|
»
»
« Return to Thread: very very very slow computing
Re: very very very slow computing
by Rob Mahurin
::
Rate this Message:
On Feb 6, 2009, at 11:04 AM, Depo wrote:
> Hi everybody,
>
> I was just trying making some simple operations on a large matrix
> (10, 3000) and I faced off o very slow computational time comparing
> to the simple operation I was attempting to do. Looking for the
> problem I saw that even copying same values to a new matrix was a
> computationally huge operation!!!!!!!!!!!!!
>
> That is the code:
>
> clear new_sign_difference; #to ensure it is emty
> i=1; k=1;
> while (k <= n_rows) #I have used while statement instead of for
> because some says it is faster...
> while (i <= n_cols)
> new_sign_difference(k,i)=sign_difference(k,i) #reassign same
> values to a new matrix
> disp(i);
> disp(k);
> i++;
> endwhile
> k++; i=1;
> endwhile
This code does the same thing as the statement
new_sign_difference = sign_difference
which copies the entire matrix at once, very quickly.
If you have some expression you need to do element-by-element, use
"vectorize" to make sure you haven't missed any *,/,^ operators.
If you have some code that doesn't vectorize (hey, it happens),
define matrices for your results before starting so that your loop
doesn't waste time reallocating memory when the matrix grows:
new_sign_difference = zeros(size(sign_difference));
for (i = 1:numel(sign_difference))
new_sign_difference(i) = [something]
> the matrix sign_difference has size 10x3'000, so this is a 30'000-
> operation loop. It takes up to 10 minutes!
>
> If I only visualize on screen matrix' values it takes 1
> second............................., here is the code:
>
>
> clear new_sign_difference;
> i=1; k=1;
> while (k <= n_rows) while (i <= n_cols)
> sign_difference(k,i) #only to screen, without writing on a new matrix
> [...]
Sounds like your problem was repeatedly re-allocating memory.
Allocate, clear, allocate a little more every time turns a 30,000
operation loop to a (30,000)^2 operation loop.
Cheers,
Rob
--
Rob Mahurin
Department of Physics and Astronomy
University of Tennessee 865 207 2594
Knoxville, TN 37996 rob@...
_______________________________________________
Help-octave mailing list
Help-octave@...
https://www-old.cae.wisc.edu/mailman/listinfo/help-octave
> Hi everybody,
>
> I was just trying making some simple operations on a large matrix
> (10, 3000) and I faced off o very slow computational time comparing
> to the simple operation I was attempting to do. Looking for the
> problem I saw that even copying same values to a new matrix was a
> computationally huge operation!!!!!!!!!!!!!
>
> That is the code:
>
> clear new_sign_difference; #to ensure it is emty
> i=1; k=1;
> while (k <= n_rows) #I have used while statement instead of for
> because some says it is faster...
> while (i <= n_cols)
> new_sign_difference(k,i)=sign_difference(k,i) #reassign same
> values to a new matrix
> disp(i);
> disp(k);
> i++;
> endwhile
> k++; i=1;
> endwhile
This code does the same thing as the statement
new_sign_difference = sign_difference
which copies the entire matrix at once, very quickly.
If you have some expression you need to do element-by-element, use
"vectorize" to make sure you haven't missed any *,/,^ operators.
If you have some code that doesn't vectorize (hey, it happens),
define matrices for your results before starting so that your loop
doesn't waste time reallocating memory when the matrix grows:
new_sign_difference = zeros(size(sign_difference));
for (i = 1:numel(sign_difference))
new_sign_difference(i) = [something]
> the matrix sign_difference has size 10x3'000, so this is a 30'000-
> operation loop. It takes up to 10 minutes!
>
> If I only visualize on screen matrix' values it takes 1
> second............................., here is the code:
>
>
> clear new_sign_difference;
> i=1; k=1;
> while (k <= n_rows) while (i <= n_cols)
> sign_difference(k,i) #only to screen, without writing on a new matrix
> [...]
Sounds like your problem was repeatedly re-allocating memory.
Allocate, clear, allocate a little more every time turns a 30,000
operation loop to a (30,000)^2 operation loop.
Cheers,
Rob
--
Rob Mahurin
Department of Physics and Astronomy
University of Tennessee 865 207 2594
Knoxville, TN 37996 rob@...
_______________________________________________
Help-octave mailing list
Help-octave@...
https://www-old.cae.wisc.edu/mailman/listinfo/help-octave
« Return to Thread: very very very slow computing
| Free embeddable forum powered by Nabble | Forum Help |
