r - data.table join and j-expression unexpected behavior

Question

In R 2.15.0 and data.table 1.8.9:

d = data.table(a = 1:5, value = 2:6, key = "a")

d[J(3), value]
#   a value
#   3     4

d[J(3)][, value]
#   4

I expected both to produce the same output (the 2nd one) and I believe they should.

In the interest of clearing up that this is not a J syntax issue, same expectation applies to the following (identical to the above) expressions:

t = data.table(a = 3, key = "a")
d[t, value]
d[t][, value]

I would expect both of the above to return the exact same output.

So let me rephrase the question - why is (data.table designed so that) the key column printed out automatically in d[t, value]?

Update (based on answers and comments below): Thanks @Arun et al., I understand the design-why now. The reason the above prints the key is because there is a hidden by present every time you do a data.table merge via the X[Y] syntax, and that by is by the key. The reason it's designed this way seems to be the following - since the by operation has to be performed when merging, one might as well take advantage of that and not do another by if you are going to do that by the key of the merge.

Now that said, I believe that's a syntax design flaw. The way I read data.table syntax d[i, j, by = b] is

take d, apply the i operation (be that subsetting or merging or whatnot), and then do the j expression "by" b

The by-without-by breaks this reading and introduces cases one has to think about specifically (am I merging on i, is by just the key of the merge, etc). I believe this should be the job of the data.table - the commendable effort to make data.table faster in one particular case of the merge, when the by is equal to the key, should be done in an alternative way (e.g. by checking internally if the by expression is actually the key of the merge).

Answer 1

Edit number Infinity: Faq 1.12 exactly answers your question: (Also useful/relevant is FAQ 1.13, not pasted here).

1.12 What is the difference between X[Y] and merge(X,Y)?
X[Y] is a join, looking up X's rows using Y (or Y's key if it has one) as an index. Y[X] is a join, looking up Y's rows using X (or X's key if it has one) as an index. merge(X,Y)1 does both ways at the same time. The number of rows of X[Y] and Y[X] usually dier; whereas the number of rows returned by merge(X,Y) and merge(Y,X) is the same. BUT that misses the main point. Most tasks require something to be done on the data after a join or merge. Why merge all the columns of data, only to use a small subset of them afterwards?
You may suggest merge(X[,ColsNeeded1],Y[,ColsNeeded2]), but that takes copies of the subsets of data, and it requires the programmer to work out which columns are needed. X[Y,j] in data.table does all that in one step for you. When you write X[Y,sum(foo*bar)], data.table automatically inspects the j expression to see which columns it uses. It will only subset those columns only; the others are ignored. Memory is only created for the columns the j uses, and Y columns enjoy standard R recycling rules within the context of each group. Let's say foo is in X, and bar is in Y (along with 20 other columns in Y). Isn't X[Y,sum(foo*bar)] quicker to program and quicker to run than a merge followed by a subset?

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Old answer which did nothing to answer the OP's question (from OP's comment), retained here because I believe it does).

When you give a value for j like d[, 4] or d[, value] in data.table, the j is evaluated as an expression. From the data.table FAQ 1.1 on accessing DT[, 5] (the very first FAQ) :

Because, by default, unlike a data.frame, the 2nd argument is an expression which is evaluated within the scope of DT. 5 evaluates to 5.

The first thing, therefore, to understand is, in your case:

d[, value] # produces a "vector"
# [1] 2 3 4 5 6

This is not different when the query for i is a basic indexing like:

d[3, value] # produces a vector of length 1
# [1] 4

However, this is different when i is by itself a data.table. From data.table introduction (page 6):

d[J(3)] # is equivalent to d[data.table(a = 3)]

Here, you are performing a join. If you just do d[J(3)] then you'd get all columns corresponding to that join. If you do,

d[J(3), value] # which is equivalent to d[J(3), list(value)]

Since you say this answer does nothing to answer your question, I'll point where the answer to your "rephrased" question, I believe, lies: ---> then you'd get just that column, but since you're performing a join, the key column will also be output'd (as it's a join between two tables based on the key column).

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Edit: Following your 2nd edit, If your question is why so?, then I'd reluctantly (or rather ignorantly) answer, Matthew Dowle designed so to differentiate between a data.table join-based-subset and a index-based-subsetting operation.

Your second syntax is equivalent to:

d[J(3)][, value] # is equivalent to:

dd <- d[J(3)]
dd[, value]

where, again, in dd[, value], j is evaluated as an expression and therefore you get a vector.

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To answer your 3rd modified question: for the 3rd time, it's because it is a JOIN between two data.tables based on the key column. If I join two data.tables, I'd expect a data.table

From data.table introduction, once again:

Passing a data.table into a data.table subset is analogous to A[B] syntax in base R where A is a matrix and B is a 2-column matrix. In fact, the A[B] syntax in base R inspired the data.table package.