How can I make this code run faster?

4 Ansichten (letzte 30 Tage)
Conrado Neto
Conrado Neto am 25 Aug. 2020
Kommentiert: Conrado Neto am 26 Aug. 2020
This code takes about 30s to run, and I will need to run it several times
Can anyone help me make it faster?
here is the code:
sum_Load = rand(1,1000); % For demonstration purposes
Pseudo_Load_Factor = rand(500,1000); % For demonstration purposes
Disp_Factor = rand(500,1000); % For demonstration purposes
M = 1000;
N = 500;
Disp_Inter = zeros(M,N);
for j = 1:M
Load_Inter = sum_Load(j);
for i = 1:N
idx = find(cummin(Pseudo_Load_Factor(i,:))<Load_Inter & cummax(Pseudo_Load_Factor(i,:))>Load_Inter,1,'first');
if isempty(idx)
Disp_Inter(j,i) = 0.10*7000;
else
Disp_Inter(j,i) = interp1(Pseudo_Load_Factor(i,(idx-1):idx),Disp_Factor(i,(idx-1):idx),Load_Inter);
end
end
end
Thanks a lot!!
  3 Kommentare
Adam Danz
Adam Danz am 25 Aug. 2020
You could also use the profiler to look into which sections are relatively slower.
Conrado Neto
Conrado Neto am 25 Aug. 2020
In my codes sum_load, Pseudo_Load_Factor and Disp_Factor are obtained/imported from .dat files.
I've updated the question with a running code, using rand for those arrays

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Bruno Luong
Bruno Luong am 25 Aug. 2020
Bearbeitet: Bruno Luong am 25 Aug. 2020
M = 1000;
N = 500;
P = 1000;
sum_Load = rand(1,M); % For demonstration purposes
Pseudo_Load_Factor = rand(N,P); % For demonstration purposes
Disp_Factor = rand(N,P); % For demonstration purposes
Disp_Inter = zeros(M,N);
for j = 1:M
Load_Inter = sum_Load(j);
A = Pseudo_Load_Factor-Load_Inter;
B = A(:,1:end-1).*A(:,2:end) < 0;
[Bm,idx] = max(B,[],2);
idxl = (1:N)'+(idx-1)*N;
idxr = idxl+N;
PLFR = Pseudo_Load_Factor(idxr);
w = (PLFR-Load_Inter)./(PLFR-Pseudo_Load_Factor(idxl));
DPI = w.*Disp_Factor(idxl) + (1-w).*Disp_Factor(idxr);
DPI(Bm==0) = 700;
Disp_Inter(j,:) = DPI;
end
  5 Kommentare
Bruno Luong
Bruno Luong am 25 Aug. 2020
Yes it's the firstpoint (the idx returned by MAX command is the first one). Otherwise mine result won't match yours.
Conrado Neto
Conrado Neto am 26 Aug. 2020
perfect, thanks

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Steven Lord
Steven Lord am 25 Aug. 2020
I have some off the top of my head suggestions that come from the structure of the code, but if you explain in words what Disp_Inter represents and/or the underlying problem you're trying to solve we may be able to suggest functions to help improve your algorithm and/or its implementation at a higher level.
%snip
for j = 1:M
Load_Inter = sum_Load(j);
for i = 1:N
idx = find(cummin(Pseudo_Load_Factor(i,:))<Load_Inter & cummax(Pseudo_Load_Factor(i,:))>Load_Inter,1,'first');
An obvious low-hanging fruit: since Pseudo_Load_Factor doesn't change inside the nested loops, compute the cumulative minimum and cumulative maximum of them once before entering the loop and index into those precomputed matrices inside the loop.
if isempty(idx)
Disp_Inter(j,i) = 0.10*7000;
Preallocate Disp_Inter to this constant value before the loop (though I'd use 700 rather than multiplying the integer value 7000 by the double precision approximation to one tenth.) That way you only need to change the value if idx is not empty.
%snip the rest of the code
  1 Kommentar
Conrado Neto
Conrado Neto am 25 Aug. 2020
Bearbeitet: Conrado Neto am 25 Aug. 2020
Sure
Disp_inter is a vector with values of displacements
each Disp_inter has a corresponding Pseudo_Load_Factor value
(It is a load x displacement curve)
Then, separately, I have another vector, sum_Load, with values of load that I need to find the corresponding displacement in this load displacement curve
since sum_Load may not correspond exactly to Pseudo_load_Factor, I interpolate to find the corresponding displacement.
look at the following figure:
I hope I was able to explain clearly, if not, let me know
thanks

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Produkte


Version

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by