You are correct - the effect is the same as '{."0' which you can see if you
substitute this: the expression returns the same result. It's probably a
matter of how the expression was arrived at or how the coder is used to
working, though I would maintain that the rank-zero take is a clearer
expression of what is intended.
On Wed, May 13, 2009 at 10:10 PM, gary ng <[email protected]> wrote:
> On Wed, May 13, 2009 at 6:53 PM, bill lam <[email protected]> wrote:
>
> > On Wed, 13 May 2009, gary ng wrote:
> > > Are there any hidden 'gem' or reason to say '{.&>' rather than '({."0)'
> > as
> > > to me the only reason for the &> is to modify the rank of {. ?
> >
> > If the argument is un-boxed, why would you use {.&> ? What effect did
> > you expect from that > ?
> >
>
> That is what puzzled me. This is the original code I came across :
>
> (,|....@}:)(,.|....@}:"1)({.&>~#{.i:@#)(65}.a.)(>:@i.{.[)'E'
>
> from http://www.craigmurphy.com/blog/?p=1417#comment-66863
>
> It does save a pair of '()' but that seems to be not a good practice as it
> doesn't use the 'primary' functionality of '>' which is to unbox
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>
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
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