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使用一个数据集上一列中的日期设置一年范围向前,以在 R 中与 id 匹配的不同数据集上查找日期

如何解决使用一个数据集上一列中的日期设置一年范围向前,以在 R 中与 id 匹配的不同数据集上查找日期

我有两个数据集,(dt1) 一个带有“开始”日期,每个 ID 最多两个条目(因为这些是 L 或 R 眼手术的条目)和一个 (dt2) 具有多个日期的第二个数据库在开始日期之前和之后。这些不仅限于眼科手术,还包括任何其他医疗保健就诊。我想在我的数据集中所有 id 的手术 L 和 R 的开始日期的一年内寻找一个事件。如果他们有结果,我想通过偏向性来匹配他们。如果他们没有,那么在一年内最后一次访问时没有侧向匹配(lat)。先前的结果只是开始日期之前的结果或“事件”的数量,先前结果的总和。

id   lat year status       date
1    le   18      1 2018-07-06
1    re   11      1 2011-04-12
2    le   15      0 2015-01-10
2    re   11      0 2011-07-20
3    NA   10      1 2010-02-18
3 bilat   13      1 2013-09-26



    id   lat outcome       date year
 1:  1    NA       0 2015-07-06   15
 2:  1    le       0 2019-04-03   19
 3:  1    le       1 2019-04-30   19
 4:  1    re       1 2011-07-14   11
 5:  1    re       1 2015-09-10   15
 6:  1    re       1 2008-07-14    8
 7:  2    NA       0 2015-11-10   15
 8:  2    re       0 2012-04-23   12
 9:  2    NA       0 2015-02-18   15
10:  2    57       0 2008-12-01    8
11:  3    57       0 2014-01-15   14
12:  3    NA       0 2014-02-21   14
13:  3 bilat       1 2014-02-28   14

我希望决赛桌看起来像这样

id lat year status       date outcome   end_date prior_outcome
1  le   18      1 2018-07-06       1 2019-04-30             3
1  re   11      1 2011-04-12       1 2011-07-14             1
2  le   15      0 2015-01-10       0 2015-11-10             0
2  re   11      0 2011-07-20       0 2012-04-23             0
3  NA   10      1 2010-02-18       0       <NA>             0
3 bilat 13      1 2013-09-26       1 2014-02-28             0

这里是数据集的代码

 dates <- as.Date(c("2018-07-06","2011-04-12","2015-01-10","2011-07-20","2010-02-18","2013-09-26"))
dt1 <- data.table(id=c(1,1,2,3,3),lat=c("le","re","le","NA","bilat"),year= c(18,11,15,10,13),status=status <- c(1,1),date= dates)

dates2 <- as.Date(c('2015-07-06','2019-04-03','2019-04-30','2011-07-14','2015-09-10','2008-07-14','2015-11-10','2012-04-23','2015-02-18','2008-12-01','2014-01-15','2014-02-21','2014-02-28' ))
dt2 <- data.table(id=c(1,lat=c("NA","57",outcome = c(0,date= dates2,year= c(15,19,08,12,14,14))
                          

我尝试过这样的事情,但它在原始集合中不起作用,因为我在开始日期之前得到结果,所以我假设代码可能在这个小数据集中起作用,但实际上,这是不正确的不知何故。

#left join dt1 and dt2
dt1_dt2 <- left_join( dt2,dt1,by= "id",suffix=c("2event","1op"))

#filter does with outcome after date1 
dt1_dt2$tdiff = difftime(dt1_dt2$date2event,dt1_dt2$date1op,units= "days")
dt1_dt2 = dt1_dt2 %>% filter(outcome== 1) %>% 
  filter(tdiff <= 365,tdiff >= 0)
         

#then match on the closest date since farthest was not supported 
setDT(dt1)
setDT(dt1_dt2)
setkey(dt1,id,lat) #key set to match
dt3 <- dt1_dt2[,date2,by =.(id,lat2),roll ="nearest"] #how can I keep all variables?
dt3 = unique(dt3,by = c("id","date2","lat2"))

解决方法

更新答案

# construct complete data
da <- merge(dt1,dt2,by=c("id","lat"),all.x = TRUE,suffixes = c("_start","_end"))
# select desired columns => add tdiff ==> replace outcome == NA with 0
da2 <- da[,.(id,lat,year_start,status,date_start,outcome,date_end)][,tdiff := as.numeric( difftime(date_end,units= "days"))][,outcome:=fifelse(is.na(outcome),outcome)]

# add flag to show whether the part of da2 (id,lat) also appears in dt2
setkey(da2,id,lat)
setkey(dt2,lat)
da2[,flag:=FALSE][dt2,flag:=TRUE]

# get desired result
dt_desired <- da2[0 <= tdiff & tdiff <= 365 | lat == "NA" | is.na(date_end)]
rows <- dt_desired[flag==FALSE & outcome == 0]
# fill with last event's date_end within one year
dt_desired[flag==FALSE & outcome == 0]$date_end <- dt2[,.SD,keyby=.(id,date)][,.SD[.N],by=id][rows,date]
dt_desired[as.numeric( difftime(date_end,units= "days")) > 365]$date_end <- NA

