Question

我有一个数据集，其中包含studyid，year和两个标志：事件和流行。我希望事件标记为true之后，所有年份的变量都为TRUE（1）（并且事件变量只能为一次）。 case_when和lag似乎是完美的组合，但是如果将事件在N年设置为1，则仅在N + 1中将流行设置为TRUE，而在N + 1中将流行设置为0。这不是预期的行为。

这是示例代码：

library(tidyverse)

# make a fake dataset
testdat <- tribble(
  ~studyid, ~datestring, ~incident,
  "1", "2000-01-01", 0,
  "1", "2001-01-01", 1,
  "1", "2002-01-01", 0,
  "1", "2003-01-01", 0,
  "2", "2003-01-01", 0,
  "2", "2004-01-01", 1,
  "2", "2005-01-01", 0,
  "2", "2006-01-01", 0
) %>% mutate(
  prevalent = 0,
  date = lubridate::ymd(datestring)
) %>% group_by(studyid) %>% 
  arrange(studyid, date) %>% 
  mutate(prevalent = case_when(
    #logic is, if prevalent in year N-1, the prevalent in year N
    # if incident in year N-1, then prevalent in year N
    # otherwise not prevalent (because never incident)
    dplyr::lag(prevalent, 1L)==1 ~1,
    dplyr::lag(incident, 1L)==1 ~1,
    TRUE ~ 0
  ) #close case_when
  ) #close mutate
testdat

输出为：

# A tibble: 8 x 5
# Groups:   studyid [2]
  studyid datestring incident prevalent date      
  <chr>   <chr>         <dbl>     <dbl> <date>    
1 1       2000-01-01        0         0 2000-01-01
2 1       2001-01-01        1         0 2001-01-01
3 1       2002-01-01        0         1 2002-01-01
4 1       2003-01-01        0         0 2003-01-01
5 2       2003-01-01        0         0 2003-01-01
6 2       2004-01-01        1         0 2004-01-01
7 2       2005-01-01        0         1 2005-01-01
8 2       2006-01-01        0         0 2006-01-01
>

所需的输出是：

studyid=1, year=2003  prevalent ==1 (not 0)
studyid=2, year=2006  prevalent ==1 (not 0)

我怀疑这与case_when与dplyr :: lag交互的方式有关。如何改善逻辑/语法以获得所需的结果？

非常感谢，

Answer 1

您正在寻找类似上一次结转的观测值，例如zoo::na.locf或tidyr::fill，但是我将使用类似以下的简单内容：

library(dplyr)
testdat %>% 
   mutate(date = lubridate::ymd(datestring)) %>% group_by(studyid) %>% 
   arrange(studyid, date) %>% mutate(prevalent=cumsum(lag(incident,default = 0)==1))

# A tibble: 8 x 5
# Groups:   studyid [2]
  studyid datestring incident date       prevalent
  <chr>   <chr>         <dbl> <date>         <int>
1 1       2000-01-01        0 2000-01-01         0
2 1       2001-01-01        1 2001-01-01         0
3 1       2002-01-01        0 2002-01-01         1
4 1       2003-01-01        0 2003-01-01         1
5 2       2003-01-01        0 2003-01-01         0
6 2       2004-01-01        1 2004-01-01         0
7 2       2005-01-01        0 2005-01-01         1
8 2       2006-01-01        0 2006-01-01         1

dplyr在case_when和滞后内的行为

1 个答案: