Re: how to find the first element for each row which satisfy the condition - KX Learning Hub

Laura

Administrator
May 16, 2022 at 12:00 am
```
q)u 
isMatched index id 
------------------------- 
1100b 0 1 2 3 a b c d 
1100b 1 0 2 3 a b c d 
0011b 0 1 2 3 a b c d 

q)fw:first where@ / composition 
q)select index:index@'j,id:id@'j from update j:fw each isMatched from u 
index id 
-------- 
0 a 
1 a 
2 c
```
Or you might prefer the same thing as vector operations:
```
q)u[`index`id]@':fw each u`isMatched 
0 1 2 
a a c 

q)ts:10000 select index:index@'j,id:id@'j from update j:fw each isMatched from u 
65 2832 

q)ts:10000 u[`index`id]@':fw each u`isMatched 
21 1376
```
In the vector expression, fw each u`isMatched corresponds to fw each isMatched from u. But instead of forming column j of a new temporary table, it is applied direct to the index and id columns of u. Notice how Each Left : takes Index At Each @' as its argument to derive binary function Index At Each Each Left.

Your sublists are short; if they were longer you might prefer faster ?'[;1b] to fw each:

q)ts:10000 select index:index@’j,id:id@’j from update j:?'[;1b] isMatched from u

61 3056

q)ts:10000 u[`index`id]@’:u[`isMatched]?’

1b

16 1568