I am ignoring the 'linked' in 'linked list'.
I think you are saying that f has natural rank 1 1: one row of x and one
row of y.
Define
f =: 4 : 0
And then use
x f"1 1/ y
/ is equivalent to ("(lr,_))
so this is
x f"1 1"1 _ y
If I were you I would take whatever time it takes to see how that
sentence operates.
After you understand it, you could go back and write
f =: 4 : 0"1 1 ( or f =: {{ ... }}"1 1 )
and
x f/ y
Henry Rich
On 2/22/2021 4:46 PM, Emir U wrote:
I'm trying to table a function run with various parameters against data. The
function is in the form: x f y, where x are P parameters in a linked list, and
y is the data which has N rows. f has to be explicitly defined because its
complicated.
Say X is Q different sets of parameters. I want to do:
X f/ y
and get a Q x N table where each cell represents the application of one
parameter set to one data row.
Even if a use:
f=: 4 : 0 '0'
which definitely returns a scalar, the table verb doesn't seem to think its
rank 0 and won't behave as I'd like it to.
I'd be grateful for some help on this.
Emir
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