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 <[email protected]>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 < > [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]> > 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. > > > > > > > >
_______________________________________________ datatable-help mailing list [email protected] https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help
