Accumulators - Access additional list / column

[RV]

I have a table created like below.

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

c2 is calculated Column (value from c1 or prev c2 is used based on evaluation).

prev value is taken as 0 for first row / if not available

Below are the calculations.

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

What I have till now

update c2: 0{?[y>x;y;0N]}\c1 from t

How can I OR the condition prev c < prev c2 along with this? Can column c be accessed? Something like below.

update c2: 0{?[y>x|0^c <x;y;x]}\c1 from t   

[rocuinneagain · Accepted solution]

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

[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

[RV]

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]

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

[RV]

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 · Accepted solution]

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

[RV · Accepted solution]

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

[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.

[RV]

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

[binitafelicity]

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