Here’s another, hopefully simpler, example of my problem with @
]a=:?>:i.6
0 1 1 3 1 5
]a=:=a
1 0 0 0 0 0
0 1 1 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1
]b=:?>:i.6
0 1 0 3 1 3
a #@# b
1 3 1 1
There must be some way to get this result without @
Here are some of the things I have tried:
NB. x u@v y ↔ u x v y (vocabulary definition of @
#a#b
4
NB. ([: f g)"({. g b. 0) <=> f@g (someone on the forum provided this in
an earlier struggle I had with @ )
a(([:##)"({.# b. 0))b
4
Both of these definitions appear to describe @:
a #@:# b
4
Are changes ever made to the definitions in the vocabulary when they are
unclear or incorrect?
Linda
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