A trick for coercing x and y to the same length in
x u y
is
x u/@,: y
Works if default fill is suitable.
Henry Rich
On 4/9/2011 6:35 AM, Steven Taylor wrote:
> Hi,
>
> as an exercise, I decided to make a simple "like" operator
>
> like=:([:*./([=])+.('?'E.]))
> NB. in english: get membership flags where '?' is in the set y
> NB. OR that with the comparison flags
> NB. *./ insert-AND to get the result
>
> the problem with this "like" operator is that x and y portions must be the
> same length. For example:
> 'abcaabc' like 'abc?abc'
> 1
> 'Abcaabc' like 'abc?abc'
> 0
> 'abcaabc' like 'abc?ab'
> |length error --- NB. fair enough that's how equals '=' is defined
>
> Is there an easy way to do guard conditionally, but to do this tacitly? i.e.
> the lengths are different, so return fail/zero?
> ideas so far:
> * email the list without fully exploring the possibilities ;)
> * look at J's regex library (did this a little)
> * implement something like "stretch=: [$],($,:@{:)" but with an unlikely
> non-ASCII character (hack so *bad*).
> * convert characters to numbers. move elements to a 2D array pre-filled
> with null whose right bound is the max of #x,#y (i.e.>/ (#x,#y). Continue
> on from there. create various verbs + flatten it afterwards to get tacit
> version. (not done yet).
>
> thanks,
> -Steven
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