R语言使用dplyr按指定AOI分类汇总时长并计算占比

核心逻辑是在分组汇总前，先对AOI字段做重编码，将所有A/B/C的取值统一替换为X，再按新的AOI分组计算即可，完整可运行代码如下：

library(tidyverse)

# 构造示例数据集
d <- structure(list(Line = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 
2L), Speaker = c("A", "A", "A", "A", "A", "B", "B", "B", "B", 
"B", "B"), Utterance = c("hi there", "hi there", "hi there", 
"hi there", "hi there", "how're ya", "how're ya", "how're ya", 
"how're ya", "how're ya", "how're ya"), Gaze_by = c("A", "A", 
"A", "A", "A", "B", "B", "B", "B", "B", "B"), AOI = c("C", "*", 
"B", "*", "C", "*", "A", "*", "A", "*", "C"), dur = c(100L, 25L, 
30L, 50L, 144L, 50L, 22L, 33L, 100L, 150L, 999L)), row.names = c(NA, 
-11L), class = c("tbl_df", "tbl", "data.frame"))

# 数据处理代码
d %>%
  # 重编码AOI列：取值为*的保留原值，其余A/B/C统一替换为X
  mutate(AOI = if_else(AOI == "*", "*", "X")) %>%
  # 按指定字段分组
  group_by(Speaker, Utterance, Gaze_by, AOI) %>%
  summarise(dur = sum(dur), .groups = "drop_last") %>%
  # 计算各AOI分类的时长占比，保留1位小数
  mutate(prop = round(dur/sum(dur)*100, 1))

运行以上代码即可得到你需要的输出结果：

# A tibble: 4 × 6
# Groups:   Speaker, Utterance, Gaze_by [2]
  Speaker Utterance Gaze_by AOI     dur  prop
  <chr>   <chr>     <chr>   <chr> <int> <dbl>
1 A       hi there  A       *        75  21.5
2 A       hi there  A       X       274  78.5
3 B       how're ya B       *       233  17.2
4 B       how're ya B       X      1121  82.8

内容的提问来源于stack exchange，提问作者Chris Ruehlemann