Good night!
Third day with q / kdb +.
I took some time after work to dedicate myself to my studies.
I am still at the beginning, but today I decided to study based on a challenge.
Simple challenge:
Given a list, add the even numbers and add the odd numbers.
Notice the (beginner’s) reasoning I used:
q) l:(1 4 6 9 3 5 2 7 8)
q) 5 mod 2
1
q) 4 mod 2
0
q)l mod 2
1 0 0 1 1 1 0 1 0
q)where l mod 2
0 3 4 5 7
q)where 0= l mod 2
1 2 6 8
q)l[where l mod 2]
1 9 3 5 7
q)l[where 0= l mod 2]
4 6 2 8
q)sum l[where l mod 2]
25
q)sum l[where 0= l mod 2]
20
Then, I thought as follows:
q)group l mod 2
1| 0 3 4 5 7
0| 1 2 6 8
q) l[group l mod 2]
1| 1 9 3 5 7
0| 4 6 2 8
q) sum each l[group l mod 2]
1| 25
0| 20
q)value sum each l[group l mod 2]
25 20
How to do it more optimally?
Best regards,
Geraldo
Not necessarily more optimal, but an introduction to boolean multiplication:
q)sum l*l mod 2
ajay1
October 15, 2020, 8:55am
3
wsum
q)l wsum l mod 2
for this specific case of x mod 2 on ints, i think this is cheaper
isOdd:{0<signum
Attila
October 15, 2020, 11:28am
5
nice
0<0W*
seems to be enough for positives
and
0<0W*abs@
for handling negatives too
ajay1
October 15, 2020, 11:30am
6
Agreed…
There is another way (can be faster if there is a better way to get least significant bit)
isOdd: (last vs[0b]@)’
ajay1
October 15, 2020, 12:46pm
7
Agreed…
There is another way (can be faster if there is a better way to get least significant bit)
isOdd: (last vs[0b]@)’
Attila
October 15, 2020, 8:22pm
8
SJT1
October 15, 2020, 8:25pm
9
You are trying to work your vector muscle; maybe an iterator as well?
{sum each x where each 1 not\x mod 2}til 9
or taking any of the isOdd alternatives
sum each x where each 1 not\isOdd x
1 f\x is the answer to: How many Zen monks does it take to change a light bulb? (Two: one to change it, and one not to change it.)
Stephen Taylor | Librarian | Kx | +44 7713 400852 | stephen@kx.com
Guys, good afternoon!
I’m speechless.
Thank you all so much for the tips and patience with my simple questions.
Thank you very much for the solutions presented.
I just got home and saw the amount of help I received.
It’s been a great learning experience for me.
I will take each solution and analyze each detail.
Terry Lynch, Ajay Rathore, Charles Skelton, Attila and Stephen Taylor. Thank you!
Best Regards.
Geraldo.
Em quinta-feira, 15 de outubro de 2020 às 04:48:28 UTC-3, TerryLynch escreveu:
Not necessarily more optimal, but an introduction to boolean multiplication:
q)sum l*l mod 2
Haha .. multiplying infinities.
Similarly to Ajay’s I tried
q)\t r1:{x mod 2} til 1000000
It’s interesting that you run into these omissions (like no fast binary logical operators) during these experiments but rarely during actual work.
One other idea:0W x} til 1000000
>> 1 f\x is the answer to: How many Zen monks does it take to change a light bulb? (Two: one to change it, and one not to change it.)
:)
Thank you!
I really liked all the solutions.
For those starting out, like me, this solution is the easiest to understand:
q){sum each x where each 1 not \ x mod 2} (2 3 5 8 9)
17 10
Best Regards
Geraldo
Em quinta-feira, 15 de outubro de 2020 às 16:25:21 UTC-3, ste...@kx.com escreveu:
You are trying to work your vector muscle; maybe an iterator as well?
{sum each x where each 1 not\x mod 2}til 9
or taking any of the isOdd alternatives
sum each x where each 1 not\isOdd x
1 f\x is the answer to: How many Zen monks does it take to change a light bulb? (Two: one to change it, and one not to change it.)
Stephen Taylor | Librarian | Kx | +44 7713 400852 | ste…@kx.com
ajay1
October 15, 2020, 9:46pm
13
mod should ideally be a primitive.