检查组中是否有任何日期在该组的特定时间间隔内

时间:2019-04-22 19:28:18

标签: r dplyr lubridate

我想创建一个新变量,该变量指示visit_date是否在ID列出的日期范围内

我已使用此代码进行逐行比较,但我想扩展此代码并将ID的所有行与该ID列出的所有间隔行进行比较

df <- df %>%
  group_by(id) %>%
  mutate(between_any = ifelse((visit_date >= start & visit_date <= end), 1))

我也尝试过创建一个间隔变量,并在变异前使用crossing(visit_date,interval),但是我无法对日期对象使用Crossing。

以下是一些示例数据:

df <- data.frame(id = c("a","a","a","a","a","b","b","b"),
                 visit_date = c("2001-08-22","2001-09-21","2001-10-30","2001-11-10","2001-12-20","2002-12-22", "2003-04-30","2003-05-10"),
                 start = c(NA,"2001-09-21",NA,"2001-11-10",NA,"2002-12-22", "2003-04-30",NA),
                 end = c(NA, "2001-11-01",NA,"2001-11-10",NA,"2002-12-22","2003-06-01",NA))

> df
id visit_date    start        end
a 2001-08-22       <NA>       <NA>
a 2001-09-21 2001-09-21 2001-11-01
a 2001-10-30       <NA>       <NA>
a 2001-11-10 2001-11-10 2001-11-10
a 2001-12-20       <NA>       <NA>
b 2002-12-22 2002-12-22 2002-12-22
b 2003-04-30 2003-04-30 2003-06-01
b 2003-05-10       <NA>       <NA>

我想要的输出如下:

id visit_date      start       end   between_any
a 2001-08-22       <NA>       <NA>      0
a 2001-09-21 2001-09-21 2001-11-01      1
a 2001-10-30       <NA>       <NA>      1
a 2001-11-10 2001-11-10 2001-11-10      1
a 2001-12-20       <NA>       <NA>      0
b 2002-12-22 2002-12-22 2002-12-22      1
b 2003-04-30 2003-04-30 2003-06-01      1
b 2003-05-10       <NA>       <NA>      1

谢谢!

3 个答案:

答案 0 :(得分:3)

in_range包中的

data.table函数可以做到这一点...

library(data.table)

df <- df %>%
  group_by(id) %>%
  mutate(between_any = as.numeric((inrange(visit_date, start, end))))

#> df
#  id visit_date      start        end between_any
#1  a 2001-08-22       <NA>       <NA>           0
#2  a 2001-09-21 2001-09-21 2001-11-01           1
#3  a 2001-10-30       <NA>       <NA>           1
#4  a 2001-11-10 2001-11-10 2001-11-10           1
#5  a 2001-12-20       <NA>       <NA>           0
#6  b 2002-12-22 2002-12-22 2002-12-22           1
#7  b 2003-04-30 2003-04-30 2003-06-01           1
#8  b 2003-05-10       <NA>       <NA>           1

以data.table形式...

dt <- setDT(df)      
dt[, between_any := inrange(visit_date, start, end), 
     by = id]

答案 1 :(得分:2)

我的回答不像我想要的那样“漂亮”,但是它可以带您找到想要去的地方。

我首先将您的日期转换为日期:

library(lubridate)
library(dplyr)
library(tibble)
library(tidyr)
library(purrr)

df <- data.frame(id = c("a","a","a","a","a","b","b","b"),
                 visit_date = c("2001-08-22","2001-09-21","2001-10-30","2001-11-10","2001-12-20","2002-12-22", "2003-04-30","2003-05-10"),
                 start = c(NA,"2001-09-21",NA,"2001-11-10",NA,"2002-12-22", "2003-04-30",NA),
                 end = c(NA, "2001-11-01",NA,"2001-11-10",NA,"2002-12-22","2003-06-01",NA)) %>%
  mutate_at(-1,as.Date)

> df
  id visit_date      start        end
1  a 2001-08-22       <NA>       <NA>
2  a 2001-09-21 2001-09-21 2001-11-01
3  a 2001-10-30       <NA>       <NA>
4  a 2001-11-10 2001-11-10 2001-11-10
5  a 2001-12-20       <NA>       <NA>
6  b 2002-12-22 2002-12-22 2002-12-22
7  b 2003-04-30 2003-04-30 2003-06-01
8  b 2003-05-10       <NA>       <NA>

接下来,我为每个组创建一个间隔列表:

df_intervals <- df %>% 
  mutate_at(-1,as.Date) %>%
  filter(!is.na(start),
         !is.na(end)) %>%
  mutate(interval = start %--% end) %>%
  select(id,interval) %>%
  group_by(id)

> df_intervals
# A tibble: 4 x 2
# Groups:   id [2]
  id    interval                      
  <fct> <S4: Interval>                
1 a     2001-09-21 UTC--2001-11-01 UTC
2 a     2001-11-10 UTC--2001-11-10 UTC
3 b     2002-12-22 UTC--2002-12-22 UTC
4 b     2003-04-30 UTC--2003-06-01 UTC

最后,我将间隔数据与基于id的原始数据结合起来,并在间隔内搜索visit_date

df_output <- df %>% as.tbl() %>%
  inner_join(df_intervals) %>%
  mutate(between_any = map2_lgl(visit_date,interval,~ .x >= int_start(.y) & .x <= int_end(.y))) %>%
  group_by(id,visit_date,start,end) %>%
  summarise(between_any = as.numeric(any(between_any)))

> df_output
# A tibble: 8 x 5
# Groups:   id, visit_date, start [8]
  id    visit_date start      end        between_any
  <fct> <date>     <date>     <date>           <dbl>
1 a     2001-08-22 NA         NA                   0
2 a     2001-09-21 2001-09-21 2001-11-01           1
3 a     2001-10-30 NA         NA                   1
4 a     2001-11-10 2001-11-10 2001-11-10           1
5 a     2001-12-20 NA         NA                   0
6 b     2002-12-22 2002-12-22 2002-12-22           1
7 b     2003-04-30 2003-04-30 2003-06-01           1
8 b     2003-05-10 NA         NA                   1

答案 2 :(得分:0)

另一种可能是:

df %>% 
 rowid_to_column() %>%
 full_join(df %>%
            filter(!is.na(start) & !is.na(end)) %>%
            mutate(interval = interval(ymd(start), ymd(end))) %>%
            select(id, interval), by = c("id" = "id")) %>%
 group_by(rowid, id) %>%
 summarise(between_any = max(ymd(visit_date) %within% interval * 1)) %>%
 left_join(df %>%
            rowid_to_column(), by = c("rowid" = "rowid",
                                      "id" = "id")) %>%
 ungroup() %>%
 select(-rowid)
  id    between_any visit_date start      end       
  <fct>       <dbl> <fct>      <fct>      <fct>     
1 a               0 2001-11-08 <NA>       <NA>      
2 a               1 2001-09-21 2001-09-21 2001-11-01
3 a               1 2001-10-30 <NA>       <NA>      
4 a               1 2001-11-10 2001-11-10 2001-11-10
5 a               0 2001-12-20 <NA>       <NA>      
6 b               1 2002-12-22 2002-12-22 2002-12-22
7 b               1 2003-04-30 2003-04-30 2003-06-01
8 b               1 2003-05-10 <NA>       <NA> 

在这里,首先创建interval变量,然后基于“ id”执行完全连接。其次,它检查“ visit_date”是否在每个“ id”和“ rowid”的间隔内。最后,它将结果与原始数据结合起来。