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From: Jan Simon on 24 Jul 2010 15:54
Dear Matt,

> Result=bsxfun(@eq,B,permute(A(:),[3,2,1]));
> then each slice Result(:,:,i) will be the the result of find(B==A(i));

Nice. For a vector, RESHAPE might be a little bit simpler:
Result = bsxfun(@eq, B, reshape(A, 1, 1, numel(A)));

Jan
From: Matt J on 24 Jul 2010 16:41
"Jan Simon" <matlab.THIS_YEAR(a)nMINUSsimon.de> wrote in message <i2fgcr$1oi$1(a)fred.mathworks.com>...
> Dear Matt,
>
> > Result=bsxfun(@eq,B,permute(A(:),[3,2,1]));
> > then each slice Result(:,:,i) will be the the result of find(B==A(i));
>
> Nice. For a vector, RESHAPE might be a little bit simpler:
> Result = bsxfun(@eq, B, reshape(A, 1, 1, numel(A)));
=============

Good point, Jan. The short test below shows that reshape() is a fair bit faster.

>> A=rand(1e7,1); tic; permute(A,[3,2,1]);toc; tic; reshape(A,1,1,[]);toc
Elapsed time is 0.166652 seconds.
Elapsed time is 0.004647 seconds.
From: Jan Simon on 24 Jul 2010 17:32
Dear Matt J,

> >> A=rand(1e7,1); tic; permute(A,[3,2,1]);toc; tic; reshape(A,1,1,[]);toc
> Elapsed time is 0.166652 seconds.
> Elapsed time is 0.004647 seconds.

Which Matlab?

1.5GHz Pentium-M, Matlab 2009a, 512MB, WinXP-32bit:
A = rand(1e7, 1);
clear('y'); tic; y = permute(A, [3,2,1]); toc
>> 0.35 sec
clear('y'); tic; y = reshape(A, 1, 1, []); toc
>> 0.000439 sec
clear('y'); tic; y = reshape(A, 1, 1, numel(A)); toc
>> 0.000019

But if I put the lines in a function (more realistic, and of course repeated 1000 times...), RESHAPE(1, 1, []) is only 30% slower than RESHAPE(1, 1, NUMEL(A)).
Reshape is much faster than PERMUTE, because it replies a shared data copy, while PERMUTE duplicates the data. See: "format debug".

@guyz Smarty: I hope you are not too much frightend by this discussion. For [1x3] vectors the timing between RESHAPE and PERMUTE is absolutely meaningless! My for loop is simple, Matt's BSXFUN is efficient

Kind regards, Jan
From: Matt J on 24 Jul 2010 17:55
"Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i2fed5$s68$1(a)fred.mathworks.com>...
>
> If you do
>
> Result=bsxfun(@eq,B,permute(A(:),[3,2,1]));
>
> then each slice Result(:,:,i) will be the the result of find(B==A(i));

Should be "will be the resulf of B==A(i)"
From: guyz Smarty on 24 Jul 2010 20:25
Hi all,

Thank you very much to all of you for your quick response and help.

@ Nic Roberts
I know that the problem I presented here can be easily solved with FOR loops. But I have to do calculations for a matrix containing more than 100000 rows. In that case, loops are not a good choice(I suppose).

Therefore, I was looking for a solution without using any FOR loops.

@someone
The use of Cell array is really a new thing for me. I will keep that in mind and try to use that in future.

@Matt
Thank you. You gave me the direction that I wanted to follow.

But, as you too would agree, bsxfun is not a very easy function to comprehend. So, I want to learn more about this. However, I have managed to get rid of my problem as of now using bsxfun.

@Jan
No..No.. I am not frightened at all with all those discussions. In fact, I got to learn many things here. The use of cell array, pre-allocation of variables and bsxfun functions are just a few to name for.

So the credit goes to all of you.

Thanks once again to all..
====================================================
"Jan Simon" <matlab.THIS_YEAR(a)nMINUSsimon.de> wrote in message <i2fm4k$pov$1(a)fred.mathworks.com>...
> Dear Matt J,
>
> > >> A=rand(1e7,1); tic; permute(A,[3,2,1]);toc; tic; reshape(A,1,1,[]);toc
> > Elapsed time is 0.166652 seconds.
> > Elapsed time is 0.004647 seconds.
>
> Which Matlab?
>
> 1.5GHz Pentium-M, Matlab 2009a, 512MB, WinXP-32bit:
> A = rand(1e7, 1);
> clear('y'); tic; y = permute(A, [3,2,1]); toc
> >> 0.35 sec
> clear('y'); tic; y = reshape(A, 1, 1, []); toc
> >> 0.000439 sec
> clear('y'); tic; y = reshape(A, 1, 1, numel(A)); toc
> >> 0.000019
>
> But if I put the lines in a function (more realistic, and of course repeated 1000 times...), RESHAPE(1, 1, []) is only 30% slower than RESHAPE(1, 1, NUMEL(A)).
> Reshape is much faster than PERMUTE, because it replies a shared data copy, while PERMUTE duplicates the data. See: "format debug".
>
> @guyz Smarty: I hope you are not too much frightend by this discussion. For [1x3] vectors the timing between RESHAPE and PERMUTE is absolutely meaningless! My for loop is simple, Matt's BSXFUN is efficient
>
> Kind regards, Jan
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