在r语言中，如何识别多个患者ID的时间间隔是否重叠？

Question

我有几个诊所的入院和出院数据，并需要确定是否存在两个或更多患者在同一诊所以及相同病原体DNA菌株下的重叠入院情况。使用R处理。

只有5个变量：研究ID、入院日期、出院日期、DNA菌株和诊所。每行是一个单独的访问，这意味着 "ID" 变量可能会重复。我需要确定有多少"重叠"，也就是说，相同病原体DNA的患者在同一时间在同一诊所。

这似乎是一个简单的问题，在这个网站上也被广泛讨论过。当每个ID都是唯一的时，我能够识别重叠。我的具体问题是如何处理重复的ID。在下面的代码中，我提供了样本数据和我的代码，在每个ID都是唯一的情况下可以工作（通过将代码第一行的seq(1:20)),each=2)更改为seq(1:40)),each=1)，您可以快速测试此代码）。如果我使用重复的ID运行此代码，则标记所有访问为重叠= 1，而不管它实际上是否重叠的访问。

我知道重叠日期间隔是一个广泛讨论的主题，所以请相信我已经彻底查看了我能找到的所有内容，除了这最后一部分。非常感谢您的任何建议。

谢谢！

#Set globals
set.seed(8126) 
library(lubridate); library(data.table)

#Example Data
have<-data.frame(rep(paste0("k",seq(1:20)),each=2),sample(seq(as.Date('2015/01/01'), as.Date('2020/01/01'), by="day"), 40))
names(have)<-c("id","datein")
have$dateout<-have$datein+40 #arbitrarily add 40 days to admission date
have$dnastrain<-as.vector(replicate(20,rep(sample(c("Type 1","Type 2","Type 3"),1),each=2)))
have$clinic<-sample(c("A","B","C","D"),40, replace=TRUE)


#Code that works if each ID is unique
setDT(have); setkey(have,datein, dateout) #Convert to DT and set date in/out as keys
overlaps<-unique(foverlaps(have, have)[id!=i.id & dnastrain==i.dnastrain & clinic==i.clinic, id]) #find overlaps

have[,`:=`(overlap=0)][id %in% overlaps, overlap:=1][order(datein)] #Identify overlaps

- jpsmith

Answer 1

这是一个dplyr方法。

have$rownum <- 1:nrow(have)

 crossover <- have %>% 
   inner_join(have, by = c("clinic", "dnastrain"), suffix=(c(".a",".b"))) %>% 
   filter(datein.a<=dateout.b & 
          dateout.a>=datein.b &
          id.a!=id.b & 
          rownum.a < rownum.b
          )


crossover
  id.a   datein.a  dateout.a dnastrain clinic rownum.a id.b   datein.b  dateout.b rownum.b
1   k1 2017-11-02 2017-12-12    Type 3      B        1  k18 2017-10-03 2017-11-12       35
2  k10 2016-02-15 2016-03-26    Type 3      A       19  k13 2016-03-22 2016-05-01       26
3  k17 2017-08-06 2017-09-15    Type 2      C       33  k20 2017-09-02 2017-10-12       39

- Patrick

Answer 2

我发现可以使用purrr来以略微不同的方式完成它：

participant_in <- have$datein
participant_out <- have$dateout

# For each participant, create a vector of days at which they were in a clinic
days <- map2(participant_in, participant_out, ~ seq(from = ymd(.x), to = ymd(.y), by = 'days'))
ids <- map2(have$id, days, ~ rep(.x, length(.y)))

days <- days %>% reduce(c)
ids <- ids %>% unlist()

participant_tib <- tibble(id = ids, day = days) %>%
  left_join(have, by = 'id') %>%
  select(-datein, -dateout) 

participant_tib %>%
  group_by(day, dnastrain, clinic) %>%
  count() %>%
  arrange(desc(n))

通过上述方法，您可以检查在同一诊所中有几位参与者携带相同菌株的日期。从那里，您可以过滤出n > 1的日期，并检查这些日期的参与者ID。

- Adam B.

