The perils of benchmarking
Joel Reymont
phone numbers and wanted to take
just the first 8 digits of each.
We could create the list like this:
q)l:10000000 10#99?"0123456789"
Here, 99?... creates a vector of 99
random characters from the set [0-9]
and 10#... takes 10 characters from
each generated "phone number".
10000000 ... repeats it that many times.
A simple 8#l will just gives us the first
8 elements of the list which is not what
we want.
q)f1:{x@\:til 8}
q)f2:{8#/:x}
q)f3:{8#'x}
The above 3 solutions will give us the
first 8 digits of every phone number
in the list.
Which solution is the fastest, though?
The list is very very long specifically
to make benchmarking easier.
q)\t f1 l
906
\t here gives us execution time for f1
in milliseconds. We proceed to time f2
and f3.
q)\t f2 l
738
q)\t f3 l
738
Looks like f1 is much slower but is it?
Lets run the benchmark several times...
q)\t f1 l
624
q)\t f2 l
735
-joel
sa
KDB+ 2.4 2008.03.13 Copyright (C) 1993-2008 Kx Systems
w32/ 2()core 2046MB sa jax 172.16.6.139 EXPIRE 2009.01.01 s...@nsl.com
#42171
q)\t 8#'1000 10000#0
156
q)\t 8#'1000 10000#0
62
q)
the best way to benchmark is to take averages over several runs:
C:\q>q
KDB+ 2.4 2008.03.13 Copyright (C) 1993-2008 Kx Systems
w32/ 2()core 2046MB sa jax 172.16.6.139 EXPIRE 2009.01.01 s...@nsl.com
#42171
q)\t do[10;8#'1000 10000#0]
640
q)
Joel Reymont
I agree with you but f1
is still the fastest!
q)\t do[10;f1 l]
7749
q)\t do[10;f2 l]
8906
q)\t do[10;f3 l]
8894
On Apr 6, 2008, at 4:07 PM, sa wrote:
>
> the best way to benchmark is to take averages over several runs:
>
> C:\q>q
> KDB+ 2.4 2008.03.13 Copyright (C) 1993-2008 Kx Systems
> w32/ 2()core 2046MB sa jax 172.16.6.139 EXPIRE 2009.01.01 s...@nsl.com
> #42171
>
> q)\t do[10;8#'1000 10000#0]
> 640
> q)
Christian Langreiter
> Which solution is the fastest, though?
depending on what you'll do to the data afterwards, you can also flip l:
q)l0:flip l
then:
q)\t l0[til 8]
0
;-)