I would like to join string in kdb but didn't work well. This is the data:
tab:([]service:`CS`CS.US`CS.US_ABC;y:1 2 3)
`CS 1
`CS.US 2
`CS.US_ABC 3
I would like to add :0 and :primary depending on the given parameter. 0 is working now
update service:`$(((string[service],'(":"))),'("C"$string 0)) from tab
If I would like the data to become
`CS:primary 1
`CS.US:primary 2
`CS.US_ABC:primary 3
and the primary is either string or symbol, how could I join?
I am parameterizing the 0 and primary.
Currently, 0 works as follows
update service:`$(((string[service],'(":"))),'( "0")) from tab
but "primary" is not working
update service:`$(((string[service],'(":"))),'( "primary")) from tab
If you want primary to be a parameter rather than a fixed string, the following will work (primary is "no" in this example):
q)update {`$string[y],\:":",x}[primary;]service from tab
service y
--------------
CS:no 1
CS.US:no 2
CS.US_ABC:no 3
If primary is a fixed string then you can place it inside the lambda in lieu of "x" and replace "y" with "x", yielding the following:
q)update {`$string[x],\:":","primary"}service from tab
service y
-------------------
CS:primary 1
CS.US:primary 2
CS.US_ABC:primary 3
Your query gives you a length error:
q)tab:([]service:`CS`CS.US`CS.US_ABC;y:1 2 3)
q)update service:`$(((string[service],'(":"))),'( "primary")) from tab
'length
[0] update service:`$(((string[service],'(":"))),'( "primary")) from tab
^
This happens because ,' (concatenate each) expects vectors of equal length on both sides, but gets a table column size (3) vector on the left and a character vector of length 7 on the right. Notice what happens when you pass 3 characters:
q)update service:`$(((string[service],'(":"))),'( "pri")) from tab
service y
-------------
CS:p 1
CS.US:r 2
CS.US_ABC:i 3
Each row gets a different suffix. What you want is to use ,\: (concatenate each-left):
q)update service:`$(((string[service],'(":"))),\:( "primary")) from tab
service y
-------------------
CS:primary 1
CS.US:primary 2
CS.US_ABC:primary 3
Why does it work for "0"? It works because "0" is not a vector but a character scalar
q)type "0"
-10h
q)type "primary"
10h
and with a scalar on the right, ,' works the same as .\::
q)"ab",'"0"
"a0"
"b0"
q)"ab",\:"0"
"a0"
"b0"
Finally, your query will run faster if you first prepend ":" to the suffix and then append the result to each service:
q)update service:`$(string[service],\:":","primary") from tab
service y
-------------------
CS:primary 1
CS.US:primary 2
CS.US_ABC:primary 3
q)update service:`$(((string[service],'(":"))),'(count[i]#enlist "primary"))
from tab
service y
-------------------
CS:primary 1
CS.US:primary 2
CS.US_ABC:primary 3
