How can we do some calculations using last row within a group in data.table in R? -

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i have 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 calculate overdue days today's date within each group of 'id' , 'cond', trying difference of days between last date in each group , sys.date. desired output ;

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 achieve following code: 

sample <- sample[ , overdue := sys.date() - date[.n], = c('id','cond')]

but getting following output, value 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 not sure, how can restrict code calculations last row , not recycle. sure there ways this, appreciated.

you make table of overdue values , rows belong in:

bycols    = c("id","cond") newcoldt2 = dt[, sys.date() - date[.n], = 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 (arguably uglier) one-liner version:

dt[j(unique(dt[, bycols, with=false])),    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)]  # know how fread()?

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