I reran the tests on a 64 bit system with 8 gigabytes of memory and verified
that paging did not occur in this environment. The results confirm that 32 bit
paging was the culprit in my previous results.
a =: ? 100000000 $ 100
(100) 6!:2 'b=.+/ a'
0.259717
(100) 6!:2 'b=.(+/ % #) a'
0.207758
(100) 6!:2 'b=.(+/) a'
0.259719
--
David Mitchell
Roger Hui wrote:
> The size of a is such that paging probably came into play,
> whence you can throw the timings out the window.
> (The timing difference between +/a and (+/)a really
> demonstrated that; there shouldn't be any difference
> betwween the two.) Moreover, unless you really are
> doing integers, floating point numbers give a more
> meaningful answer. Thus:
>
> x=: 1e7 ?...@$ 100
> 100 ts '+/x'
> 0.0297878 1024
> 100 ts '(+/%#)x'
> 0.0166005 1216
> 100 ts '(+/)x'
> 0.0298247 1024
>
> - No difference between +/x and (+/)x
> - +/x has some extra code to handle integer overflow;
> (+/ %#) x "knows" that the answer is floating point
> and doesn't have the integer overflow code.
>
> x=: 1e7 ?...@$ 0
> 100 ts '+/x'
> 0.0192155 1024
> 100 ts '(+/%#)x'
> 0.0191934 1216
> 100 ts '(+/)x'
> 0.0202687 1024
>
> No difference among the three on a floating point argument.
>
>
>
> ----- Original Message -----
> From: David Mitchell <[email protected]>
> Date: Wednesday, May 20, 2009 8:31
> Subject: Re: [Jprogramming] +/ slower than (+/ % #)/
> To: Programming forum <[email protected]>
>
>> bill lam wrote:
>>> On Wed, 20 May 2009, Steven Taylor wrote:
>>>> 6!:2 'b=.+/ a'
>>>> 0.240794
>>>> 6!:2 '(+/ % #) a'
>>>> 0.13628
>>> I think that 'a' is already cached when executed the tacit.
>> You can
>>> compare
>>> (100) 6!:2 'b=.+/ a'
>>> (100) 6!:2 '(+/ % #) a'
>>>
>> Some fixed tests:
>>
>> a =: ? 100000000 $ 100
>> (100) 6!:2 'b=.+/ a'
>> 0.441664432211
>> (100) 6!:2 'b=.(+/ % #) a'
>> 0.24710853195
>> (100) 6!:2 'b=.(+/) a'
>> 0.375218973361
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