r - 计算两个或多个时间点的分数变化

Question

我有一个非常简单的问题:想象一个有 6 个项目的认知测试，每个项目都在 0-10 的范围内单独评分现在假设有 5 名患者在一天中的三个(或更多)不同时间参加了这项测试。

df <- tibble(
  id = rep(1:5, each = 3),
  time = rep(c("morning", "evening", "night"), each = 1, times = 5),
  i1 = round(runif(15, 0, 10)),
  i2 = round(runif(15, 0, 10)),
  i3 = round(runif(15, 0, 10)),
  i4 = round(runif(15, 0, 10)),
  i5 = round(runif(15, 0, 10)),
  i6 = round(runif(15, 0, 10))
)

现在，我如何计算一天中不同时间每个患者的每个项目的分数变化？tidyverse 方法受到高度赞赏。

P.S: 输出最好是这样的:

df <- tibble(
  id = rep(1:5, each = 3),
  time = rep(c("morning", "evening", "night"), each = 1, times = 5),
  i1 = round(runif(15, 0, 10)),
  i2 = round(runif(15, 0, 10)),
  i3 = round(runif(15, 0, 10)),
  i4 = round(runif(15, 0, 10)),
  i5 = round(runif(15, 0, 10)),
  i6 = round(runif(15, 0, 10)),
  diff_morning_evening = rep(NA, 15),
  diff_morning_night = rep(NA, 15),
  diff_evening_night = rep(NA, 15),
)

Answer 1

一种以紧凑方式计算变化的方法:

• 第一行:早上 - 晚上
• 第二行:上午 - 晚上
• 第 3 行:晚上 - 晚上

df %>% 
  group_by(id) %>% 
  mutate(across(i1:i6, ~ ifelse(is.na(.x - lag(.x)), 
                           lead(.x, 2) - .x, .x - lag(.x)), 
           .names="{.col}_chg")) %>% 
  ungroup()
# A tibble: 15 × 14
      id time       i1    i2    i3    i4    i5    i6 i1_chg i2_chg i3_chg i4_chg
   <int> <chr>   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
 1     1 morning     1     2     4     9     5     6      7     -1      1     -8
 2     1 evening     5     0     4     8     9     4      4     -2      0     -1
 3     1 night       8     1     5     1     4     5      3      1      1     -7
 4     2 morning     7     7     4     3     2     9     -5     -1      4      5
 5     2 evening     8     9     1     2     4     5      1      2     -3     -1
 6     2 night       2     6     8     8    10     2     -6     -3      7      6
 7     3 morning     9     6     6     1     5     5     -8     -1      2      0
 8     3 evening     3     2     8     1     3    10     -6     -4      2      0
 9     3 night       1     5     8     1     3     3     -2      3      0      0
10     4 morning     7     2     9     1     5     8      1      1     -6      0
11     4 evening     3     5     9     5     6     3     -4      3      0      4
12     4 night       8     3     3     1     3     2      5     -2     -6     -4
13     5 morning     4     1     3     7     1     0      4      6      4      2
14     5 evening     7     2     7     7     5     2      3      1      4      0
15     5 night       8     7     7     9     8     4      1      5      0      2
# … with 2 more variables: i5_chg <dbl>, i6_chg <dbl>

求和特殊行方法

df %>% 
  group_by(id) %>% 
  summarize(across(i1:i6, ~ 
    .x[which(time == "morning")] + .x[which(time == "night")]))
# A tibble: 5 × 7
     id    i1    i2    i3    i4    i5    i6
  <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     9     3     9    10     9    11
2     2     9    13    12    11    12    11
3     3    10    11    14     2     8     8
4     4    15     5    12     2     8    10
5     5    12     8    10    16     9     4