Accessing previous calculated value from a column in current row

[jason_fealy]

https://code.kx.com/q/ref/accumulators/#binary-values

q)select c1,c2:0{(x+10+y)%2}c1 from t 
c1 c2 
----------- 
1 5.5 
2 8.75 
3 10.875 
4 12.4375 
5 13.71875 
6 14.85938 
7 15.92969 
8 16.96484 
9 17.98242

[vkennedy]

Hello,

To modify the table in place, pass it by name (i.e. `t).

9. Queries q-sql – Q for Mortals (kx.com)

q)t:([]c1: 1 2 3 4 5 6 7 8 9) 
q)update c2:((prev(c1)+10) + c1)%2 from `t `t 
q)update c2:((prev(c2)+10) + c1)%2 from `t `t 
q)t 
c1 c2 
-------- 
1 2 
3 9.75 
4 10.75 
5 11.75 
6 12.75 
7 13.75 
8 14.75 
9 15.75

 

[RVR]

Thanks for the reply vkennedy.

Actually I don’t want to update the existing table at all. Sorry for confusing with update keyword. (But I learnt something new here Thanks).

Please consider below case. The select fails here.

t:([]c1: 1 2 3 4 5 6 7 8 9)select c1, c2:((prev(c2)+10) + c1)%2 from t

How can I use the previous calculated value of c2 in the current row calculation? For the first row in the table there is no prev(c2); so that could be considered as 0. From the second row, it should consider the previous calculated value of c2.

 

[vkennedy]

Hi,

Is this the result you are looking for?

q)select c2:(0^(prev(c2)+10) + c1)%2 from update c2:((0^prev(c1)+10) + c1)%2 from t 
c2 
----- 
0 
6.25 
9.75 
10.75 
11.75 
12.75 
13.75 
14.75 
15.75

In this case, you need to explicitly fill in the zero as the result of prev on the first item of a list is null.

Appendix A. Built-in Functions – Q for Mortals (kx.com)

[RVR]

Hi,

I have added the calculations below.

c1	c2	Calculation
1	5.5	=((0+10) + 1)/2	=((prev(c2)+10)+c1)/2
2	8.75	=((5.5+10) + 2)/2
3	10.875	=((8.75+10) + 3)/2
4	12.4375	=((10.875+10) + 4)/2
5	13.71875	=((12.4375+10) + 5)/2
6	14.859375	=((13.71875+10) + 6)/2
7	15.9296875	=((14.859375+10) + 7)/2
8	16.96484375	=((15.9296875+10) + 8)/2

c2 column values is the expected result.

[RVR]

Wow!! Thanks Jason & vkennedy.