I would write the original as:
z ([ #~ 0 1 0 -:"1 {) 1 0 1 0 1 0
3 4 5
Then to reverse the arguments you could do one of the following:
1 0 1 0 1 0 ([ #~ 0 1 0 -:"1 {)~ z
3 4 5
1 0 1 0 1 0 (] #~ 0 1 0 -:"1 {~) z
3 4 5
On Sun, Jun 4, 2017 at 1:45 PM, Michael Rice <[email protected]> wrote:
> z =: 2 3 $ i.6
> z
> 0 1 2
> 3 4 5
> z { 1 0 1 0 1 0
> 1 0 1
> 0 1 0
> (-: & 0 1 0) z { 1 0 1 0 1 0
> 0
> (-: & 0 1 0)"1 z { 1 0 1 0 1 0
> 0 1
> ((-: & 0 1 0)"1 z { 1 0 1 0 1 0) # z
> 3 4 5
> z #~ ((-: & 0 1 0)"1 z { 1 0 1 0 1 0)
> 3 4 5
> z #~ ((-: & 0 1 0)"1 (1 0 1 0 1 0 {~ z))
> 3 4 5
> z #~ ((-: & 0 1 0)"1 ({~ & z) 1 0 1 0 1 0)
> 3 4 5
> z (#~ ((-: & 0 1 0)"1 @: ({~ & z))) 1 0 1 0 1 0
> 3 4 5
>
> Now write the function so the left paramenter is 1 0 1 0 1 0
> and the right parameter is z
>
> ((-: & 0 1 0)"1 @: ({~ & z) 1 0 1 0 1 0) # z
> 3 4 5
>
> 1 0 1 0 1 0 (# ((-: & 0 1 0)"1 @: ({~ & z))) z
> |index error
> | 1 0 1 0 1 0 (#((-:&0 1 0)"1@:({~&z)))z
>
> (1 0 1 0 1 0) (# ((-: & 0 1 0)"1 @: ({~ & z))) z
> |index error
> | (1 0 1 0 1 0) (#((-:&0 1 0)"1@:({~&z)))z
>
> 1 0 1 0 1 0 (# (({~ & z) @:~ (-: & 0 1 0)"1))) z
> |syntax error
> | 1 0 1 0 1 0(#( ({~&z)@:~(-:&0 1 0)"1)))z
>
> (1 0 1 0 1 0) (# (({~ & z) @:~ (-: & 0 1 0)"1))) z
> |syntax error
> | (1 0 1 0 1 0)(#( ({~&z)@:~(-:&0 1 0)"1)))z
>
> How would it be written?
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