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r - Add a row by reference at the end of a data.table object

In this question the data.table package creator explains why rows cannot be inserted (or removed) by reference in the middle a data.table yet. He also points out that such operations could be possible at end of the table. Could you show a code to perfome this action? It would be the "by reference" version of

a<- data.table(id=letters[1:2], var=1:2)
> a
   id var
1:  a   1
2:  b   2
> rbind(a, data.table(id="c", var=3))
   id var
1:  a   1
2:  b   2
3:  c   3

thanks.

EDIT:

since a proper solution is not possible yet, which of the following is better (if internally different, not sure) either from a speed and memory usage perpective?

rbind(a, data.table(id="c", var=3))

rbindlist(list(a,  data.table(id="c", var=3)))

are there eventually other (better) methods?

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To answer your edit, just run a benchmark:

a = data.table(id=letters[1:2], var=1:2)
b = copy(a)
c = copy(b) # let's also just try modifying same value in place
            # to see how well changing existing values does
microbenchmark(a <- rbind(a, data.table(id="c", var=3)),
               b <- rbindlist(list(b,  data.table(id="c", var=3))),
               c[1, var := 3L],
               set(c, 1L, 2L, 3L))
#Unit: microseconds
#                                                  expr     min        lq    median        uq      max neval
#          a <- rbind(a, data.table(id = "c", var = 3)) 865.460 1141.2585 1357.1230 1539.4300 6814.492   100
#b <- rbindlist(list(b, data.table(id = "c", var = 3))) 260.440  325.3835  445.4190  522.8825 1143.930   100
#                                   c[1, `:=`(var, 3L)] 482.147  626.5570  778.3135  904.3595 1109.539   100
#   ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?set(c, 1L, 2L, 3L) ? 2.339 ? ?5.677 ? ?7.5140 ? ?9.5170 ? 19.033 ? 100

rbindlist is clearly better than rbind. Thanks to Matthew Dowle pointing out the problems with using [ in a loop, I added another benchmark with set.

From the above your best options are using rbindlist, or sizing the data.table to begin with and then just populating the values (you can also use a similar strategy to std::vector in C++, and double the size every time you run out of space, if you don't know the size of the data to begin with, and then once you're done filling it in, delete the extra rows).


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