Hello,
I’d like to count rows with like sided numbers in a row and then if there’s an opposite side reset the count.
q)t:(number:“I”$())
number streak
2 1
3 2
4 3
-2 1
-7 2
1 1
4 2
2 3
Thanks, All
q)select number,streak:{1+(x;0)y}[1;differ signum number] from tab
number streak
2 1
3 2
4 3
-2 1
-7 2
1 1
4 2
2 3
Cheers, Thanks Terry
Very nice!<o:p></o:p>
<o:p> </o:p>
Added to the qidioms page, where all good concise code deserves to be: http://code.kx.com/wiki/Qidioms#q_idioms<o:p></o:p>
<o:p> </o:p>
From: personal-kdbplus@googlegroups.com [mailto:personal-kdbplus@googlegroups.com] On Behalf Of Terry Lynch
Sent: Friday, August 14, 2015 11:52 AM
To: Kdb+ Personal Developers
Subject: Re: [personal kdb+] Count a streak of numbers with a reset?<o:p></o:p>
<o:p> </o:p>
q)select number,streak:{1+(x;0)y}[1;differ signum number] from tab<o:p></o:p>
number streak<o:p></o:p>
-------------<o:p></o:p>
2 1<o:p></o:p>
3 2<o:p></o:p>
4 3<o:p></o:p>
-2 1<o:p></o:p>
-7 2<o:p></o:p>
1 1<o:p></o:p>
4 2<o:p></o:p>
2 3<o:p></o:p>
<o:p> </o:p>
On Fri, Aug 14, 2015 at 2:00 PM, Roni Hoffman <hoffmanroni@gmail.com> wrote:<o:p></o:p>
Hello,<o:p></o:p>
<o:p> </o:p>
I’d like to count rows with like sided numbers in a row and then if there’s an opposite side reset the count.<o:p></o:p>
<o:p> </o:p>
q)t:(number:“I”$())<o:p></o:p>
<o:p> </o:p>
<o:p> </o:p>
number streak<o:p></o:p>
2 1<o:p></o:p>
3 2<o:p></o:p>
4 3<o:p></o:p>
-2 1<o:p></o:p>
-7 2<o:p></o:p>
1 1<o:p></o:p>
4 2<o:p></o:p>
2 3 <o:p></o:p>
<o:p> </o:p>
<o:p> </o:p>
Thanks, All<o:p></o:p>
<o:p> </o:p>
–
Submitted via Google Groups
Thanks David!
In fact, I think “differ x>0” achieves the same thing as “differ signum x” (with less work) so this would be slightly better
select number,streak:{1+(x;0)y}[1;differ number>0] from tab
Terry
<0a6f01d0d6c7$a60a6740$f21f35c0$@aquaq.co.uk>
To: Terry Lynch , Kdb+ Personal Developers
It depends on he wants to treat 0 - should that=E2=80=8E reset the count or have it continue the previous series or have it increase? Or does 0 not appear in the dataset?
|
From: Terry Lynch
Sent: Friday, August 14, 2015 13:00
To: Kdb+ Personal Developers
Reply To: personal-kdbplus@googlegroups.com
Subject: Re: FW: [personal kdb+] Count a streak of numbers with a reset?
|
Thanks David!
In fact, I think "differ x>0" achieves the same thing as "differ signum x" (with less work) so this would be slightly better
select number,streak:{1+(x;0)y}\[1;differ number>0] from tab
Terry
Another option, certainly not as elegant as Terry’s response, but figured it was worthwhile sharing for performance purposes:
(assumes no nulls, but can be easily modified to get around that)
q)\S 10
q)n:-5+(`int$1e7)?10
q)\ts {1+(x;0)y}[1;differ signum n]
6334 378104240
q)\ts raze sums each not null where[differ signum n] cut n
3796 474005520
Heres another one:
q)\ts a:{1+(x;0)y}[1;differ signum n]
3677 445213216
q)\ts b:{1+t - maxs (t:til count x)*differ signum x}n
181 536871520
q)a~b
1b
Jonny
Good logic.
Nice one Jonny, this is the optimal solution.
However, it’s great and all that but I’m afraid to say I’ve got you beat :)
q)\ts b:{1+t - maxs (t:til count x)*differ signum x}n
200 536871520
q)k)\ts c:{1+t-|(t:!#x)*~~':(x>0)-x<0}n
199 536871520
/Thread closed/
I bow to your superiority Terry!
Sent from my iPhone