Thanks Thomas. I started working from your original answer and came up with:
groups=:[: (}."1 each
wrote:
> Sorry, I hit send too early. My complete answer is:
>
> t=. ;: ;._2 ( 0 : 0 )
>
> c1 p1 0.25
>
> c1 p2 0.35
>
> c2 p1 0.25
>
> c2 p2 0.35
>
> c3 p1 0.25
>
> c3 p2 0.35
>
> c3 p3 0.45
>
>
Sorry, I hit send too early. My complete answer is:
t=. ;: ;._2 ( 0 : 0 )
c1 p1 0.25
c1 p2 0.35
c2 p1 0.25
c2 p2 0.35
c3 p1 0.25
c3 p2 0.35
c3 p3 0.45
c4 p1 0.25
)
]key=. (}."1
wrote:
> You could key the first column on the values.
>
> (}."1 On Jul 25, 2014 7:07 AM, "Joe Bogner" wr
You could key the first column on the values.
(}."1 wrote:
> Given the following data:
>
> t =: ;: ;._2 (0 : 0)
> c1 p1 0.25
> c1 p2 0.35
> c2 p1 0.25
> c2 p2 0.35
> c3 p1 0.25
> c3 p2 0.35
> c3 p3 0.45
> )
>
>
> c1 has two rows (p1 0.25) and (p2 0.35)
> c2 has two rows (p1 0.25) and (p2 0.35)
>
Given the following data:
t =: ;: ;._2 (0 : 0)
c1 p1 0.25
c1 p2 0.35
c2 p1 0.25
c2 p2 0.35
c3 p1 0.25
c3 p2 0.35
c3 p3 0.45
)
c1 has two rows (p1 0.25) and (p2 0.35)
c2 has two rows (p1 0.25) and (p2 0.35)
c3 has three rows (p1 025), (p2 0.35), (p3 0.45)
How can I identify that c1 and c2 have t
