Accumulators – Access additional list / column

[RVR]

Thanks rocuinneagain,

Works for most scenarios. I have added one more row at the end where I get a different result than expected. The value for c2 in last row should be 4 instead of 3.

t:([]c: 30 40 25 20 4 4 1; c1: 10 20 5 25 5 4 3)

 

c	c1	c2		c1>prev c2	OR	prev c < prev c2	?	c1	prev c2
30	10	10		10>0 (True)	OR	No need of evaluation	?	10
40	20	20		20>10(True)	OR	No need of evaluation	?	20
25	5	20		5>20(False)	OR	40<20(False)	?		20
20	25	25		25>20(True)	OR	No need of evaluation	?	25
4	5	5		5>25(False)	OR	20<25(True)	?	5
4	4	4		4>5(False)	OR	4<5(True)	?	4
1	3	4		3>4(False)	OR	4<4(False)	?		4

 

update c2:fills ?[(c1>prev c1) or c<prev c1;c1;0N] from t 
c c1 c2 
-------- 
30 10 10 
40 20 20 
25 5 20 
20 25 25 
4  5  5 
4  4  4 
1  3  3

 

 

[rocuinneagain]

Looking at your expected c2 this logic may be what you are looking for:

 

q)update c2:fills ?[(c1>prev c1) or c<prev c1;c1;0N] from t 
c c1 c2 
-------- 
30 10 10 
40 20 20 
25 5 20 
20 25 25 
4  5  5 
4  4  4

 

[rocuinneagain]

changing c to prev[c] looks to be what was missing

 

q)update c2:fills ?[(c1>prev c1) or prev[c]<prev c1;c1;0N] from t 
c c1 c2 
-------- 
30 10 10 
40 20 20 
25 5 20 
20 25 25 
4  5  5 
4  4  4 
1  3  4

 

[rocuinneagain]

Yes in that case an accumulator is needed. One method you could choose would be to pass a table through to accumulate while also allowing you to look back to previous rows:

q)update c2:1_@[;`c2]{y[`c2]:enlist $[(y[`c1][0]>last x[`c2]) or ((last x[`c])<last x[`c2]);y[`c1][0];last x`c2];x,y}/[enlist each {(1#0#x),x}update c2:0 from `c`c1#t] from t 
c c1 c2 
---------- 
30 10 10 
40 20 20 
25 5  20 
20 25 25 
4  5  5 
4  4  4 
4.5 3 4 
4.5 3.5 4

[RVR]

I was trying something like below. I am not sure if it works in all scenarios.

I will validate your solution as well and revert soon.

Thanks much for your patience. KX community support and response is just excellent

update c2: {?[((y>x)|(z<x));y;x]}[0;c1;0^prev c] from t 
c c1 c2 
---------- 
30 10 10 
40 20 20 
25 5  20 
20 25 25 
4  5  5 
4  4  4 
4.5 3 4 
4.5 3.5 4

 

 

[RVR]

Somehow not being able to use the previous calculated value of c2 directly is causing fallout for different combination of data. Last value of c2 will be 4 as per the logic but we get 3.5

t:([]c: 30 40 25 20 4 4 4.5 4.5; c1: 10 20 5 25 5 4 3 3.5)

 

c	c1	c2		c1>prev c2	OR	prev c < prev c2	?	c1	prev c2
30	10	10		10>0 (True)	OR	No need of evaluation	?	10
40	20	20		20>10(True)	OR	No need of evaluation	?	20
25	5	20		5>20(False)	OR	40<20(False)	?		20
20	25	25		25>20(True)	OR	No need of evaluation	?	25
4	5	5		5>25(False)	OR	20<25(True)	?	5
4	4	4		4>5(False)	OR	4<5(True)	?	4
4.5	3	4		3>4(False)	OR	4<4(False)	?		4
4.5	3.5	4		3.5>4(False)	OR	4.5<4(False)	?		4

 

update c2:fills ?[(c1>prev c1) or prev[c]<prev c1;c1;0N] from t 
c c1 c2 
----------- 
30 10 10 
40 20 20 
25 5  20 
20 25 25
4  5  5 
4  4  4 
4.5 3 4 
4.5 3.5 3.5

 

[rocuinneagain]

Yes that’s much cleaner – as you only ever compute one value c2 and only look back 1 step. this will do exactly as you need.

[binitafelicity]

From the Primary Accumulator list, select 0. From the First list, select Increment primary. From the Second list, select Use primary.Defines a custom accumulator operator. Accumulators are operators that maintain their state (e.g. totals, maximums, minimums,  click test   and related data) as documents . If you have access to Accumulator manager, you can access the Accumulator … To locate a record, you can sort each column alphabetically, …

[RVR]

Thanks for confirming. I checked both solutions and the result column matches exactly for my bigger data set.