How to speed up the following code. I need to optimize it for fastest performance in Matlab 7.0 version

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code
  • TIR_KK & WV_KK lat_TIR, lon_TIR are 1400x1400 matrices.
  • V1,V2,W1,W2 are scalars
  • lat_BOX(m,n),lon_BOX(m,n) are scalars.m,n have values 1:300
  • the problem here is to convert a 1400x1400 matrix into a 300x300 matrix.
  • the constraint is to apply nanmean to the set using latitude and longitude
  • values which are also 1400x1400 matrices.
cnt = 0; TIR_KK = 0;WV_KK = 0;
for i = V1:V2
for j = W1:W2
if lat_BOX(m,n) >= latTIR(i,j) && latTIR(i,j) > lat_BOX(m+1,n) && lon_BOX(m,n) <= lonTIR(i,j) && lonTIR(i,j)< lon_BOX(m,n+1)
cnt = cnt+1;
TIR_KK(cnt) = TIR_K(i,j);
WV_KK(cnt) = WV_K(i,j);
% fprintf ('\n %f\t%f',latTIR(i,j),lonTIR(i,j))
end
end
end
CNT_check(m,n) = cnt;
TIR_K_box(m,n) = nanmean(TIR_KK);
WV_K_box(m,n) = nanmean(WV_KK);
  1 comentario
Cedric
Cedric el 3 de Mayo de 2014
Editada: Cedric el 3 de Mayo de 2014
Could you describe what you want to achieve? Is it summarizing e.g. a raster on a coarser grid? If so, is the grid regular, irregular?

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Respuestas (2)

Justin
Justin el 2 de Mayo de 2014
What is lat_BOX and lon_BOX and where are m and n defined? Also, what is the concept of what the conditional statements are applying?
Since the calculation of each value in the output arrays is based on a conditional statement it may be difficult to avoid for loops. You could try to apply the conditional statements to the entire latTIR and lonTIR arrays. Then you could use the results logical values to index you TIR_K and WV_K and take the mean values of those directly.
For example:
x = [1 2 3 4 5 6];
y = [7; 6; 5; 4; 3; 2];
removeIndex = x<3;
y(removeIndex) = [];
value = nanmean(y);
  1 comentario
Santino M
Santino M el 3 de Mayo de 2014
Thanks Justin for the prompt response. I have added the clarifications to the question. The Speed up of code would be done if I can get a matrix output of indices of a larger matrix using "find" command

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Jan
Jan el 3 de Mayo de 2014
Editada: Jan el 3 de Mayo de 2014
Pre-allocation is essential, when larger arrays are created. So try:
max_n = (V2-V2+1) * (W2-W1+1);
TIR_KK = zeros(1, max_n);
WV_KK = zeros(1, max_n);
A nex idea is the vectorization mentioned by Justin already:
index = (lat_BOX(m,n) >= latTIR(V1:V2,W1:W2)) & ...
(latTIR(V1:V2,W1:W2) > lat_BOX(m+1,n)) & ...
(lon_BOX(m,n) <= lonTIR(V1:V2,W1:W2)) & ...
(lonTIR(V1:V2,W1:W2) < lon_BOX(m,n+1));
TIR_K_box(m,n) = nanmean(TIR_K(index));
WV_K_box(m,n) = nanmean(WV_K(index));
By the way, this looks nicer also, such that it faster to debug.
Perhaps using temporary arrays is slightly faster:
T1 = latTIR(V1:V2,W1:W2);
T2 = lonTIR(V1:V2,W1:W2);
index = (lat_BOX(m,n) >= T1) & ...
(T1 > lat_BOX(m+1,n)) & ...
(lon_BOX(m,n) <= T2) & ...
(T2 < lon_BOX(m,n+1));
TIR_K_box(m,n) = nanmean(TIR_K(index));
WV_K_box(m,n) = nanmean(WV_K(index));

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