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 > > > <[email protected] (mailto:[email protected])> 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 > > > > > <[email protected] (mailto:[email protected])> 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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