What you need is actually the (?) vector conditional operator:
https://code.kx.com/q/ref/vector-conditional/ ? accepts a boolean list
as the first argument but $ accepts a boolean atom. Think of ? as a
boolean mask.
You can use the drop (_) operator, where n is a number to drop. n_list
To remove from the end of the list: -n_list neg[n]_list If you want to
remove nulls, you can do something like: list except 0N You can use the
drop with each (') operator to remov...
The range of generate for symbols is only from abcdefghijklmnop as
documented here: https://code.kx.com/q/ref/deal/#roll-and-deal As to why
is that so, I have no idea as well. But you can use something like this
to solve your problem: q){[num;len] ne...
You could do something like this: q)fix:{.Q.fmt'[x+1+count each string
floor y;x;y]} q)select time,sym,fix[1]price from trade time sym price
------------------------ 09:30:00.000 a "10.8" 09:31:00.000 a "11.8"
09:32:00.000 a "13.2" 09:30:00.000 b "10...
Yes, it is creating a table schema of specific type for the cols. So
when you are manipulating the table, it throws a type error if the type
is not of that defined in the schema.
https://code.kx.com/q/ref/insert/#type More information here:
https://c...
