You could have generalised your original 'c' using 'rank' as follows:
c=: 4 : '(!(x+y)) % (x+y)' NB. As you had below ...
a=:1 2 3 4 5
b=:1 2 3
a c"0 _ b NB. Apply c to each 'scalar item' of
a, against 'whole' of b
1 2 6
2 6 24
6 24 120
24 120 720
120 720 5040
On 07/08/2008, at 5:07 AM, Ian Gorse wrote:
...
If I was to extend it slightly, I have this
c=: 4 : '(!(x+y)) % (x+y)'
a=: 1 2 3 4 5
and the result I want are
a c 1
1 2 6 24 120
a c 2
2 6 24 120 720
a c 3
6 24 120 720 5040
...
c2=: 4 : '(!(a+/b)) % (a+/b)'
NB. Third Approach
a c2 b
1 2 6
2 6 24
6 24 120
24 120 720
120 720 5040
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