[datatable-help] changing data.table by-without-by syntax to require a "by"

[Eduard Antonyan]

Arun, from my previous email:

"Take 'dt' and apply 'i' and return 'j' (for any 'i' and 'j') by 'b':
  dt[i, j, by = b] <-> dt[i][, j, by = b] in general, but also dt[i, j, by
= b] if 'i' is not a join, and can also be dt[i, j, by = b] if 'i' is a
join in some cases but not others

Take 'dt' and apply 'i' and return j, applying cross-apply/by-without-by
(will do cross-apply only when 'i' is a join):
  dt[i, j, each.i = TRUE] <-> dt[i, j]"

Together with the default being each.i=FALSE, you can see that the answer
to your question will be:

DT1[DT2, sum(y), each.i = FALSE, allow.cartesian = TRUE] <-> DT1[DT2,
allow.cartesian=TRUE][, sum(y)], i.e.
[1] 21

and
DT1[DT2, sum(y), each.i = TRUE, allow.cartesian = TRUE] <-> DT1[DT2,
sum(y), allow.cartesian=TRUE], i.e.
   x V1
1: 1  6
2: 2  9
3: 1  6



On Wed, May 1, 2013 at 5:23 PM, Arunkumar Srinivasan
<aragorn168b at gmail.com>wrote:

