Ok,
'a b c' =: 1 2 3 NB. multiple assignments. this is easy
NB. therefore 'age spec astig tear'=:
NB. is easy for me to understand
NB. but the expression i.&.>3 2 2 2
NB. has me confused.
I read up on &. and we are dealing with the monadic case...
so:
u&.v y IS THE SAME AS vi u v y
which basically means 3 verbs get applied in sequence to y
so:
i.&.>y IS THE SAME AS vi i. > y
where vi is the obverse of >
Is there a programmatic way to get the obverse?
I'm guessing it is < NB. box
And why is the obverse applied to 0-cells of y?
So I think it actually is not to hard to figure out what this is doing from a
holistic standpoint, but
(< i. >) 3 2 2 2
is not returning the same thing as
(i.&.>)3 2 2 2
which means I am making a mistake somewhere...
Help appreciated.
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