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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