> eddi,
>
> sorry again, I am confused a bit now.
>
> DT1 <- data.table(x=c(1,1,1,2,2), y=1:5))
> DT2 <- data.table(x=c(1,2,1))
> setkey(DT1, "x")
>
> What's the intended result for `DT1[DT2, sum(y), allow.cartesian = TRUE,
> .join = FALSE]` ? c(6,9,6) or 21?
>
>
> Arun
>
> On Thursday, May 2, 2013 at 12:20 AM, Arunkumar Srinivasan wrote:
>
>  Sorry the proposed result was a wrong paste in the last message:
>
>     # proposed way and the result:
>     DT1[DT2, sum(y), .join = FALSE]
>     [1] 6 9 6
>
> And the last part that it *should* be a data.table is quite obvious then.
>
> Arun
>
> On Thursday, May 2, 2013 at 12:16 AM, Arunkumar Srinivasan wrote:
>
>  Eduard,
>
> Great. That explains me the difference between `drop` and `.join` here.
> Even though I don't *need* this feature (I can't recall the last time when
> I use a `data.table` for `i` and had to reduce the function, say, sum).
> But, I think it can only better the usage.
>
> However, there's one point *I think* would still disagree with @eddi here,
> not sure.
>
>     DT1 <- data.table(x=c(1,1,1,2,2), y=1:5)
>     DT2 <- data.table(x=c(1,2,1))
>     setkey(DT1, "x")
>
>     # proposed way and the result:
>     DT1[DT2, sum(y), .join = FALSE]
>     [1] 21
>
>
> So far nice. However, the operation `DT1[DT2, sum(y), .join = TRUE]`
> *should* result in a `data.table` output as follows (it's even more clearer
> now that .join is set to TRUE, meaning it's a data.table join):
>
>    x V1
> 1: 1  6
> 2: 2  9
> 3: 1  6
>
> Basically, `.join = TRUE` is the current functionality unchanged and nice
> to be default (as Matthew hinted).
>
> Arun
>
> On Tuesday, April 30, 2013 at 5:03 PM, Eduard Antonyan wrote:
>
> Arun,
>
> Yes, DT1[DT2, y, .JOIN = FALSE] would do the same as DT1[DT2][, y] does
> currently.
> No, DT1[DT2, y, .JOIN=FALSE], will NOT do a by-without-by, which is
> literally a 'by' by each of the rows of DT2 that are in the join (thus
> each.i! - the operation 'y' will be performed for each of the rows of 'i'
> and then combined and returned). There is no efficiency issue here that I
> can see, but Matthew can correct me on this. As far as I understand the
> efficiency comes into play when e.g. the rows of 'i' are unique, and after
> the join you'd like to do a 'by' by those, then DT1[DT2][, j, by =
> key(DT1)] would be less efficient since the 'by' could've already been done
> while joining.
>
> DT1[DT2, .JOIN=FALSE] would be equivalent to both current and future
> DT1[DT2] - in this expression there is no by-without-by happening in either
> case.
>
> The purpose of this is NOT for j just being a column or an expression that
> gets evaluated into a signal column. It applies to any j. The extra
> 'by-without-by' column is currently output independently of how many
> columns you output in your j-expression, the behavior is very similar as to
> when you specify a by=., except that the 'by' happens by a very special
> expression, that only exists when joining two data-tables and that
> generally doesn't exist before or after the join.
>
> Hope this answers your questions.
>
>
> On Tue, Apr 30, 2013 at 8:48 AM, Arunkumar Srinivasan <
> aragorn168b at gmail.com> wrote:
>
>  Eduard, thanks for your reply. But somethings are unclear to me still.
> I'll try to explain them below.
>
> First I prefer .JOIN (or cross.apply) just because `each.i` seems general
> (that it is applicable to *every* i operation, which as of now seems
> untrue). .JOIN is specific to data.table type for `i`.
>
> From what I understand from your reply, if (.JOIN = FALSE), then,
>
>     DT1[DT2, y, .JOIN = FALSE] <=> DT1[DT2][, y]
>
> Is this right? It's a bit confusing because I think you're okay with
> "by-without-by" and I got the impression from Sadao that he finds the
> syntax of "by-without-by" unaccessible/advanced for basic users. So, just
> to clarify, here the DT1[DT2, y, .JOIN=FALSE] will still do the
> "by-without-by" and then result in a "vector", right?
>
> Matthew explains in the current documentation that DT1[DT2][, y] would
> "join" all columns of DT1 and DT2 and then subset. I assume the
> implementation underneath is *not* DT1[DT2][, y] rather the result is an
> efficient equivalence. Then, that of course seems alright to me.
>
> If what I've told so far is right, then the syntax `DT1[DT2, .JOIN=FALSE]`
> doesn't make sense/has no purpose to me. At least I can't think of any at
> the moment.
>
> To conclude, IMHO, if the purpose of `.JOIN` is to provide the same as
> DT1[i, j] for DT1[DT2, j] (j being a column or an expression that results
> in getting evaluated as a scalar for every group in the current
> by-without-by syntax), then, I find this is covered in `drop = TRUE/FALSE`.
> Correct me if I am wrong. But, one could do: `DT1[DT2, j, drop=TRUE]`
> instead of `DT1[DT2, j, .JOIN=FALSE]` and DT1[i, j, drop=FALSE] instead of
> DT1[i, list(x,y)].
>
> If you/anyone believes it's wrong, I'd be all ears to clarify as to what's
> the purpose of `drop` then (and also how it *doesn't* suit here as compared
> to .JOIN).
>
> Arun
>
> On Tuesday, April 30, 2013 at 2:54 PM, Eduard Antonyan wrote:
>
> Arun,
>
> If the new boolean is false, the result would be the same as without it
> and would be equal to current behavior of d[i][, j]. If it's true, it will
> only have an effect if i is a join (I think each.i= fits slightly better
> for this description than .join=) - this will replicate current underlying
> behavior. If you think the cross-apply is something that could work not
> just for i being a data-table but other things as well, then it would make
> perfect sense to implement that action too when the bool is true.
>
> On Apr 30, 2013, at 2:58 AM, Arunkumar Srinivasan <aragorn168b at gmail.com>
> wrote:
>
> (The earlier message was too long and was rejected.)
> So, from the discussion so far, I see that Matthew is nice enough to
> implement `.JOIN` or `cross.apply`. I've a couple of questions. Suppose,
>
>     DT1 <- data.table(x=c(1,1,2,3,3), y=1:5, z=6:10)
>     setkey(DT1, "x")
>     DT2 <- data.table(x=1)
>     DT1[DT2, y, .JOIN=TRUE] # I guess the syntax is something like this. I
> expect here the same output as current DT1[DT2, y]
>
> The above syntax seems "okay". But my first question is what is
> `.JOIN=FALSE` supposed to do under these two circumstances? Suppose,
>
>     DT1 <- data.table(x=c(1,1,2,3,3), y=1:5, z=6:10)
>     setkey(DT1, "x")
>     DT2 <- data.table(x=c(1,2,1), w=c(11:13))
>     # what's the output supposed to be for?
>     DT1[DT2, y, .JOIN=FALSE]
>     DT1[DT2, .JOIN = FALSE]
>
> Depending on this I'd have to think about `drop = TRUE/FALSE`. Also, how
> does it work with `subset`?
>
>     DT1[x %in% c(1,2,1), y, .JOIN=TRUE] # .JOIN is ignored?
>  Is this supposed to also do a "cross-apply" on the logical subset? I
> guess not. So, .JOIN is an "extra" parameter that comes into play *only*
> when `i` is a `data.table`?
>
> I'd love to have some replies to these questions for me to take a stance
> on `.JOIN`. Thank you.
>
> Best,
> Arun.
>
>
>
>
>
>
>
>
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