It's not surprising at all if you know a little bit about how the verbs
in question work. Dyad (,) creates a new array and copies its arguments
into it. Thus computing (a,b,c) requires the steps:
- Copy b and c into new array t1 (2e6 copy operations)
- Copy a and t1 into new array t2 (3e6 copy operations).
In contrast, dyad (;) is almost free, since it just creates an array of
pointers to its arguments. Monad (;) computes the length of its result,
allocates an array of that length, and then copies everything in, for a
total of 3e6 copies. Provided we can't resize a to hold all the values
(which would allow us to only copy b and c), this is optimal with J's
array layout.
The observed ratio of measurements for the two operations, in both space
and time, is pretty close to (3%5) as we would expect from the above.
Marshall
On Sun, Jan 31, 2016 at 08:35:01PM -0500, Jose Mario Quintana wrote:
> That is interesting, I was independently testing the same expression ;A ;
> B ; C vs A , B , C with different sample nouns and it seems leaner and
> meaner:
>
> st=. (, */&.:>@:(1 2&{))@:(] ; 7!:2@:] ; 6!:2)
>
> 'A B C'=. i.3 1000000
>
> 111 st&>'A , B , C' ; ';A ; B ; C'
> ┌──────────┬────────┬─────────┬─────────┐
> │A , B , C │50333184│0.0335228│1.68731e6│
> ├──────────┼────────┼─────────┼─────────┤
> │;A ; B ; C│33556608│0.0198092│664728 │
> └──────────┴────────┴─────────┴─────────┘
>
> (A , B , C) -: i.3000000
> 1
> (;A ; B ; C) -: i.3000000
> 1
>
> Using your sample nouns it is not as dominant but remains dominant
> nevertheless:
>
> A=:?~1000000
> B=:0.1+?~1000000
> C=:?~1000000
>
> 111 st&>'A , B , C' ; ';A ; B ; C'
> ┌──────────┬────────┬─────────┬─────────┐
> │A , B , C │67110400│0.0496758│3.33377e6│
> ├──────────┼────────┼─────────┼─────────┤
> │;A ; B ; C│50333824│0.0360027│1.81216e6│
> └──────────┴────────┴─────────┴─────────┘
>
> (A , B , C) -: (;A ; B ; C)
> 1
>
>
> JVERSION
> Installer: j602a_win.exe
> Engine: j803/2014-10-19-11:11:11
> Library: 6.02.023
>
>
>
> On Sun, Jan 31, 2016 at 7:05 PM, Raul Miller <[email protected]> wrote:
>
> > Oops, I meant:
> >
> > timespacex ';a;b;c'
> > 0.012332 5.03338e7
> > timespacex ';a;b;c'
> > 0.011768 5.03338e7
> > (;a;b;c)-:a,b,c
> > 1
> >
> > Efficiency characteristics are the same, result (the most important
> > part) is different.
> >
> > --
> > Raul
> >
> > On Sun, Jan 31, 2016 at 7:04 PM, Raul Miller <[email protected]>
> > wrote:
> > > If I define:
> > > a=:?~1000000
> > > b=:0.1+?~1000000
> > > c=:?~1000000
> > >
> > > I get:
> > > timespacex 'a,b,c'
> > > 0.016585 6.71104e7
> > > timespacex '>a;b;c'
> > > 0.012863 5.0334e7
> > > timespacex '>a;b;c'
> > > 0.011867 5.0334e7
> > > timespacex 'a,b,c'
> > > 0.015703 6.71104e7
> > >
> > > So it looks like >a;b;c is slightly more efficient than a,b,c, but
> > > it's nowhere close to a factor of 2, so I think I'd ignore this issue
> > > in most contexts.
> > >
> > > --
> > > Raul
> > >
> > >
> > > On Sun, Jan 31, 2016 at 5:21 PM, Henry Rich <[email protected]>
> > wrote:
> > >> You have 3 large lists a, b, c (1000000 atoms each). You want to join
> > them
> > >> into one long list. What is the best way to do this?
> > >>
> > >> Henry Rich
> > >> ----------------------------------------------------------------------
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> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> >
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