r - R 中的日期汇总

Question

我有一个如下所示的数据集:

    ID  FromDate    ToDate  SiteID  Cost
    1   8/12/2014   8/31/2014   12  245.98
    1   9/1/2014    9/7/2014    12  269.35
    1   10/10/2014  10/17/2014  12  209.98
    1   11/22/2014  11/30/2014  12  309.12
    1   12/1/2014   12/11/2014  12  202.14
    2   8/16/2014   8/21/2014   12  109.35
    2   8/22/2014   8/24/2014   14  44.12
    2   9/25/2014   9/29/2014   12  98.75
    3   9/15/2014   9/30/2014   23  536.27
    3   10/1/2014   10/31/2014  12  529.87
    3   11/1/2014   11/30/2014  12  969.55
    3   12/1/2014   12/12/2014  12  607.35

我希望它看起来像:

    ID  FromDate    ToDate  SiteID  Cost
    1   8/12/2014   9/7/2014    12  515.33
    1   10/10/2014  10/17/2014  12  209.98
    1   11/22/2014  12/11/2014  12  511.26
    2   8/16/2014   8/21/2014   12  109.35
    2   8/22/2014   8/24/2014   14  44.12
    2   9/25/2014   9/29/2014   12  98.75
    3   9/15/2014   9/30/2014   23  536.27
    3   10/1/2014   12/12/2014  12  2106.77

如您所见，如果有延续，则日期会累积，并且成本按 ID 和 SiteID 相加。为了帮助某人理解复杂性，如果日期间隔有延续，但 SiteID 发生变化，则它是一个单独的行。如果日期间隔没有延续，则它是一个单独的行。我如何在 R 中做到这一点？此外，我有超过 100,000 个个人 ID。那么最有效的方法/包是什么？

Answer 1

这可能会

df %>% 
  mutate(gr = cumsum(FromDate-lag(ToDate, default=1) != 1)) %>% 
  group_by(gr, ID, SiteID) %>% 
  summarise(FromDate = min(FromDate), 
            ToDate   = max(ToDate), 
            cost     = sum(Cost))


     gr    ID SiteID   FromDate     ToDate    cost
  (int) (int)  (int)     (date)     (date)   (dbl)
1     1     1     12 2014-08-12 2014-09-07  515.33
2     2     1     12 2014-10-10 2014-10-17  209.98
3     3     1     12 2014-11-22 2014-12-11  511.26
4     4     2     12 2014-08-16 2014-08-21  109.35
5     4     2     14 2014-08-22 2014-08-24   44.12
6     5     2     12 2014-09-25 2014-09-29   98.75
7     6     3     23 2014-09-15 2014-09-30  536.27
8     6     3     12 2014-10-01 2014-12-12 2106.77

与 data.table

library(data.table)
setDT(df)
df[, gr := cumsum(FromDate - shift(ToDate, fill=1) != 1),
   ][, list(FromDate=min(FromDate), ToDate=max(ToDate), cost=sum(Cost)), by=.(gr, ID, SiteID)]



   gr ID SiteID   FromDate     ToDate    cost
1:  1  1     12 2014-08-12 2014-09-07  515.33
2:  2  1     12 2014-10-10 2014-10-17  209.98
3:  3  1     12 2014-11-22 2014-12-11  511.26
4:  4  2     12 2014-08-16 2014-08-21  109.35
5:  4  2     14 2014-08-22 2014-08-24   44.12
6:  5  2     12 2014-09-25 2014-09-29   98.75
7:  6  3     23 2014-09-15 2014-09-30  536.27
8:  6  3     12 2014-10-01 2014-12-12 2106.77