select literal string from a table

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1

How do I do this:

select colA,enlist “” from tableA

or

select colA,enlist “abc” from tableA

I get a length error in both cases. 

2

this works, but there is probably a shorter way of defining d :

q)select x,b:3,c:`abc,d:(count t)#enlist"ab" from t:(x:1 2 3)
x b c d

1 3 abc “ab”
2 3 abc “ab”
3 3 abc “ab”

ta, jack

3

A bit more idiomatic to use count[i], but otherwise I dont know of anything shorter.

Cheers,
? Attila

4

very nice thankyou

5

much nicer - but there is a speed difference (at least on my ancient box):

q)\t t:(x:til 1000000);do[100;(d:enlist"abc") cross t]
1393
q)\t t:(x:til 1000000);do[100;select x,d:(count t)#enlist “abc” from t]
780

but i don’t understand why i get different counts here:

q)count select x,d:(count t)#enlist “abc” from t
1000000
q)count select d:(count t)#enlist “abc” from t
1

6

q checks only the first expression in the select to derive shape for the result.

q)count select d:count[i]#1 from(til 10)

1
q)count select d:count[i]#1,x from([]til 10)
1
q)count select x,d:count[i]#1 from([]til 10)
10

maybe you want to use update for such a scenario anyway (it can derive the shape from the target)

q)count update d:count[i]#1 from (til 10)

10

7

Another solution, takes around the same time 

q)\t t:(x:til 1000000);do[100;select x,b:(count t)#enlist “abc” from t]
1658
q)\t t:(x:til 1000000);do[100;t^(b:count[t]#enlist"abc")]
1625

8

q)\t count[t]#enlist"abc"

dominates

,’ is less work than ^

t,'(b:count[t]#enlist"abc")