Looking over the other responses here, I do not think I have anything
new to add. But perhaps another pass over the issues will help?
Here's your original sentence with its result:
(]j.1:)^:<10
10 10j1 10j2 10j3 10j4 10j5 10j6 10j7 10j8 10j9
Here is how you might get those same numbers using (]j.1:) without ^:
(] j.1:) 10
10j1
(] j.1:) 10j1
10j2
(] j.1:) 10j2
10j3
(] j.1:) 10j3
10j4
(] j.1:) 10j4
10j5
(] j.1:) 10j5
10j6
(] j.1:) 10j6
10j7
(] j.1:) 10j7
10j8
(] j.1:) 10j8
10j9
(You could also do that all on one line, but ^: suggests a different
way of going about doing things.)
The relevant dictionary page for (]j.1:)^:< is
http://www.jsoftware.com/help/dictionary/d202v.htm which says:
u ^: v y ↔ u^:( v y) y
And that's a reference to
http://www.jsoftware.com/help/dictionary/d202n.htm which says:
u^:(<m)
↔ u^:(i.m) y
There are, of course, other ways to get the same results.
Using the dictionary page, we'd get
(] j.1:)^:(i.10) 10
And, as you noted, (] j. 1:) gives the same results here as (j. 1:)
Thinking this through, the simplest approach might be:
10 j. i.10
But does any of this actually help you?
Thanks,
--
Raul
On Mon, Mar 7, 2016 at 9:09 PM, Linda A Alvord <[email protected]> wrote:
> Several days ago this expression was included. Could you please explain how
> it creates the result.
>
>
>
> (]j.1:)^:<10
>
> 0j1 0j2 0j3 0j4 0j5 0j6 0j7 0j8 0j9 0j10
>
>
>
> Thanks, Linda
>
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