data.table 1.8.1.: "DT1 = DT2" is not the same as DT1 = copy(DT2)?
data.table, r
Solution
Yes. This is expected behaviour, and well documented.
Since `data.table` uses references to the original object to achieve modify-in-place, it is very fast.
For this reason, if you really want to copy the data, you need to use `copy(DT)`
From the documentation for `?copy`:
The data.table is modified by reference, and returned (invisibly) so it can be used in compound statements; e.g., `setkey(DT,a)[J("foo")]`. If you require a copy, take a copy first (using `DT2=copy(DT)`). `copy()` may also sometimes be useful before `:=` is used to subassign to a column by reference. See `?copy`.
See also this question : Understanding exactly when a data.table is a reference to vs a copy of another
Problem
I've noticed some inconsistent (inconsistent to me) behaviour in data.table when using different assignment operators. I have to admit I never quite got the difference between "=" and copy(), so maybe we can shed some light here. If you use "=" or "<-" instead of copy() below, upon changing the copied data.table, the original data.table will change as well. Please execute the following commands and you will see what I mean ``` library(data.table) example(data.table) DT x y v 1: a 1 42 2: a 3 42 3: a 6 42 4: b 1 4 5: b 3 5 6: b 6 6 7: c 1 7 8: c 3 8 9: c 6 9 DT2 = DT ``` now i'll change the v column of DT2: ``` DT2[ ,v:=3L] x y v 1: a 1 3 2: a 3 3 3: a 6 3 4: b 1 3 5: b 3 3 6: b 6 3 7: c 1 3 8: c 3 3 9: c 6 3 ``` but look what happened to DT: ``` DT x y v 1: a 1 3 2: a 3 3 3: a 6 3 4: b 1 3 5: b 3 3 6: b 6 3 7: c 1 3 8: c 3 3 9: c 6 3 ``` it changed as well. so: changing DT2 changed the original DT. not so if I use copy(): ``` example(data.table) # reset DT DT3 <- copy(DT) DT3[, v:= 3L] x y v 1: a 1 3 2: a 3 3 3: a 6 3 4: b 1 3 5: b 3 3 6: b 6 3 7: c 1 3 8: c 3 3 9: c 6 3 DT x y v 1: a 1 42 2: a 3 42 3: a 6 42 4: b 1 4 5: b 3 5 6: b 6 6 7: c 1 7 8: c 3 8 9: c 6 9 ``` is this behaviour expected?