_count <- sum(group_df$Counts)
  yo_count <- sum(group_df$Counts[group_df$levels %in% c("Y", "O")])
  data.frame(ID = unique(group_df$ID), freq = yo_count / total_count * 100)
}))
rownames(result) <- NULL

内容的提问来源于stack exchange，提问作者Marwah Al-kaabi