i, := and Cartesian product (another allow.cartesian case)

[matthieugomez]

DT=data.table(
  id = c(1, 1),
  date = c(1992, 1991),
  value = c(4.1, 4.5),
  key = "id"
)

DT[DT, a := 1]
"Error in vecseq(f__, len__, if (allow.cartesian || notjoin) NULL else as.integer(max(nrow(x),  : 
  Join results in 4 rows; more than 2 = max(nrow(x),nrow(i)). Check for duplicate key values in i, each of which join to the same group in x over and over again. If that's ok, try including `j` and dropping `by` (by-without-by) so that j runs for each group to avoid the large allocation. If you are sure you wish to proceed, rerun with allow.cartesian=TRUE. Otherwise, please search for this error message in the FAQ, Wiki, Stack Overflow and datatable-help for advice."

DT[DT, a := 1, allow.cartesian = TRUE]

does not throw any error even though it does not do a cartesian product.

Update: Also, I'm not clear on what happens when multiple rows in i corresponds to a row in X, as in

DT1 <- unique(DT)
DT1[DT, a := i.value]

It seems to match to the last row in i in this example. Is it documented somewhere?

[eantonya]

This is because DT[DT] has 4 rows, but this specific error is pretty annoying and I bump against it all the time. The allow.cartesian error should not be there when doing an assignment, as the result is never larger than DT. Just a warning should be enough.

Changing the label to feature request.

[matthieugomez]

Great. Could you answer my second point too (you may not have seen it as I added it later on)

[eantonya]

I have no idea what's going on there :)

My best guess is that since you try to assign to a single line the value c(1,2), it first assigns a=1, then loops and assigns a=2, so you're left with the second value at the end.

[arunsrinivasan]

One way to comprehend what's going on is to realise that, even for x[i] where i is a data.table, under-the-hood, the matching row numbers of x is first obtained using binary search. And then a subset using those row numbers is returned.

So when you do:

DT1 = unique(DT)
DT1[DT, a := 1L]

internally, it first computes the matching rows in x for i, which can also be seen using:

DT1[DT, which=TRUE]
# [1] 1 1

That would transform your earlier code to:

DT1[c(1L,1L), a := 1L]

which is quite straightforward to understand as to what's happening, and is correct behaviour. Take for example:

x <- 1:5
x[c(1L,1L)] <- 10L

If you'd instead done:

DT1[DT, a := c(2,4)]
#    id date value a
# 1:  1 1992   4.1 4

and is correct once again.

However, on your first point, even though i has duplicate values (and therefore the error message is right), the fact that the end result will never be greater than nrow(x) warrants for that error message to be a bug.

I'll add this to the other allow.cartesian issues, set for fixing in the next release.

[matthieugomez]

Thanks. Basically you're saying that it is due to the fact that, in base R,

 l=seq(1,10)
 l(c(1,1))=c(3,4)

modifies the first element by 4, not 3, right? I did not know about that. Thanks!

[rsaporta]

The error is coming from the call to vecseq at line 520 in data.table.r
Namely:

    if (is.data.table(i)) {
        .
        .
        if (mult=="all") {
            if (!byjoin) {
                irows = if (allLen1) f__ else vecseq(f__,len__,if(allow.cartesian || notjoin) NULL else as.integer(max(nrow(x),nrow(i))))
                .
                .