Partitioning

[Skip Cave]

What verb 's' can do the following partitioning?

Skip Cave
Cave Consulting LLC

[Elijah Stone]

I suppose something like this could work

n=. i.10
t=. 2 5
(t,_) {.&.> (0,+/\t) <@}."0 _ n
┌───┬─────────┬─────┐
│0 1│2 3 4 5 6│7 8 9│
└───┴─────────┴─────┘

or, maybe nicer:

n <;.1~ 1 (0,+/\t)} 0"_1 n
┌───┬─────────┬─────┐
│0 1│2 3 4 5 6│7 8 9│
└───┴─────────┴─────┘

On Sun, 28 Apr 2024, 'Skip Cave' via forum wrote:

> What verb 's' can do the following partitioning?
>
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> Skip Cave
> Cave Consulting LLC
>
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[Alex Shroyer]

Another approach:
   2 5 {{ (I.x,(#y)-+/x)</.y }} i.10
┌───┬─────────┬─────┐

│0 1│2 3 4 5 6│7 8 9│
└───┴─────────┴─────┘

[Raul Miller]

It's probably worth noting that this problem statement is ambiguous.

I think it's clear that we want to treat the left argument as the
running sum +/\0,x but the significance of the right argument is
unclear to me.

I think it's safe to assume that the right argument is small (fewer
than a million elements and we're not overly concerned about time).
But the the value 2 happens to match the position 2 and I am not
completely sure which was intended. Nor am I certain whether the right
argument is intended to be a sorted sequence or a set of unique values
or something else.

Meanwhile, J provides several different partitioning facilities. We
have (;.) which partitions sequences, and there's (/.) which
partitions sets (and we can "roll our own" using structural
operations).

So... anyways, I'll propose two different 's' verbs - spos and sval:

NB. partition by position in y
spos=: {{ (1 ((#~ (#y)&>) +/\0,x)}(#y)#0) <;.1 y }}
spos=: {{ ((i.#y) e. +/\0, x) <;.1 y }}

NB. partition by value in y
sval=: {{ (+/(+/\x) </ y) </. y }}

And, of course, there are other approaches...

FYI,

--
Raul

On Sun, Apr 28, 2024 at 6:10 PM 'Skip Cave' via forum
<fo...@jsoftware.com> wrote:
>
> What verb 's' can do the following partitioning?
>
>
>
>

> Skip Cave
> Cave Consulting LLC
>
>

[LdBeth]

My take on it:

c=.(#@]~:@{.[#>:@i.@#@[) <;.1 ]

2 5 c i. 10


A break down:

a=.[#>:@i.@#@[
b=.#@]~:@{.a
c=.b <;.1 ]

2 5 a i. 10
1 1 2 2 2 2 2

2 5 b i. 10
1 0 1 0 0 0 0 1 0 0

LdBeth


>>>>> In <CAJ8Lg_f+661eA5OxhXHr5CbR...@mail.gmail.com>

>>>>> "'Skip Cave' via forum" <fo...@jsoftware.com> wrote:

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> What verb 's' can do the following partitioning?

[Skip Cave]

Alex had the right idea!

pv=:{{(I.x,(#y)-+/x)</.y}}

]n=.?15#20

9 12 3 14 3 13 18 8 12 17 17 14 19 9 18

Skip Cave
Cave Consulting LLC