1. Speed test
- Posted by rforno at tutopia.com
Sep 01, 2001
This is a multi-part message in MIME format.
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charset="iso-8859-1"
The enclosed file tests the speed of five different methods of doing the
same thing. According to the results, it seems that using "parallel"
sequences is faster than using sequences of sequences, when possible,
although it might be less clean.
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name="Test.ex"
filename="Test.ex"
constant M = 1000
sequence x, y, a, b, c, d
procedure one(integer n)
x = rand(repeat(repeat(M, 4), n))
for i = 1 to n do
x[i][1] += x[i][2] - x[i][3] + x[i][4]
end for
end procedure
procedure two(integer n)
a = rand(repeat(M, n))
b = rand(repeat(M, n))
c = rand(repeat(M, n))
d = rand(repeat(M, n))
for i = 1 to n do
a[i] += b[i] - c[i] + d[i]
end for
end procedure
procedure three(integer n)
a = rand(repeat(M, n))
b = rand(repeat(M, n))
c = rand(repeat(M, n))
d = rand(repeat(M, n))
a += b - c + d
end procedure
procedure four(integer n)
x = rand(repeat(repeat(M, 4), n))
for i = 1 to n do
y = x[i]
y[1] += y[2] - y[3] + y[4]
x[i] = y
end for
end procedure
procedure five(integer n)
x = rand(repeat(repeat(M, 4), n))
for i = 1 to n do
y = x[i]
x[i][1] += y[2] - y[3] + y[4]
end for
end procedure
procedure test()
integer n, k
atom t0
k = 500
n = 1000
t0 = time()
for i = 1 to k do
one(n)
end for
? time() - t0
t0 = time()
for i = 1 to k do
two(n)
end for
? time() - t0
t0 = time()
for i = 1 to k do
three(n)
end for
? time() - t0
t0 = time()
for i = 1 to k do
four(n)
end for
? time() - t0
t0 = time()
for i = 1 to k do
five(n)
end for
? time() - t0
end procedure
test()
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2. Re: Speed test
Yes, I often use parallel lists too. Vertical slicing becomes an issue then.
Pity we don't have any neat ways of do that yet.
----- Original Message -----
From: <rforno at tutopia.com>
To: "EUforum" <EUforum at topica.com>
Subject: Speed test
>
> The enclosed file tests the speed of five different methods of doing the
> same thing. According to the results, it seems that using "parallel"
> sequences is faster than using sequences of sequences, when possible,
> although it might be less clean.
>
>
>
>
3. Re: Speed test
- Posted by rforno at tutopia.com
Sep 02, 2001
I'm afraid I don't understand what do you call "vertical slicing". If one
uses parallel sequences, let's say x, y, z, then there is only one way to
refer to each element of these, namely x[i], etc. Please give an example of
what you mean.
Regards.
----- Original Message -----
From: "Derek Parnell" <ddparnell at bigpond.com>
To: "EUforum" <EUforum at topica.com>
Subject: Re: Speed test
>
> Yes, I often use parallel lists too. Vertical slicing becomes an issue
then.
> Pity we don't have any neat ways of do that yet.
>
> ----- Original Message -----
> From: <rforno at tutopia.com>
> To: "EUforum" <EUforum at topica.com>
> Sent: Saturday, September 01, 2001 1:37 PM
> Subject: Speed test
>
>
> > The enclosed file tests the speed of five different methods of doing the
> > same thing. According to the results, it seems that using "parallel"
> > sequences is faster than using sequences of sequences, when possible,
> > although it might be less clean.
> >
> >
>
>
4. Re: Speed test
Hello Ricardo,
> I'm afraid I don't understand what do you call
> "vertical slicing". If one uses parallel sequences,
> let's say x, y, z, then there is only one way to
> refer to each element of these, namely x[i], etc.
> Please give an example of what you mean.
This case is described in refman.doc file.
See please below:
--program vert.ex
sequence a,c
a={{01, 02, 03, 04}, --> row 1
{05, 06, 07, 08}, --> row 2
{09, 10, 11, 12}, --> row 3
{13, 14, 15, 16}} --> row 4
-- 1 2 3 4 <-- columns 1..4
? a[3][2..3] -- this is a simple expression for any row
function b(sequence s, integer n)
sequence v
v=repeat(0,length(s))
for i=1 to length(s) do
v[i]=s[i][n]
end for
return v
end function
c=b(a,3) -- but to take a column you must write
? c [3..4] -- your own function.
-- this is a 'vertical slicing problem'
-- in Euphoria
function d(sequence s, integer n, integer m, integer l)
sequence v
v=repeat(0,length(s))
for i=1 to length(s) do
v[i]=s[i][n]
end for
return v[m..l]
end function
? d(a,3,3,4)-- this complicated expression for columns
-- has 10 elements
-- but simple one for rows has same 10 elements,
-- see above
-- end of program
Maybe, these functions above are useful too, use if so.
Regards,
Igor Kachan
kinz at peterlink.ru
