R：在重复测量设计中计算单位的整洁方法[重复]答案

【问题标题】：R: Tidy way to count units in a repeated measures design [duplicate]R：在重复测量设计中计算单位的整洁方法[重复]
【发布时间】：2021-05-27 07:52:43
【问题描述】：

这是来自虚构实验 (N=40) 的小型数据集。该实验有 2 个条件，每个条件都有可变数量的受试者（总共 8 个）。每个主题都被多次观察，因此与可变数量的行相关联。

dat <- structure(list(subject = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 
6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L), 
    condition = c("a", "a", "a", "a", "a", "a", "a", "a", "a", 
    "a", "b", "b", "a", "a", "a", "a", "a", "a", "a", "b", "b", 
    "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "a", 
    "a", "a", "a", "a", "a", "a", "a"), DV = c(2.81157687969627, 
    -0.842120813381446, 0.581945736602951, 0.338837761518314, 
    1.89265238800308, 1.61748828762215, 1.50241281473164, -0.371722939264336, 
    2.34943581573083, 1.9748530958824, 0.362129637270942, 1.8277964140968, 
    1.70637518431997, 1.12865681599091, 3.05142782728916, 0.622010892882544, 
    2.00560122425538, -0.447121746565671, 1.15358864340752, 2.12585003262731, 
    1.52184076917827, -0.50606450477134, 0.345547956000384, 1.04829010205181, 
    3.0328567780456, 0.443519707656065, -0.57901488419535, 1.26806312350003, 
    2.47565945691539, 1.27802539397507, 1.47560146605553, -0.563842875341247, 
    -1.61470314081307, 0.293947258804903, 2.39827092020247, 2.05934478059775, 
    0.171958205176952, 1.62183818483135, 1.03045239398212, -0.0228550910766967
    )), row.names = c(NA, -40L), class = c("tbl_df", "tbl", "data.frame"
))

我想高效而整洁地添加一个列，计算每个条件下的主题数。目前，我正在使用这段复杂的代码：

dat %>%
  group_by(subject, condition) %>%
  nest() %>%
  group_by(condition) %>%
  nest() %>%
  mutate(n = map_dbl(data, nrow)) %>%
  unnest(data) %>% 
  unnest(data)

有没有更好的方法来做到这一点？

【问题讨论】：

标签： r tidyverse

【解决方案1】：

也许，n_distinct 会有所帮助

library(dplyr)
dat %>% 
    group_by(condition) %>% 
    mutate(n = n_distinct(subject))

注意：通过执行多个group_by，列顺序会有所不同。如果我们使列顺序相同，并且arrange(subject, condition)、all.equal 将为两者返回 TRUE

【讨论】：