R在data.table中部分填充NA

Question

I have the following data.table:

dt <- data.table(date=rep(c(2014,2013), each=4), price=c(3.14, 1.45, 3.4 ,5.1, 1, 2.3, 2.79, 3), brand=rep(c("Mercedes", "Audi"), each=4), num=c(3,6,7,8,3,5,9,12), seller=rep(c("gregory", "dan"), each=4))

Resulting in:

   date price    brand num  seller
1: 2013  1.00     Audi   3     dan
2: 2013  2.30     Audi   5     dan
3: 2013  2.79     Audi   9     dan
4: 2013  3.00     Audi  12     dan
5: 2014  3.14 Mercedes   3 gregory
6: 2014  1.45 Mercedes   6 gregory
7: 2014  3.40 Mercedes   7 gregory
8: 2014  5.10 Mercedes   8 gregory

My target is now to have this:

    date num price    brand  seller
 1: 2013   3  1.00     Audi     dan
 2: 2013   5  2.30     Audi     dan
 3: 2013   6    NA     Audi     dan
 4: 2013   7    NA     Audi     dan
 5: 2013   8    NA     Audi     dan
 6: 2013   9  2.79     Audi     dan
 7: 2013  12  3.00     Audi     dan
 8: 2014   3  3.14 Mercedes gregory
 9: 2014   5    NA Mercedes gregory
10: 2014   6  1.45 Mercedes gregory
11: 2014   7  3.40 Mercedes gregory
12: 2014   8  5.10 Mercedes gregory
13: 2014   9    NA Mercedes gregory
14: 2014  12    NA Mercedes gregory

I first add lines for the missing num for every date:

setkey(dt, date, num)
dtt<-dt[CJ(unique(date), unique(dt[,num]))]

Giving this first step:

    date num price    brand  seller
 1: 2013   3  1.00     Audi     dan
 2: 2013   5  2.30     Audi     dan
 3: 2013   6    NA       NA      NA
 4: 2013   7    NA       NA      NA
 5: 2013   8    NA       NA      NA
 6: 2013   9  2.79     Audi     dan
 7: 2013  12  3.00     Audi     dan
 8: 2014   3  3.14 Mercedes gregory
 9: 2014   5    NA       NA      NA
10: 2014   6  1.45 Mercedes gregory
11: 2014   7  3.40 Mercedes gregory
12: 2014   8  5.10 Mercedes gregory
13: 2014   9    NA       NA      NA
14: 2014  12    NA       NA      NA

And then:

dtt[date==2013, c("brand","seller"):=list("Audi","dan")]
dtt[date==2014, c("brand","seller"):=list("Mercedes","gregory")]

Gives the wanted result.

However:

1 - the last piece of code is awfull.

2 - I would like to make a generic function (or a join) because I have lots of different dates and columns to replace/keep the NA's in my real data.table.

It seems simple but I am stuck!

Answer 1

How about:

require(data.table) ## 1.9.2
setkey(dt, num)
nums = unique(dt$num)
dt[, list(price=.SD[J(nums)]$price, brand=brand[1L], 
          num=nums, seller=seller[1L]), by=date]
#     date price    brand num  seller
#  1: 2014  3.14 Mercedes   3 gregory
#  2: 2014    NA Mercedes   5 gregory
#  3: 2014  1.45 Mercedes   6 gregory
#  4: 2014  3.40 Mercedes   7 gregory
#  5: 2014  5.10 Mercedes   8 gregory
#  6: 2014    NA Mercedes   9 gregory
#  7: 2014    NA Mercedes  12 gregory
#  8: 2013  1.00     Audi   3     dan
#  9: 2013  2.30     Audi   5     dan
# 10: 2013    NA     Audi   6     dan
# 11: 2013    NA     Audi   7     dan
# 12: 2013    NA     Audi   8     dan
# 13: 2013  2.79     Audi   9     dan
# 14: 2013  3.00     Audi  12     dan

or alternatively:

dt[, c(.SD[J(nums), list(price=price)], brand=brand[1L], 
           seller=seller[1L]), by=date]

where the order of columns will be different.

---

In 1.9.3 , this'll be much more efficient (in terms of both syntax and speed), because we don't have to join and return all the columns:

## 1.9.3
dt[, list(price=.SD[J(nums), price], brand=brand[1L], 
          num=nums, seller=seller[1L]), by=date]

.SD[J(nums), price] will result in a vector, as opposed to a data.table in previous versions and will not perform an implicit by (by-without-by) and will therefore be faster as well.

Have a look at under the new FRs implemented (points 1 and 2) for v1.9.3 here for details.

HTH

Answer 2

You could use the roll argument to fill the NA 's with nearest values. The problem is that will also fill the price , but that's easy to remedy:

setkey(dt, date, num)

dt[CJ(unique(date), unique(num)), roll = 'nearest'][!dt, price := NA][]
#    date price    brand num  seller
# 1: 2013  1.00     Audi   3     dan
# 2: 2013  2.30     Audi   5     dan
# 3: 2013    NA     Audi   6     dan
# 4: 2013    NA     Audi   7     dan
# 5: 2013    NA     Audi   8     dan
# 6: 2013  2.79     Audi   9     dan
# 7: 2013  3.00     Audi  12     dan
# 8: 2014  3.14 Mercedes   3 gregory
# 9: 2014    NA Mercedes   5 gregory
#10: 2014  1.45 Mercedes   6 gregory
#11: 2014  3.40 Mercedes   7 gregory
#12: 2014  5.10 Mercedes   8 gregory
#13: 2014    NA Mercedes   9 gregory
#14: 2014    NA Mercedes  12 gregory

I think this should be much faster than the .SD[...] solution.