A problem with 1 _ _1 -: i.b.0 is that the effective rank of i. is
actually never less than zero.
In other words: positive (and zero) ranks are taken literally.
Negative ranks are not taken literally (and always wind up being
non-negative in effect when used).
Thanks,
--
Raul
On Tue, Aug 8, 20
I don’t quite see what you’re pointing out. You must’ve
misunderstood me, or more probably I have misunderstood
you.
To clarify:
what I meant was that I was surprised because I expected
(-r) -: v”(-r)b.0
and therefor
(-r) -: u@(v”(-r))b.0
and by extension u to be applied to each result
of v ap
If negative rank passed through compounds, consider what the result of
]@(#"_1) i.2 3
would be.
Henry Rich
On Aug 8, 2017 00:49, "Raul Miller" wrote:
> Negative rank is a convention - it means the rank is relative to the
> noun rank. I'm not sure what a rank less than 0 would mean otherwise.
Negative rank is a convention - it means the rank is relative to the
noun rank. I'm not sure what a rank less than 0 would mean otherwise.
Thanks,
--
Raul
On Mon, Aug 7, 2017 at 7:15 PM, Louis de Forcrand wrote:
> I see. So negative ranks are sort of placeholders, and are replaced by
> positiv
I see. So negative ranks are sort of placeholders, and are replaced by
positive (effective) ranks internally during evaluation?
Because it could just announce its rank to be negative, and not actually
calculate the effective rank until it is really needed (a lazier effective rank
evaluation if you
The rank of +"_1 is infinite because the derived verb has to see the
full ranks of its arguments to figure out what rank to use for the
inner verb.
In other words, -"_1 in -"_1 i.3 3 has an effective rank of 1, but in
-"_ i.3 it has an effective rank of 0.
Since it can't know what rank to use unt
Reread the dictionary and I learned the correct interpretation of a negative
rank.
A negative rank is complementary: u"(-r) y is equivalent to u"(0>.(#$y)-r)"_
y .
I had the misconception that it should be u"(0>.(#$y)-r) y .
> On Aug 7, 2017, at 4:02 PM, Louis de Forcrand wrote:
>
> The s
The same observations go for all operators which depend on their argument’s
rank, such as @, &, or &. :
<@(,"_1)~ i.3
┌───┐
│0 0│
│1 1│
│2 2│
└───┘
Louis
> On 07 Aug 2017, at 16:55, Louis de Forcrand wrote:
>
> Yes I guess it could be replaced by
>
> u”_1”_1 _
>
> I would guess it is the i
Yes I guess it could be replaced by
u”_1”_1 _
I would guess it is the intended behavior as b.0 reports
infinite rank, but that is what I find strange.
Louis
> On 07 Aug 2017, at 16:47, Xiao-Yong Jin wrote:
>
> I remembered I had problem with this and ended up specifying both left and
> right
I remembered I had problem with this and ended up specifying both left and
right rank.
JVERSION
Engine: j806/j64avx/darwin
Beta-4: commercial/2017-06-27T12:55:06
Library: 8.06.03
Platform: Darwin 64
Installer: J806 install
InstallPath: /applications/j64-806
Contact: www.jsoftware.com
+"_1 _
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