Why is it necessary to have 'g'
repeating the process on my machine which is obviously slower gives an
insignifcant time difference.
20 timer 's=:/:~x [ /:x'
0.172937
20 timer 's=:/:~x [ g=:/:x'
0.173672
Is it because storage is necessary and it is just as fast to store to a
named location rather than to some temporary storage?
Space differences seem small.
Don Kelly
On 05/03/2014 12:11 PM, Joe Bogner wrote:
Yes, the grade is done regardless. Here's my reasoning: In the
incumbent version, s is taken from the grade with from { which is
slower than just resorting.
It seems like the difference between sorting and selecting by indexes.
I am surprised if that's the answer though because selecting by index
should be fast since it's a contiguous array.
jtifrom looks more complicated than jtsortc but I haven't been able to
figure out how jtsortc works.
On Wed, Mar 5, 2014 at 2:50 PM, Roger Hui <rogerhui.can...@gmail.com> wrote:
x=: a.{~ 1e7 ?@$ 256
timer=: 6!:2
10 timer 's=:x{~g=:/:x'
0.136961
10 timer 's=:/:~x [ g=:/:x'
0.0765459
10 timer '/:~x'
0.012887
10 timer 'x{~g'
0.065751
10 timer '/:x'
0.0606987
On Wed, Mar 5, 2014 at 11:45 AM, Roger Hui <rogerhui.can...@gmail.com>wrote:
That's my alternative faster expression as well. But the more interesting
question is, why is it faster? Since we do the grade in both cases, the
comparison is between /:~x and g{x (or x{~g) with g pre-computed. The
answer does not depend knowledge specific to J.
On Wed, Mar 5, 2014 at 11:38 AM, Joe Bogner <joebog...@gmail.com> wrote:
Sorting and grading separately seems faster
timer=: 6!:2
x=:(1e7 $ 26?26) { 'abcdefghijklmnopqrstuvwxyz'
NB. incumbent
timer 's=: x{~g=: /:x'
0.0914002
NB. alternate
timer 'S=: /:~x[G=: /:x'
0.0668677
s-:S
1
G-:g
1
I am speculating that sorting does it in place? which is faster than
the selection from the grade
On Wed, Mar 5, 2014 at 2:02 PM, Raul Miller <rauldmil...@gmail.com>
wrote:
Hmm...
G=:a.i.S=:/:~x
is faster.
But while s-:S, g and G are different.
So I'm drawing a blank here, on how to make the grade.
Thanks,
--
Raul
On Wed, Mar 5, 2014 at 1:52 PM, Roger Hui <rogerhui.can...@gmail.com>
wrote:
Suppose x is a long vector of characters and you need both its sort
and its
grade. Can you do it faster than s=: x{~g=: /:x ?
Posed this way, the answer is of course yes. But how, and why is it
faster?
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