我们如何使用 R 中 data.table 中组中的最后一行进行一些计算？

Question

I have this data.table:

sample:

id cond date
1  A1   2012-11-19
1  A1   2013-05-09
1  A2   2014-09-05
2  B1   2015-03-05
2  B1   2015-07-06
3  A1   2015-02-05
4  B1   2012-09-26
4  B1   2015-02-05
5  B1   2012-09-26

I want to calculate overdue days from today's date within each group of 'id' and 'cond', so I am trying to get the difference of days between the last date in each group and sys.date. Desired output is ;

id cond date        overdue
1  A1   2012-11-19  NA
1  A1   2013-05-09  832 
1  A2   2014-09-05  348
2  B1   2015-03-05  NA 
2  B1   2015-07-06  44
3  A1   2015-02-05  195
4  B1   2012-09-26  NA 
4  B1   2015-02-05  195
5  B1   2012-09-26  1057

I tried to achieve this by following code:

sample <- sample[ , overdue := Sys.Date() - date[.N], by = c('id','cond')]

But I am getting following output, where it the value are recycling:

id cond date        overdue
1  A1   2012-11-19  832
1  A1   2013-05-09  832 
1  A2   2014-09-05  348
2  B1   2015-03-05  44 
2  B1   2015-07-06  44
3  A1   2015-02-05  195
4  B1   2012-09-26  195 
4  B1   2015-02-05  195
5  B1   2012-09-26  1057

I am not sure, how can I restrict my code to just do calculations for the last row and not recycle. I am sure there would be ways to do this, help is appreciated.

Answer 1

You could make a table of overdue values and the rows they belong in:

bycols    = c("id","cond")
newcolDT2 = DT[, Sys.Date() - date[.N], by = bycols]

DT[newcolDT2, overdue := V1, on = bycols, mult = "last"]
#    id cond       date   overdue
# 1:  1   A1 2012-11-19   NA days
# 2:  1   A1 2013-05-09  832 days
# 3:  1   A2 2014-09-05  348 days
# 4:  2   B1 2015-03-05   NA days
# 5:  2   B1 2015-07-06   44 days
# 6:  3   A1 2015-02-05  195 days
# 7:  4   B1 2012-09-26   NA days
# 8:  4   B1 2015-02-05  195 days
# 9:  5   B1 2012-09-26 1057 days

This is the (arguably uglier) one-liner version:

DT[J(unique(DT[, ..bycols])), 
  overdue := Sys.Date() - date, on = bycols, mult = "last"]

---

Data:

DT <- data.table(read.table(header=TRUE,text="id cond date
1  A1   2012-11-19
1  A1   2013-05-09
1  A2   2014-09-05
2  B1   2015-03-05
2  B1   2015-07-06
3  A1   2015-02-05
4  B1   2012-09-26
4  B1   2015-02-05
5  B1   2012-09-26"))[, date := as.IDate(date)]

# anyone know how to do this with fread()?

Answer 2

First, extract the rows you're interested in, then assign the values:

rows = DT[, .I[.N], by = .(id, cond)]$V1
DT[rows, overdue := Sys.Date() - date]

DT
#   id cond       date   overdue
#1:  1   A1 2012-11-19   NA days
#2:  1   A1 2013-05-09  832 days
#3:  1   A2 2014-09-05  348 days
#4:  2   B1 2015-03-05   NA days
#5:  2   B1 2015-07-06   44 days
#6:  3   A1 2015-02-05  195 days
#7:  4   B1 2012-09-26   NA days
#8:  4   B1 2015-02-05  195 days
#9:  5   B1 2012-09-26 1057 days