Answer 3

你可以使用join而不是%in%：

setDT(have, key=c("datein","dateout"))
overlaps <- unique(foverlaps(have, have)[clinic==i.clinic, dnastrain==i.dnastrain & id!=i.id])

cols <- copy(names(have))
have[, overlap := 0][
    overlaps, on=cols, overlap := 1][
        order(datein)] 

输出：

     id     datein    dateout dnastrain clinic overlap
 1:  k6 2015-01-01 2015-02-10    Type 1      D       0
 2:  k6 2015-01-20 2015-03-01    Type 1      D       0
 3:  k9 2015-04-09 2015-05-19    Type 2      B       1
 4: k11 2015-04-12 2015-05-22    Type 2      B       1
 5: k10 2015-04-19 2015-05-29    Type 3      C       0
 6: k10 2015-07-03 2015-08-12    Type 3      A       0
 7:  k4 2015-08-06 2015-09-15    Type 1      C       0
 8: k16 2015-08-26 2015-10-05    Type 2      A       0
 9: k13 2016-05-20 2016-06-29    Type 3      D       1
10: k19 2016-05-26 2016-07-05    Type 3      D       1
11: k15 2016-06-23 2016-08-02    Type 3      A       0
12:  k7 2016-06-29 2016-08-08    Type 3      B       0
13: k18 2016-07-18 2016-08-27    Type 2      B       0
14: k13 2016-08-11 2016-09-20    Type 3      A       0
15:  k1 2016-09-24 2016-11-03    Type 3      C       0
16: k18 2016-09-29 2016-11-08    Type 2      B       0
17: k12 2017-03-13 2017-04-22    Type 3      B       0
18:  k7 2017-03-24 2017-05-03    Type 3      C       0
19: k14 2017-05-28 2017-07-07    Type 3      B       0
20:  k3 2017-06-05 2017-07-15    Type 2      B       0
21: k17 2017-07-06 2017-08-15    Type 3      A       0
22: k17 2017-09-08 2017-10-18    Type 3      B       1
23:  k2 2017-09-30 2017-11-09    Type 3      B       1
24: k15 2017-10-07 2017-11-16    Type 3      D       0
25:  k8 2018-01-01 2018-02-10    Type 1      C       0
26:  k5 2018-07-07 2018-08-16    Type 2      B       0
27: k20 2018-07-18 2018-08-27    Type 1      A       0
28:  k5 2018-08-30 2018-10-09    Type 2      B       0
29: k16 2018-09-22 2018-11-01    Type 2      D       0
30: k14 2018-09-24 2018-11-03    Type 3      B       0
31:  k2 2018-11-02 2018-12-12    Type 3      A       1
32: k12 2018-11-13 2018-12-23    Type 3      A       1
33: k19 2018-11-21 2018-12-31    Type 3      D       0
34:  k3 2018-11-30 2019-01-09    Type 2      A       0
35:  k1 2018-12-14 2019-01-23    Type 3      A       1
36:  k9 2019-03-13 2019-04-22    Type 2      D       0
37:  k8 2019-07-18 2019-08-27    Type 1      C       1
38: k20 2019-08-10 2019-09-19    Type 1      C       1
39: k11 2019-08-30 2019-10-09    Type 2      B       0
40:  k4 2019-09-04 2019-10-14    Type 1      B       0
     id     datein    dateout dnastrain clinic overlap

数据：

set.seed(8126) 
library(data.table)

have<-data.frame(rep(paste0("k",seq(1:20)),each=2),sample(seq(as.Date('2015/01/01'), as.Date('2020/01/01'), by="day"), 40))
names(have)<-c("id","datein")
have$dateout<-have$datein+40 #arbitrarily add 40 days to admission date
have$dnastrain<-as.vector(replicate(20,rep(sample(c("Type 1","Type 2","Type 3"),1),each=2)))
have$clinic<-sample(c("A","B","C","D"),40, replace=TRUE)

- chinsoon12

Answer 4

我刚学会了一种使用 dplyr 和 purrr 的解决方案。

library(dplyr)
library(purrr)
library(lubridate)


have %>% 
  mutate(interval=interval(datein,dateout)) %>%
  group_by(dnastrain,clinic) %>% 
  mutate(overlap = purrr::map_int(row_number(),
  ~+any(datein[.x] %within% interval[-.x] | dateout[.x] %within% interval[-.x])))

- captcoma