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    <p>I am not sure if I understand the problem well, and I also just started to read the docs of <em>data.table</em> library, but I think if you would like to get the columns of <em>y</em> and also do something to those by the columns of <em>a</em>, you might try something like:</p> <pre><code>&gt; x[y,a*y] foo boo [1,] 5 50 [2,] 8 44 [3,] 9 36 [4,] 8 26 [5,] 5 14 </code></pre> <p>Here, you get back the columns of <em>y</em> multiplied by the <em>a</em> column of <em>x</em>. If you want to get <em>x</em>'s <em>foo</em> multiplied by <em>y</em>'s <em>boo</em>, try:</p> <pre><code>&gt; y[,x*boo] foo a [1,] 10 50 [2,] 22 44 [3,] 36 36 [4,] 52 26 [5,] 70 14 </code></pre> <hr> <p><strong>After editing:</strong> thank you @Prasad Chalasani making the question clearer for me.</p> <p>If simple merging is preferred, then the following should work. I made up a more complex data to see the actions deeper:</p> <pre><code>x &lt;- data.table( foo = 1:5, a=20:24, zoo = 5:1 ) y &lt;- data.table( foo = 1:5, b=30:34, boo = 10:14) setkey(x, foo) setkey(y, foo) </code></pre> <p>So only an extra column was added to each data.table. Let us see <code>merge</code> and doing it with <code>data.tables</code>:</p> <pre><code>&gt; system.time(merge(x,y)) user system elapsed 0.027 0.000 0.023 &gt; system.time(x[,list(y,x)]) user system elapsed 0.003 0.000 0.006 </code></pre> <p>From which the latter looks a lot faster. The results are not identical though, but can be used in the same way (with an extra column of the latter run):</p> <pre><code>&gt; merge(x,y) foo a zoo b boo [1,] 1 20 5 30 10 [2,] 2 21 4 31 11 [3,] 3 22 3 32 12 [4,] 4 23 2 33 13 [5,] 5 24 1 34 14 &gt; x[,list(x,y)] foo a zoo foo.1 b boo [1,] 1 20 5 1 30 10 [2,] 2 21 4 2 31 11 [3,] 3 22 3 3 32 12 [4,] 4 23 2 4 33 13 [5,] 5 24 1 5 34 14 </code></pre> <p>So to get <code>xy</code> we might use: <code>xy &lt;- x[,list(x,y)]</code>. To compute a one-column data.table from <code>xy$foo * xy$boo</code>, the following might work:</p> <pre><code>&gt; xy[,foo*boo] [1] 10 22 36 52 70 </code></pre> <p>Well, the result is not a data.table but a vector instead.</p> <hr> <p><strong>Update (29/03/2012):</strong> thanks for @David for pointing my attention to the fact that <code>merge.data.table</code> were used in the above examples.</p>
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    1. POR: When using data.table how do I get columns of y when I do x[y]?
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    1. COReferring to the example in my question, I want to do a join of `x` and `y`, let's call it `xy`, and then create a single-column data-frame that is equal to `xy$foo * xy$boo`.
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    2. CO@Prasad Chalasani: I edited my answer, I hope you can find something new and valuable in it.
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    3. COthanks for the details, but my question was about why the specific syntax I describe in my question is not working, contrary to what it says in the FAQ. I know that I can do it in two stages (merge, then operate on columns), but I want the `x[y, <expr>]` syntax to work *in one step* -- i.e. do the join and operate on the `x` and `y` columns in one step. This is syntactically less tedious, and possibly faster (if implemented right internally). I am dealing with 10-million row data-frames, so I'm not concerned with the timings of the small toy example above.
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