How to reduce execution time in nested for loops and struc. arrays

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I have the following code, it is part of a much larger one. The total execution time for the whole code is around 50 sec. I am trying to reduce this figure to few secs. So the first thing is I started to look for for loops to reduce.
The following part of the code is taking 33 sec. (more than half the total time) and I am only creating strucs. arrays
NOC=100;
NOBS=2;
NORB_PER_BS=5;
NOU=200 ;
for k=1:NOU
U1(k).SINR_MAX=0;
U1(k).SINR_F_AVG=0;
U1(k).v=[];
U1(k).w=[];
U1(k).SINR_IND={};
U1(k).ASSIGNED=0;
U1(k).SINR_MAX_CANDIDATES=[];
for n=1:NORB_PER_BS
for b=1:NOBS
for v=1:NOU
U(k,n,b).Q_LIST=[];
U(k,n,b).Q_LIST1=[];
U(k,n,b).Q_LIST_MEMBERS=[];
U(k,n,b).Q_MIN_MEMBERS=[];
U(k,n,b).Q_MIN=[];
SINR(k,n,b)=0;
end
end
end
end
  3 个评论
Guillaume
Guillaume 2019-3-1
编辑:Guillaume 2019-3-1
In addition to Walter's comment, is the structure array preinitialised? Or is it resized at each step of the loops (a major cause of slow-down)?
Does your real code just assign the same value for each element? In that case, why use a loop when you could just repmat a constant scalar structure?
Oh, and just notice that NOC is never used.
Mohammed Hadi
Mohammed Hadi 2019-3-1
@walter
Thanks for pointing that out. I kept editing the code and forgot to remove this part.
-------------------------------------------------------
@Gulliaume
The struc. array is not pre-initialized. I am using these loops to initialize it and make sure all are zeros.
The size of
U(k,n,b).Q_LIST=[];
U(k,n,b).Q_LIST1=[];
U(k,n,b).Q_LIST_MEMBERS=[];
U(k,n,b).Q_MIN_MEMBERS=[];
U(k,n,b).Q_MIN=[];
changes in the steps folllowing this part
Sorry for the NOC, yes you are corrent it is not used here but used in the following steps.

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Guillaume
Guillaume 2019-3-1
Your current code is equivalent to:
U = struct('SINR_MAX', num2cell(zeros(1, NOU)), 'SINR_F_AVG', 0, 'v', [], 'w', [], 'SINR_IND', {{[]}}, 'ASSIGNED', 0, 'SINR_MAX_CANDIDATES', []);
U1 = struct('Q_LIST', cell(NOU, NORB_PER_BS, NOBS), 'Q_LIST1',[], 'Q_LIST_MEMBERS', [], 'Q_MIN_MEMBERS', [], 'Q_MIN', []);
SINR = zeros(NOU, NORB_PER_BS, NOBS);
  5 个评论
Guillaume
Guillaume 2019-3-4
That really depends on what is going on in the loops. Sometimes, there's no way to gain any speed other than completely changing the algorithm.
Walter Roberson
Walter Roberson 2019-3-4
And sometimes it is several centuries of research to find a better algorithm. Or to prove that no better algorithm exists (only implementation details.)

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