Thanks. I can see your expression is
|:@:>@(i.~L:0 ])
which gives the same x i. x as suggested by Roger.
Чт, 17 дек 2015, jprogramming написал(а):
> I have
>
>
>
> nubi =. ({~ each"_ 1 ~.@|:@:>@((i."_ _1~) leaf~)) f.
>
>
>
> keyi =. (i."_ _1~)@:(|:@:>@((i."_ _1~) leaf~))f.
>
>
>
> still gives same unique indexes, just not contiguous.
>
> (i."_ _1 ])@:(|:@:>@((i."_ _1 ]) leaf ])) i
> 0 1 2 0 4 2 6 2 8 6
>
>
>
> ----- Original Message -----
> From: bill lam <[email protected]>
> To: jprogramming <[email protected]>
> Sent: Thursday, December 17, 2015 10:21 AM
> Subject: [Jprogramming] nub and key of inverted table
>
> I want an effecient method of finding nub and key of a
> inverted table (rank-1 box array with equal number of
> items in each cells). eg.
>
> ]i=. (,.0 2 1 0 2 1 1 1 2 1);10 2$'aaaabaaaabbabbbababb'
> +-+--+
> |0|aa|
> |2|aa|
> |1|ba|
> |0|aa|
> |2|ab|
> |1|ba|
> |1|bb|
> |1|ba|
> |2|ba|
> |1|bb|
> +-+--+
>
> NB. nub of i
> ifa ~. a=. afi i
> +-+--+
> |0|aa|
> |2|aa|
> |1|ba|
> |2|ab|
> |1|bb|
> |2|ba|
> +-+--+
>
> NB. key of i
> a i.~ ~.a
> 0 1 2 0 3 2 4 2 5 4
>
> NB. definition of afi and ifa
> afi
> |:@:(<"_1@>)
> ifa
> <@(>"1)@|:
>
> Although nub and key can be found using afi/ifa, it involves a
> lot of box, I worry the time and space efficency when the table
> is large. Can anyone suggest other ways to do it.
>
> --
> regards,
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