使用 group_by() 计算受试者首次测量的时间间隔

Question

我在 'long format' 中有一个数据框，对象被多次观察：

dat1 <- tribble(
  ~CODE, ~V1, ~V2, ~session, ~date,
  "1111P11", 2, 3, 1, "2020-09-01",
  "1111P11", 3, 2, 2, "2020-09-08",
  "1111P11", 1, 3, 3, "2020-09-15",
  "1111P11", 3, 4, 4, "2020-09-25",
  "2222P22", 5, 1, 1, "2020-05-15",
  "2222P22", 3, 2, 2, "2020-05-22",
  "2222P22", 1, 4, 3, "2020-05-30",
  "3333P33", 3, 4, 1, "2020-06-10",
  "3333P33", 4, 1, 2, "2020-06-17",
  "3333P33", 3, 5, 3, "2020-06-24", 
  "3333P33", 4, 2, 4, "2020-07-01",
  "3333P33", 3, 4, 5, "2020-07-10"
)
dat1$date <- date(dat$date)

我想为每个主题计算每个会话与第一个会话之间的时间间隔，结果应该是：

dat2 <- tribble(
  ~CODE, ~V1, ~V2, ~session, ~date, ~interv.1st.sess,
  "1111P11", 2, 3, 1, "2020-09-01", 0,
  "1111P11", 3, 2, 2, "2020-09-08", 7,
  "1111P11", 1, 3, 3, "2020-09-15", 14, 
  "1111P11", 3, 4, 4, "2020-09-25", 24,
  "2222P22", 5, 1, 1, "2020-05-15", 0,
  "2222P22", 3, 2, 2, "2020-05-22", 7,
  "2222P22", 1, 4, 3, "2020-05-30", 15,
  "3333P33", 3, 4, 1, "2020-06-10", 0,
  "3333P33", 4, 1, 2, "2020-06-17", 7,
  "3333P33", 3, 5, 3, "2020-06-24", 14,
  "3333P33", 4, 2, 4, "2020-07-01", 21,
  "3333P33", 3, 4, 5, "2020-07-10", 30
)

我一直在尝试用 group_by() 以某种方式解决这个问题，但没有成功。有没有一种整洁的方式（或任何其他方式）来做到这一点？

Answer 1

试试 dplyr 和 lubridate。

包含date1明确转换日期格式

library(dplyr)
library(lubridate)

dat1 %>% 
  group_by(CODE) %>% 
  mutate(date1 = ymd(date),
         diff = date1 - first(date1)) 
#> # A tibble: 12 x 7
#> # Groups:   CODE [3]
#>    CODE       V1    V2 session date       date1      diff   
#>    <chr>   <dbl> <dbl>   <dbl> <chr>      <date>     <drtn> 
#>  1 1111P11     2     3       1 2020-09-01 2020-09-01  0 days
#>  2 1111P11     3     2       2 2020-09-08 2020-09-08  7 days
#>  3 1111P11     1     3       3 2020-09-15 2020-09-15 14 days
#>  4 1111P11     3     4       4 2020-09-25 2020-09-25 24 days
#>  5 2222P22     5     1       1 2020-05-15 2020-05-15  0 days
#>  6 2222P22     3     2       2 2020-05-22 2020-05-22  7 days
#>  7 2222P22     1     4       3 2020-05-30 2020-05-30 15 days
#>  8 3333P33     3     4       1 2020-06-10 2020-06-10  0 days
#>  9 3333P33     4     1       2 2020-06-17 2020-06-17  7 days
#> 10 3333P33     3     5       3 2020-06-24 2020-06-24 14 days
#> 11 3333P33     4     2       4 2020-07-01 2020-07-01 21 days
#> 12 3333P33     3     4       5 2020-07-10 2020-07-10 30 days

^{由 reprex package (v2.0.1)}

于 2021-12-19 创建

Answer 2

这是使用 ave -

的基础 R 选项

transform(dat1, diff_in_days = as.integer(date - ave(date, CODE, 
                               FUN = function(x) x[1])))

 #     CODE V1 V2 session       date diff_in_days
#1  1111P11  2  3       1 2020-09-01            0
#2  1111P11  3  2       2 2020-09-08            7
#3  1111P11  1  3       3 2020-09-15           14
#4  1111P11  3  4       4 2020-09-25           24
#5  2222P22  5  1       1 2020-05-15            0
#6  2222P22  3  2       2 2020-05-22            7
#7  2222P22  1  4       3 2020-05-30           15
#8  3333P33  3  4       1 2020-06-10            0
#9  3333P33  4  1       2 2020-06-17            7
#10 3333P33  3  5       3 2020-06-24           14
#11 3333P33  4  2       4 2020-07-01           21
#12 3333P33  3  4       5 2020-07-10           30

Answer 3

使用data.table

library(data.table)
setDT(dat1)[, diff := date - first(date), CODE]

-输出

> dat1
       CODE V1 V2 session       date    diff
 1: 1111P11  2  3       1 2020-09-01  0 days
 2: 1111P11  3  2       2 2020-09-08  7 days
 3: 1111P11  1  3       3 2020-09-15 14 days
 4: 1111P11  3  4       4 2020-09-25 24 days
 5: 2222P22  5  1       1 2020-05-15  0 days
 6: 2222P22  3  2       2 2020-05-22  7 days
 7: 2222P22  1  4       3 2020-05-30 15 days
 8: 3333P33  3  4       1 2020-06-10  0 days
 9: 3333P33  4  1       2 2020-06-17  7 days
10: 3333P33  3  5       3 2020-06-24 14 days
11: 3333P33  4  2       4 2020-07-01 21 days
12: 3333P33  3  4       5 2020-07-10 30 days

使用 group_by() 计算受试者首次测量的时间间隔

Calculate time interval to first measurement within subjects using group_by()

r

dplyr

tidyverse

lubridate