结果

   id lat year_start status date_start outcome   date_end tdiff  flag
1:  1  le         18      1 2018-07-06       1 2019-04-03   271  TRUE
2:  1  re         11      1 2011-04-12       1 2011-07-14    93  TRUE
3:  2  le         15      0 2015-01-10       0 2015-11-10    NA FALSE
4:  2  re         11      0 2011-07-20       0 2012-04-23   278  TRUE
5:  3  57         13      0 2013-09-26       1 2014-01-15   111  TRUE
6:  3  NA         10      1 2010-02-18       0       <NA>    NA FALSE

原答案

根据你的代码,我得到的结果接近你想要的结果。请检查是否正确。

dt1_dt2 <- left_join( dt2,dt1,by= "id",suffix=c("2event","1op"))
# add column tdiff,equal to your method
dt1_dt2[,tdiff := as.numeric( difftime(date2event,date1op,units= "days") )]

# select desired columns
dt1_dt2[0 <= tdiff & tdiff <= 365,lat1op,year1op,date2event)]

我和你的结果之间的差异位于第 5 行。 从您提供的数据中,我找不到任何带有 NAend_date。 至于prior_outcome,您没有说明如何计算它。我认为这不是主要问题。

enter image description here

,

这里有一个使用非对等连接的选项:

cols <- c("outcome","end_date") 

dt1[,oneyr := date + 365L] #or in case of leap year,dt1[,oneyr := as.Date(sapply(date,function(d) seq(d,by="1 year",length.out=2L)[[2L]]),origin="1970-01-01")]

dt1[,(cols) := 
    dt2[.SD,on=.(id,date>=date,date<=oneyr),mult="first",.(x.outcome,x.date)]
]

dt1[is.na(outcome),outcome := 0L]

输出:

   id lat year status       date      oneyr outcome   end_date
1:  1  le   18      1 2018-07-06 2019-07-06       1 2019-04-03
2:  1  re   11      1 2011-04-12 2012-04-11       1 2011-07-14
3:  2  le   15      0 2015-01-10 2016-01-10       0 2015-11-10
4:  2  re   11      0 2011-07-20 2012-07-19       0 2012-04-23
5:  3  NA   10      1 2010-02-18 2011-02-18       0       <NA>
6:  3  57   13      0 2013-09-26 2014-09-26       1 2014-01-15

qn 更新后编辑。不清楚新的所需输出是什么,您可以尝试以下操作:

dt1[,oneyr := date + 365L] 

cols <- paste0("i.",names(dt1))
a1 <- dt2[dt1,c(mget(cols),.(outcome=outcome,end_date=x.date))]
setnames(a1,names(a1),gsub("^i.","",names(a1)))

a2 <- dt2[a1[is.na(outcome)],end_date=x.date))]
setnames(a2,names(a2),names(a2)))
    
setorder(rbindlist(list(a1[!is.na(outcome)],a2),use.names=TRUE),date)[]

输出:

   id lat year status       date      oneyr outcome   end_date
1:  1  re   11      1 2011-04-12 2012-04-11       1 2011-07-14
2:  1  le   18      1 2018-07-06 2019-07-06       0 2019-04-03
3:  1  le   18      1 2018-07-06 2019-07-06       1 2019-04-30
4:  2  re   11      0 2011-07-20 2012-07-19       0 2012-04-23
5:  2  le   15      0 2015-01-10 2016-01-10       0 2015-11-10
6:  2  le   15      0 2015-01-10 2016-01-10       0 2015-02-18
7:  3  NA   10      1 2010-02-18 2011-02-18      NA       <NA>
8:  3  57   13      0 2013-09-26 2014-09-26       0 2014-01-15

在所需的输出更新后编辑:

cols <- c("outcome",by=.EACHI,{
        w <- which(outcome==1L)
        if (length(w) > 0L) {
            .(outcome=outcome[w[1L]],x.date[w[1L]])
        } else {
            .(outcome=outcome[1L],x.date[1L])
        }
    }][,(1L:4L) := NULL]
]

dt1[is.na(outcome),(1L:3L) := NULL]
]

dt1[is.na(outcome),outcome := 0L][]

输出:

   id   lat year status       date      oneyr outcome   end_date
1:  1    le   18      1 2018-07-06 2019-07-06       1 2019-04-30
2:  1    re   11      1 2011-04-12 2012-04-11       1 2011-07-14
3:  2    le   15      0 2015-01-10 2016-01-10       0 2015-02-18
4:  2    re   11      0 2011-07-20 2012-07-19       0 2012-04-23
5:  3    NA   10      1 2010-02-18 2011-02-18       0       <NA>
6:  3 bilat   13      1 2013-09-26 2014-09-26       1 2014-02-28

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