您可以使用 dplyr 来获取:
创建数据
df <- data.frame(
ID=c("A","A","A","A","A","A","B","B","B","B","C","C","C","C","C","C","C"),
date=as.Date(c("28/08/2016","29/08/2016","30/08/2016","2/09/2016","3/09/2016","4/09/2016","8/08/2016","9/08/2016","10/08/2016","11/08/2016","30/11/2016","2/12/2016","3/12/2016","5/12/2016","6/12/2016","8/12/2016","9/12/2016"),format= "%d/%m/%Y"),
drug=c(2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,1,1),
score=c(3,4,4,4,4,4,3,4,3,3,4,5,1,4,4,2,2)
)
df
#> ID date drug score
#> 1 A 2016-08-28 2 3
#> 2 A 2016-08-29 1 4
#> 3 A 2016-08-30 2 4
#> 4 A 2016-09-02 2 4
#> 5 A 2016-09-03 1 4
#> 6 A 2016-09-04 2 4
#> 7 B 2016-08-08 1 3
#> 8 B 2016-08-09 2 4
#> 9 B 2016-08-10 2 3
#> 10 B 2016-08-11 1 3
#> 11 C 2016-11-30 2 4
#> 12 C 2016-12-02 1 5
#> 13 C 2016-12-03 2 1
#> 14 C 2016-12-05 1 4
#> 15 C 2016-12-06 2 4
#> 16 C 2016-12-08 1 2
#> 17 C 2016-12-09 1 2
填写缺失的行(天)
解决此类问题的一个好方法是制作行隐含地缺失的观察结果明确地缺少,就是使用tidyr::complete http://tidyr.tidyverse.org/reference/complete.html
library(dplyr)
library(tidyr)
df1 <- df %>%
group_by(ID) %>%
complete(date = seq(min(date), max(date), by = "day"))
df1
#> Source: local data frame [22 x 4]
#> Groups: ID [3]
#>
#> # A tibble: 22 x 4
#> ID date drug score
#> <fctr> <date> <dbl> <dbl>
#> 1 A 2016-08-28 2 3
#> 2 A 2016-08-29 1 4
#> 3 A 2016-08-30 2 4
#> 4 A 2016-08-31 NA NA
#> 5 A 2016-09-01 NA NA
#> 6 A 2016-09-02 2 4
#> 7 A 2016-09-03 1 4
#> 8 A 2016-09-04 2 4
#> 9 B 2016-08-08 1 3
#> 10 B 2016-08-09 2 4
#> # ... with 12 more rows
对日期进行分类
df2 <- df1 %>%
group_by(ID) %>%
mutate(day_of = drug == 1,
day_before = (lead(drug) == 1 & day_of == FALSE),
day_after = (lag(drug) == 1 & day_of == FALSE))
df2
#> Source: local data frame [22 x 7]
#> Groups: ID [3]
#>
#> # A tibble: 22 x 7
#> ID date drug score day_of day_before day_after
#> <fctr> <date> <dbl> <dbl> <lgl> <lgl> <lgl>
#> 1 A 2016-08-28 2 3 FALSE TRUE NA
#> 2 A 2016-08-29 1 4 TRUE FALSE FALSE
#> 3 A 2016-08-30 2 4 FALSE NA TRUE
#> 4 A 2016-08-31 NA NA NA NA FALSE
#> 5 A 2016-09-01 NA NA NA FALSE NA
#> 6 A 2016-09-02 2 4 FALSE TRUE NA
#> 7 A 2016-09-03 1 4 TRUE FALSE FALSE
#> 8 A 2016-09-04 2 4 FALSE NA TRUE
#> 9 B 2016-08-08 1 3 TRUE FALSE FALSE
#> 10 B 2016-08-09 2 4 FALSE FALSE TRUE
#> # ... with 12 more rows
按日类型汇总
dplyr::mutate_at
应用一个函数(在funs()
) 到所有选择的列vars()
. summarise_at
在对某些选定列进行操作方面,操作方式相同,但不是更改整个数据集的值,而是将其减少为每组一行。可以阅读更多关于 m 的内容mutate http://dplyr.tidyverse.org/reference/mutate.html, summarise http://dplyr.tidyverse.org/reference/summarise.html,以及特殊的*_at http://dplyr.tidyverse.org/reference/summarise_all.html版本。
df3 <- df2 %>%
mutate_at(vars(starts_with("day_")), funs(if_else(. == TRUE, score, NA_real_))) %>%
summarise_at(vars(starts_with("day_")), mean, na.rm = TRUE)
df3
#> # A tibble: 3 x 4
#> ID day_of day_before day_after
#> <fctr> <dbl> <dbl> <dbl>
#> 1 A 4.00 3.5 4.0
#> 2 B 3.00 3.0 4.0
#> 3 C 3.25 NaN 2.5