基于多条件计算diary14数据集活动时长的均值与标准差

你需要先按activity和day分组统计单日活动时长，再基于这些单日数据计算均值和标准差，具体代码如下：

1. 先统计每个活动在不同日期的时长 #

library(dplyr)

# 按活动类型+日期分组，计算每个组合的总时长（分钟）
activity_day_data <- diary14 %>%
  group_by(activity, day) %>%
  summarise(
    duration_min = n() * 10,  # 每个记录对应10分钟，数量×10得总分钟数
    .groups = "drop"          # 取消分组，得到扁平化的结果
  )

2. 计算每个活动的时长均值与标准差 #

基于上面的单日时长数据，再按activity分组汇总统计：

activity_stats <- activity_day_data %>%
  group_by(activity) %>%
  summarise(
    avg_duration = mean(duration_min),
    sd_duration = sd(duration_min),
    .groups = "drop"
  )

也可以一步完成，无需中间变量：

diary14 %>%
  group_by(activity, day) %>%
  summarise(duration_min = n() * 10, .groups = "drop_last") %>%
  summarise(
    avg_duration = mean(duration_min),
    sd_duration = sd(duration_min),
    .groups = "drop"
  )

额外优化：给日期添加可读性标签 #

如果需要让结果更直观，可以先把day的数字转换成对应的日期类型：

diary14 <- diary14 %>%
  mutate(day_label = case_when(
    day == 1 ~ "weekdays",
    day == 2 ~ "saturday",
    day == 3 ~ "sunday"
  ))

# 用标签分组统计
diary14 %>%
  group_by(activity, day_label) %>%
  summarise(duration_min = n() * 10, .groups = "drop_last") %>%
  summarise(
    avg_duration = mean(duration_min),
    sd_duration = sd(duration_min),
    .groups = "drop"
  )

内容的提问来源于stack exchange，提问作者Victoria