Question

我有一个带有id和状态的数据集。我需要在过去的四个季度中获得最大的状态。例如，我创建了这个表：

library(data.table)
library(zoo)

DT = 
  data.table(id=c("1","1","1","1","1","1","1","1","2"),
             date=as.yearqtr(c("2015Q1", "2014Q4", "2014Q3", "2014Q2", "2014Q1", "2013Q4","2013Q3","2013Q2","2015Q2")),
             status=c(0,0,0,0,0,0,1,0,0))

我希望在过去4个季度中获得status的最大值：

DT = 
data.table(id=c("1","1","1","1","1","1","1","1","2"),
           date=as.yearqtr(c("2015Q1", "2014Q4", "2014Q3", "2014Q2", "2014Q1","2013Q4","2013Q3","2013Q2","2015Q2")),
           status=c(0,0,0,0,0,0,1,0,0),
           Max_status=c(0,0,0,1,1,1,1,0,0))

有没有人有任何线索在data.table中整齐而快速地做到这一点？我试过这样的事情：

DT[,Date_1yPrior:=date-1]
DT[,Max_Status:=max(status[date>=Date_1yPrior & date<=Date_1yPrior]),by='id']

但j和i指的不正确。我更喜欢data.table解决方案。

提前谢谢！

Answer 1

使用zoo::rollapplyr

DT[order(id, date), max_status:=rollapplyr(status, FUN=max, width=4, fill=0, partial=TRUE), id]  
DT

#   id    date status max_status
#1:  1 2015 Q1      0          0
#2:  1 2014 Q4      0          0
#3:  1 2014 Q3      0          0
#4:  1 2014 Q2      0          1
#5:  1 2014 Q1      0          1
#6:  1 2013 Q4      0          1
#7:  1 2013 Q3      1          1
#8:  1 2013 Q2      0          0
#9:  2 2015 Q2      0          0

Answer 2

我认为这就是诀窍：

library(zoo)
library(data.table)
DT[, date_num := as.numeric(date)]
DT[, NextStatus1Q := .SD[match(date_num - .25, .SD$date_num), status], by=id]
DT[, NextStatus2Q := .SD[match(date_num - .5, .SD$date_num), status], by=id]
DT[, NextStatus3Q := .SD[match(date_num - .75, .SD$date_num), status], by=id]
DT[, NextStatus4Q := .SD[match(date_num - 1, .SD$date_num), status], by=id]
DT[,MaxStatus:=max(NextStatus1Q,NextStatus2Q,NextStatus3Q,status,0,na.rm=T),by=c('id','date')]
#DT[,MaxStatus:=max(NextStatus1Q,NextStatus2Q,NextStatus3Q,NextStatus4Q,status,0,na.rm=T),by=c('id','date')] # If you also want 4q (actually 5 quarters back)

创建新列，在其中查找1-3个季度前的值，然后选择最大值

如何在r data.table的最后4个季度中通过id获取最大值？

2 个答案: