faster "log10" command
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Hi all, Is anyone know how to use the "log10" more wisely? currently, I convert from dB to decimal using log10 but it seems to slow down the execution process. Any help would be appreciated!
Ta
[EDITED, 09-12-2011 08:13 UTC, Jan Simon] Code copied from Answer section:
Sorry guys for the late reply, and thanks for the comment. Anyway, my code look like this:
for trial = 1:1000
for a = 1:N
for b=1:N
interfere(b,a) = dBtodec(Pr) * power(dstsr(b,a), -(dBtodec(gamma))))) * ...
Li(b) * Ri(b);
end
%
sinrSUr(a) = dBtodec(Pr) * power(ds(a), -(dBtodec(gamma))))) * ...
LSUr(a) * RSUr(a)) / (dBtodec(No) + sum(interfere(:,a)));
snr(a) = (dBtodec(Pr)*(power(ds(a), (-(dBtodec(gamma))))) * ...
LSUr(a) * RSUr(a)) / (dBtodec(No));
end
end
where dBtodec is a function which is:
function [decimal] = dBtodec(x)
%converting dB to decimal
decimal = 10*log10(x);
end
hope that make sense
6 个评论
Jan
2011-12-8
If you post the relevant part of the code, we could look for other problems.
Jan
2011-12-9
Parenthesis can help to make the code easier to read, e.g. when mixing logical expressions. But "-dBtodec(gamma)" is nicer than "(-(dBtodec(gamma)))".
Jan
2011-12-9
Fangjun has commented the code:
Daniel and Jan are right. You probably don't have pre-allocation. Put the following prior to your for-loops.
interfere=zeros(N);
sinrSUr=zeros(N,1);
snr=zeros(N,1);
Parenthesis imbalance in:
interfere(b,a) = dBtodec(Pr)*power(dstsr(b,a),-(dBtodec(gamma)))))*Li(b)*Ri(b) is not valid.
log() is for Natural logarithm. You need to use log10().
OMG! Your dBtodec() calculation is also wrong!
Jan
2011-12-9
There is no log10 in the posted code?!
Please fix the problems Fangjun has shown. Replace the posted code by the cleaned version by editing the original question.
Ricky
2011-12-11
Jan
2011-12-12
@Rak: You can simply edit your question to fix this.
采纳的回答
Jan
2011-12-9
Calculating -(dBtodec(gamma)) and the other constants repeatedly wastes time. Better do it once and store the value in a temporary variable. All calculations can be performed without loops. E.g.:
snr = dBtodec(Pr) .* power(ds, -dBtodec(gamma)) .* LSUr .* RSUr ./ dBtodec(No);
2 个评论
Ricky
2011-12-11
Jan
2011-12-12
If all arrays have the same size ".*" multiplies them elementwise. If some are vectors and each element should be multiplied with all elements of a subvector fo the 2D-array, use BSXFUN, which "inflates" the vector "virtually":
x = rand(3, 3); b = rand(1, 3);
y = bsxfun(@times, x, b); % or: y = x .* b(ones(1,3), :)
更多回答（4 个）
Daniel Shub
2011-12-8
I am guessing you are not preallocating ...
Does you code look something like:
x = randn(1e7,1);
for ii = 1:length(x)
y(ii) = log10(x(ii));
end
You could replace it with
y = log10(x);
4 个评论
Jan
2011-12-8
I agree and this is exactly the point, why I asked Rak for posting the code.
Daniel Shub
2011-12-8
I think my favorite answers are when we have to guess what the question is: sometimes we get it right and sometimes we miss. Asking for more information not only can improve our accuracy it also gives a hint as to how much the OP cares. That said sometimes I like to ask for more info and sometimes I like to guess.
Jan
2011-12-8
Then I guess, that we will get a speedup of >55% if we apply our experiences on Rak's code. But even then this will *not* be an advantage: Currently Rak has waited 16 hours for the answer! It will be hard to recover this delay even with the fastest code...
Jan
2011-12-9
Damn English. Sometimes I'm too confused. While "I guess" is nonsense here, I meant "I bet". And if I had written this, I'd won. What a pitty.
Daniel Shub
2011-12-9
You may want to look at
doc db2mag
doc mag2db
doc pow2db
doc db2pow
they are not going to speed up your code, but they do the transformations in the correct direction and use the correct log base and scale factors ...
1 个评论
Jan
2011-12-9
See: http://en.wikipedia.org/wiki/Anti-pattern , Do not re-invent the square wheel. +1
Sean de Wolski
2011-12-8
On my system:
A = magic(10000); %Don't do this!
tic,log10(A);toc
Elapsed time is 1.388229 seconds.
1.39 seconds to calculate the log10 of 10000^2 elements seems pretty good, so you probably have something else slowing you down. How much memory are you using?
b = whos;
sum(b(:).bytes)
If you're using more memory than you have RAM available that's quite possibly your issue.
Ricky
2011-12-11
0 个投票
2 个评论
Ricky
2011-12-12
Daniel Shub
2011-12-12
the best way to thank people is to accept the best answer and vote for the other answers that helped you. This lets future people with similar questions and problems learn what you learned.
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