结束日期与开始日期之间的日期差异

Question

我有一个类似下面的数据.

id   from data   to date
1   2015-03-09   2015-03-14
2   2015-02-22   2015-02-24
2   2015-05-06   2015-05-17
3   2015-02-12   2015-02-16
4   2015-03-10   2015-03-16
4   2015-03-22  2015-04-07
4   2015-06-07  2015-07-07
4   2015-07-06   2015-07-07
4   2015-08-02   2015-08-07

我想创建一个单独的变量,它是按ID分组的日期和下一个日期之间的差异.所以id的第一次将是NA.我尝试了基于stackoverflow中的另一个答案的以下方法,我无法实现.

library(data.table)
chf1 = data.table(id = chf$id,from date = chf$f.date,to_date = chf$t.date)
setkey(chf1,id)
chf1[,diff:=c(NA,difftime(from_date, to_date, units = "days")),by=id]

输出看起来像

id   from_date   to_date      difference
1   2015-03-09   2015-03-14     NA
2   2015-02-22   2015-02-24     NA
2   2015-05-06   2015-05-17     71
3   2015-02-12   2015-02-16     NA
4   2015-03-10   2015-03-16     NA
4   2015-03-22  2015-04-07      6
4   2015-06-07  2015-06-10      64
4   2015-07-06   2015-07-07     26
4   2015-08-02   2015-08-07     26

Answer 1

代码中有三个问题

1)chf1$from_date,chf1$to_date得到整列,所以没有'id'分组的效果

2)difftime给出与初始列长度相同的输出.

3)由于difftime'from_date'的每个元素与'to_date'的对应元素之间存在差异,因此不需要by = id

因此,代码可以

chf1[, diff1:=difftime(from_date, to_date, units = "days")]
chf1
#   id  from_date    to_date    diff1
#1:  1 2015-03-09 2015-03-14  -5 days
##2:  2 2015-02-22 2015-02-24  -2 days
#3:  2 2015-05-06 2015-05-17 -11 days
#4:  3 2015-02-12 2015-02-16  -4 days
#5:  4 2015-03-10 2015-03-16  -6 days
#6:  4 2015-03-22 2015-04-07 -16 days
#7:  4 2015-06-07 2015-07-07 -30 days
#8:  4 2015-07-06 2015-07-07  -1 days
#9:  4 2015-08-02 2015-08-07  -5 days

基于在OP代码的说明,如果我们需要得到的"身份证"分组后"FROM_DATE"的下一个值之间的差异,使用difftime上shift与的"TO_DATE" ED"FROM_DATE"和分配(:=它'DIFF1'.

chf1[,  diff1 := difftime(shift(from_date, type = "lead"), to_date, 
                        units = "days") , by = id]
chf1
#  id  from_date    to_date   diff1
#1:  1 2015-03-09 2015-03-14 NA days
#2:  2 2015-02-22 2015-02-24 71 days
#3:  2 2015-05-06 2015-05-17 NA days
#4:  3 2015-02-12 2015-02-16 NA days
#5:  4 2015-03-10 2015-03-16  6 days
#6:  4 2015-03-22 2015-04-07 61 days
#7:  4 2015-06-07 2015-07-07 -1 days
#8:  4 2015-07-06 2015-07-07 26 days
#9:  4 2015-08-02 2015-08-07 NA days

或者可能是

chf1[, diff1 := difftime(from_date, shift(to_date), units = "days"), by = id]

数据

chf <- structure(list(id = c(1L, 2L, 2L, 3L, 4L, 4L, 4L, 4L, 4L), 
f.date = structure(c(16503, 
16488, 16561, 16478, 16504, 16516, 16593, 16622, 16649), class = "Date"), 
t.date = structure(c(16508, 16490, 16572, 16482, 16510, 16532, 
16623, 16623, 16654), class = "Date")), .Names = c("id", 
 "f.date", "t.date"), row.names = c(NA, -9L), class = "data.frame")

 chf1 = data.table(id = chf$id,from_date = chf$f.date,to_date = chf$t.date)